# The Auslander bijections: how morphisms are determined by modules

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## Abstract

Let $\mathrm{\Lambda}$ be an artin algebra. In his seminal Philadelphia Notes published in 1978, Auslander introduced the concept of morphisms being determined by modules. Auslander was very passionate about these investigations (they also form part of the final chapter of the Auslander–Reiten–Smalø book and could and should be seen as its culmination). The theory presented by Auslander has to be considered as an exciting frame for working with the category of $\mathrm{\Lambda}$-modules, incorporating all what is known about irreducible maps (the usual Auslander–Reiten theory), but the frame is much wider and allows for example to take into account families of modules—an important feature of module categories. What Auslander has achieved is a clear description of the poset structure of the category of $\mathrm{\Lambda}$-modules as well as a blueprint for interrelating individual modules and families of modules. Auslander has subsumed his considerations under the heading of “morphisms being determined by modules”. Unfortunately, the wording in itself seems to be somewhat misleading, and the basic definition may look quite technical and unattractive, at least at first sight. This could be the reason that for over 30 years, Auslander’s powerful results did not gain the attention they deserve. The aim of this survey is to outline the general setting for Auslander’s ideas and to show the wealth of these ideas by exhibiting many examples.

## Keywords

Auslander bijections Auslander–Reiten theory Right factorization lattice Morphisms determined by modules Finite length categories: global directedness Local symmetries Representation type Brauer–Thrall conjectures Riedtmann–Zwara degenerations Hammocks Kronecker quiver Quiver Grassmannians Auslander varieties Modular lattices Meet semi-lattices## Mathematics Subject Classification (2010)

Primary 16G70 18E10 18A25 18A32 16G60 Secondary 16G20 18A20 06C05 14M15 19A49 03C60## 1 Introduction

There are two basic mathematical structures: groups and lattices, or, more generally, semigroups and posets. A first glance at any category should focus the attention on these two structures: to symmetry groups (for example the automorphism groups of the individual objects), as well as to the posets given by suitable sets of morphisms, for example by looking at inclusion maps (thus dealing with the poset of all subobjects of an object), or at the possible factorizations of morphisms. In this way, one distinguishes between local symmetries and global directedness.

The present survey deals with the category $\mathrm{mod}\mathrm{\Lambda}$ of finite length modules over an artin algebra $\mathrm{\Lambda}$. Its aim is to report on the work of Auslander in his seminal Philadelphia Notes published in 1978. Auslander was very passionate about these investigations and they also form part of the final chapter of the Auslander–Reiten–Smalø book: there, they could (and should) be seen as a kind of culmination. It seems to be surprising that the feedback until now is quite meager. After all, the theory presented by Auslander has to be considered as an exciting frame for working with the category $\mathrm{mod}\mathrm{\Lambda}$, incorporating what is called the Auslander–Reiten theory (to deal with the irreducible maps), but this frame is much wider and allows for example to take into account families of modules—an important feature of a module category. Indeed, many of the concepts which are relevant when considering the categories $\mathrm{mod}\mathrm{\Lambda}$ fit into the frame! What Auslander has achieved (but he himself may not have realized it) was a clear description of the poset structure of $\mathrm{mod}\mathrm{\Lambda}$ and of the interplay between families of modules.

Auslander’s considerations are subsumed under the heading of morphisms being determined by modules, but the wording in itself seems to be somewhat misleading, and the basic definition looks quite technical and unattractive, at least at first sight. This could be the reason that for over 30 years, Auslander’s powerful results did not gain the attention they deserve.

*right equivalent*provided there are maps $h:X\to {X}^{\prime}$ and ${h}^{\prime}:{X}^{\prime}\to X$ such that $f={f}^{\prime}h$ and ${f}^{\prime}=f{h}^{\prime}$. The right equivalence class of $f$ will be denoted by $[f\u3009.$ The object studied by Auslander is the set of right equivalence classes of maps ending in $Y$, we denote this set by

It is easy to see that the poset $[\to Y\u3009$ is a lattice, thus we call it the *right factorization lattice* for $Y$.

Looking at maps $f:X\to Y$, we may (and often will) assume that $f$ is *right minimal*, thus that there is no non-zero direct summand ${X}^{\prime}$ of $X$ with $f({X}^{\prime})=0.$ Note that any right equivalence class contains a right minimal map, and if $f:X\to Y$ and ${f}^{\prime}:{X}^{\prime}\to Y$ are right minimal maps, then any $h:X\to {X}^{\prime}$ with $f={f}^{\prime}h$ has to be an isomorphism.

Of course, to analyze the poset $[\to Y\u3009$ is strongly related to a study of the contravariant $\mathrm{Hom}$-functor $\mathrm{Hom}(-,Y)$, however the different nature of these two mathematical structures should be stressed: $\mathrm{Hom}(-,Y)$ is an additive functor whereas $[\to Y\u3009$ is a poset, and it is the collection of these posets $[\to Y\u3009$ which demonstrates the global directedness.

*right C-factorization lattice*for $Y$. Since the concept of “right determination” looks (at least at first sight) technical and unattractive, let us first describe the set ${}^{C}[\to Y\u3009$ only in the important case when $C$ is a generator: in this case, ${}^{C}[\to Y\u3009$ consists of the (right equivalence classes of the right minimal) maps $f$ ending in $Y$ with kernel in $\mathrm{add}\mathit{\tau}C$ (we denote by $\mathit{\tau}=D\mathrm{Tr}$ and ${\mathit{\tau}}^{-}=\mathrm{Tr}D$ the Auslander–Reiten translations). Here is Auslander’s first main assertion:

What is the relevance? As we have mentioned, usually the lattice $[\to Y\u3009$ itself will not satisfy any chain conditions, but all the lattices ${}^{C}[\to Y\u3009$ are of finite height and often can be displayed very nicely: according to (2) we deal with the submodule lattice $\mathcal{S}M$ of some finite length module $M$ over an artin algebra (namely over $\mathrm{\Gamma}(C)=\mathrm{End}{(C)}^{\phantom{\rule{4.pt}{0ex}}\text{op}\phantom{\rule{4.pt}{0ex}}}$) and it is easy to see that any submodule lattice arises in this way. Using the Auslander bijections ${\mathit{\eta}}_{CY}$, one may transfer properties of submodule lattices to the right $C$-factorization lattices ${}^{C}[\to Y\u3009,$ this will be one of the aims of this paper. Given a submodule $U$ of $\mathrm{Hom}(C,Y)$, let $f$ be a right $C$-determined map ending in $Y$ such that ${\mathit{\eta}}_{CY}(f)=U$. The composition series of the factor module $\mathrm{Hom}(C,Y)/U$ correspond to certain factorizations of $f$ (to the “maximal $C$-factorizations”), and we may define the $C$-type of $f$ so that it is equal to the dimension vector of the module $\mathrm{Hom}(C,Y)/U$ (recall that the dimension vector of a module $M$ has as coefficients the Jordan–Hölder multiplicities of the various simple modules occurring in $M$).

Submodule lattices have interesting combinatorial features, and it seems to interesting that Auslander himself looked mainly at combinatorial properties (for example at waists in submodule lattices). But we should stress that we really are in the realm of algebraic geometry. Thus, let us assume for a moment that $\mathrm{\Lambda}$ is a $k$-algebra where $k$ is an algebraically closed field. If $M$ is a finite-dimensional $\mathrm{\Lambda}$-module, the set $\mathcal{S}M$ of all submodules of $M$ is the disjoint union of the sets ${\mathbb{G}}_{\mathbf{e}}(M)$ consisting of all submodules of $M$ with fixed dimension vector $\mathbf{e}$. It is well-known that ${\mathbb{G}}_{\mathbf{e}}(M)$ is in a natural way a projective variety, called nowadays a *quiver Grassmannian*. Given $\mathrm{\Lambda}$-modules $C$ and $Y$, the Auslander bijections draw the attention on the $\mathrm{End}{(C)}^{\phantom{\rule{4.pt}{0ex}}\text{op}\phantom{\rule{4.pt}{0ex}}}$-module $M=\mathrm{Hom}(C,Y)$, let $\mathbf{d}$ be its dimension vector and let $\mathbf{e},{\mathbf{e}}^{\prime}$ be dimension vectors with $\mathbf{e}+{\mathbf{e}}^{\prime}=\mathbf{d}.$ The quiver Grassmannians ${\mathbb{G}}_{{\mathbf{e}}^{\prime}}(\mathrm{Hom}(C,Y))$ corresponds under the Auslander bijection ${\mathit{\eta}}_{CY}$ to the set ${}^{C}{[\to Y\u3009}^{\mathbf{e}}$ of all right equivalence classes of right $C$-determined maps which end in $Y$ and have type $\mathbf{e}$. We call ${}^{C}{[\to Y\u3009}^{\mathbf{e}}$ an *Auslander variety.* These Auslander varieties have to be considered as an important tool for studying the right equivalence classes of maps ending in a given module.

We end this summary by an outline in which way the Auslander bijections (2) incorporate the existence of minimal right almost split maps: we have to look at the special case where $Y$ is indecomposable and $C=Y$ and to deal with the submodule $\mathrm{rad}(Y,Y)$ of $\mathrm{Hom}(Y,Y)$. The bijection (2) yields an element $f:X\to Y$ in ${}^{Y}[\to Y\u3009$ such that ${\mathit{\eta}}_{YY}(f)=\mathrm{rad}(Y,Y)$; to say that $f$ is right $Y$-determined means that $f$ is right almost split.

The survey is divided into three parts. Part I presents the general setting, it comprises the Sects. 2 to 10. The Sects. 11 to 15 form Part II, here we show in which way the Auslander bijections deal with families of modules. Finally, in Parts III, we discuss some special cases; these are the Sects. 16 to 18.

**I. The setting**

## 2 The right factorization lattice $[\to Y\u3009$

Let $Y$ be a $\mathrm{\Lambda}$-module. Let ${\u2a06}_{X}\mathrm{Hom}(X,Y)$ be the class of all homomorphisms $f:X\to Y$ with arbitrary modules $X$ (such homomorphisms will be said to be the homomorphisms *ending in*$Y$). We define a preorder $\u2aaf$ on this class as follows: Given $f:X\to Y$ and ${f}^{\prime}:{X}^{\prime}\to Y$, we write $f\u2aaf{f}^{\prime}$ provided there is a homomorphism $h:X\to {X}^{\prime}$ such that $f={f}^{\prime}h$ (clearly, this relation is reflexive and transitive). As usual, such a preorder defines an equivalence relation (in our setting, we call it *right equivalence*) by saying that $f,{f}^{\prime}$ are *right equivalent* provided we have both $f\u2aaf{f}^{\prime}$ and ${f}^{\prime}\u2aaff$, and it induces a poset relation $\le $ on the set $[\to Y\u3009$ of right equivalence classes of homomorphisms ending in $Y$. Given a morphism $f:X\to Y$, we denote its right equivalence class by $[f\u3009$ and by definition $[f\u3009\le [{f}^{\prime}\u3009$ if and only if $f\u2aaf{f}^{\prime}.$ As we will see in Proposition 2.2, the poset $[\to Y\u3009$ is a lattice, thus we will call it the *right factorization lattice* for $Y$.

It should be stressed that $[\to Y\u3009$ is a set, not only a class: namely, the isomorphism classes of $\mathrm{\Lambda}$-modules form a set and for every module $X$, the homomorphisms $X\to Y$ form a set; we may choose a representative from each isomorphism class of $\mathrm{\Lambda}$-modules and given a homomorphism $f:X\to Y$, then there is an isomorphism $h:{X}^{\prime}\to X$ where ${X}^{\prime}$ is such a representative, and $f$ is right equivalent to $fh.$

Recall that a map $f:X\to Y$ is said to be *right minimal* provided any direct summand ${X}^{\prime}$ of $X$ with $f({X}^{\prime})=0$ is equal to zero. If $f:X\to Y$ is a morphism and $X={X}^{\prime}\oplus {X}^{\prime \prime}$ such that $f({X}^{\prime \prime})=0$ and $f|{X}^{\prime}:{X}^{\prime}\to Y$ is right minimal, then $f|{X}^{\prime}$ is called a *right minimalisation of f.* The kernel of a right minimalisation of $f$ will be called the *intrinsic kernel of f*, it is unique up to isomorphism.

### **Proposition 2.1**

Every right equivalence class $[f\u3009$ in $[\to Y\u3009$ contains a right minimal morphism, namely $[{f}^{\prime}\u3009$, where ${f}^{\prime}$ is a right minimalisation of $f$. Given right minimal morphisms $f:X\to Y$ and ${f}^{\prime}:{X}^{\prime}\to Y$, then $f,{f}^{\prime}$ are right equivalent if and only if there is an isomorphism $h:X\to {X}^{\prime}$ such that $f={f}^{\prime}h$.

### *Proof*

Let $f:X\to Y$ be a homomorphism ending in $Y$. Write $X={X}_{1}\oplus {X}_{2}$ such that $f({X}_{2})=0$ and $f|{X}_{1}:{X}_{1}\to Y$ is right minimal. Let $u:{X}_{1}\to {X}_{1}\oplus {X}_{2}$ be the canonical inclusion, $p:{X}_{1}\oplus {X}_{2}\to {X}_{1}$ the canonical projection. Then $pu={1}_{{X}_{1}}$ and $f=fup$ (since $f({X}_{2})=0$). We see that $fu\u2aaff$ and $f=fup\u2aaffu$, thus $f$ and $fu$ are right equivalent and $fu=f|{X}_{1}$ is right minimal. If the right minimal morphisms $f:X\to Y$ and ${f}^{\prime}:{X}^{\prime}\to Y$ are right equivalent, then there are morphisms $h:X\to {X}^{\prime}$ and ${h}^{\prime}:{X}^{\prime}\to Y$ such that $f={f}^{\prime}h$ and ${f}^{\prime}=f{h}^{\prime}$. But $f=f{h}^{\prime}h$ implies that ${h}^{\prime}h$ is an automorphism, and ${f}^{\prime}={f}^{\prime}h{h}^{\prime}$ implies that $h{h}^{\prime}$ is an automorphism, thus $h,{h}^{\prime}$ have to be isomorphisms (see [5] I.2). $\square $

### *Remark*

Monomorphisms $X\to Y$ are always right minimal, and the right equivalence classes of monomorphisms ending in $Y$ may be identified with the submodules of $Y$ (here, we identify the right equivalence class of the monomorphism $f:X\to Y$ with the image of $X$).

### **Proposition 2.2**

The poset $[\to Y\u3009$ is a lattice with zero and one. Given ${f}_{1}:{X}_{1}\to Y$ and ${f}_{2}:{X}_{2}\to Y$, say with pullback ${g}_{1}:X\to {X}_{1},{g}_{2}:X\to {X}_{2}$, the meet of $[{f}_{1}\u3009$ and $[{f}_{2}\u3009$ is given by the map ${f}_{1}{g}_{1}:X\to Y$, the join of $[{f}_{1}\u3009$ and $[{f}_{2}\u3009$ is given by $[{f}_{1},{f}_{2}]:{X}_{1}\oplus {X}_{2}\to Y$.

### *Proof*

(a trivial verification) Write $f={f}_{1}{g}_{1}={f}_{2}{g}_{2}$. We have $f={f}_{1}{g}_{1}\u2aaf{f}_{1}$ and $f={f}_{2}{g}_{2}\u2aaf{f}_{2}$, thus $[f\u3009\le [{f}_{1}\u3009$ and $[f\u3009\le [{f}_{2}\u3009$. If ${f}^{\prime}:{X}^{\prime}\to Y$ is a morphism with $[{f}^{\prime}\u3009\le [{f}_{1}\u3009$ and $[{f}^{\prime}\u3009\le [{f}_{2}\u3009$, then ${f}^{\prime}\u2aaf{f}_{1}$ and ${f}^{\prime}\u2aaf{f}_{2}$, thus there are morphisms ${\mathit{\varphi}}_{i}$ with ${f}^{\prime}={f}_{i}{\mathit{\varphi}}_{i}$, for $i=1,2$ Since ${f}_{1}{\mathit{\varphi}}_{1}={f}_{2}{\mathit{\varphi}}_{2}$, the pullback property yields a morphism $\mathit{\varphi}:{X}^{\prime}\to X$ such that ${\mathit{\varphi}}_{i}={g}_{i}\mathit{\varphi}$ for $i=1,2$. Thus ${f}^{\prime}={f}_{1}{\mathit{\varphi}}_{1}={f}_{1}{g}_{1}\mathit{\varphi}=f\mathit{\varphi}$ shows that ${f}^{\prime}\u2aaff$, thus $[{f}^{\prime}\u3009\le [f\u3009.$ This shows that $[f\u3009$ is the meet of $[{f}_{1}\u3009$ and $[{f}_{2}\u3009$.

Second, denote the canonical inclusion maps ${X}_{i}\to {X}_{1}\oplus {X}_{2}$ by ${u}_{i}$, for $i=1,2$, thus $[{f}_{1},{f}_{2}]{u}_{i}={f}_{i}$ and therefore $[{f}_{i}\u3009\le [[{f}_{1},{f}_{2}]\u3009$ for $i=1,2.$ Assume that there is given a morphism $g:{X}^{\prime \prime}\to Y$ with $[{f}_{i}\u3009\le [g\u3009$ for $i=1,2$. This means that there are morphisms ${\mathit{\psi}}_{i}:{X}_{i}\to {X}^{\prime \prime}$ such that ${f}_{i}=g{\mathit{\psi}}_{i}$ for $i=1,2$. Let $\mathit{\psi}=[{\mathit{\psi}}_{1},{\mathit{\psi}}_{2}]:{X}_{1}\oplus {X}_{2}\to {X}^{\prime \prime}$ (with $\mathit{\psi}{u}_{i}={\mathit{\psi}}_{i}$). Then $[{f}_{1},{f}_{2}]=g[{\mathit{\psi}}_{1},{\mathit{\psi}}_{2}]=g\mathit{\psi}$ shows that $[{f}_{1},{f}_{2}]\u2aafg$, thus $[[{f}_{1},{f}_{2}]\u3009\le [g\u3009.$ This shows that $[[{f}_{1},{f}_{2}]\u3009$ is the join of $[{f}_{1}\u3009$ and $[{f}_{2}\u3009$.

It is easy to check that the map $0\to Y$ is the zero element of $[\to Y\u3009$ and that the identity map $Y\to Y$ is its unit element. $\square $

It should be stressed that *if*${f}_{1}:{X}_{1}\to Y$*and*${f}_{2}:{X}_{2}\to Y$*are right minimal, say with pullback*${g}_{1}:X\to {X}_{1},{g}_{2}:X\to {X}_{2}$, *then neither the map*${f}_{1}{g}_{1}$*nor the direct sum map*$[{f}_{1},{f}_{2}]:{X}_{1}\oplus {X}_{2}\to Y$*will be right minimal, in general.* Thus if one wants to work with right minimal maps, one has to right minimalise the maps in question. Here are corresponding examples:

### *Example 1*

All path algebras of quivers considered in the paper will have coefficients in an arbitrary field $k$, unless we specify some further conditions. When dealing with the path algebra of a quiver $\mathrm{\Delta}$, and $x$ is a vertex of $\mathrm{\Delta}$, we denote by $S(x)$ (or also just by $x$) the simple module corresponding to $x$, by $P(x)$ and $Q(x)$ the projective cover or injective envelope of $S(x)$, respectively.

Take as maps ${f}_{1},{f}_{2}$ the canonical projection ${f}_{1}={f}_{2}:P(b)\to S(b)$, this is a right minimal map. The pullback $U$ of ${f}_{1}$ and ${f}_{2}$ is a submodule of $P(b)\oplus P(b)$ which is isomorphic to $S(a)\oplus P(b)$. Since any map $S(a)\to S(b)$ is zero, there is no right minimal map $U\to S(b)$.

Also, the map $[{f}_{1},{f}_{2}]:P(b)\oplus P(b)\to S(b)$ is not right minimal, since we have $dim\mathrm{Hom}(P(b),S(b))=1.$

As we have seen, the poset $[\to Y\u3009$ is a lattice. What will be important in the following discussion is the fact that we deal with a meet-semilattice (these are the posets such that any pair of elements has a meet). Note that all the semilattices which we deal with turn out to be lattices, however the poset maps to be considered will preserve meets, but usually not joins, thus we really work in the category of meet-semilattices.

### **Proposition 2.3**

The lattice $[\to Y\u3009$ is modular.

### *Proof*

yields a map ${f}_{3}{p}_{3}$ such that $[{f}_{3}{p}_{3}\u3009=[{f}_{2}\u3009)\wedge [{f}_{3}\u3009,$ thus the map $f=[{f}_{3}h,{f}_{3}{p}_{3}]:{X}_{1}\oplus P\to Y$ belongs to $[{f}_{3}h\u3009\vee ([{f}_{2}\u3009)\wedge [{f}_{3}\u3009)$.

the map Open image in new window is an element of $([{f}_{3}h\u3009\vee [{f}_{2}\u3009)\wedge [{f}_{3}\u3009$.

### *Example 2*

**Failure of the chain conditions.**Here are examples which show that in general $[\to Y\u3009$

*neither satisfies the ascending nor the descending chain condition.*

*Let*$\mathrm{\Lambda}$

*be the Kronecker algebra, this is the path algebra of the quiver*The $\mathrm{\Lambda}$-modules are also called

*Kronecker modules*(basic facts concerning the Kronecker modules will be recalled in Sect. 14). Let $Y=S(b)$, the simple injective module.

Here, we can assume that all the kernels ${f}_{n}:{Q}_{n}\to {Q}_{n-1}$ are equal to $R$, where $R$ is a fixed indecomposable module of length $2$. Also, if the ground field $k$ is infinite, then there is such a chain of epimorphisms such that all the kernels are pairwise different and of length $2$. In the first case, the kernels of the maps ${f}_{1}{f}_{2}\cdots {f}_{n}$ are all indecomposable (namely of the form $R[n]$ for $n\in \mathbb{N}$), in the second, they are direct sums of pairwise non-isomorphic modules of length 2.

### *Remark*

*right automorphism group*of $f$. The classification problem for the right minimal maps ending in $Y$ is divided in this way into two problems: to determine, on the one hand, the structure of the right factorization lattice $[\to Y\u3009$ and, on the other hand, to determine $\mathrm{r}-\mathrm{Aut}(f)$ for every right minimal map $f$ ending in $Y$. This provides a nice separation of the local symmetries and the global directedness, as mentioned at the beginning of the paper.

## 3 Morphisms determined by modules: Auslander’s First Theorem

*right C-determined*(or right determined by $C$) provided the following condition is satisfied: given any morphism ${f}^{\prime}:{X}^{\prime}\to Y$ such that ${f}^{\prime}\mathit{\varphi}$ factors through $f$ for all $\mathit{\varphi}:C\to {X}^{\prime}$, then ${f}^{\prime}$ itself factors through $f$. Thus one deals with the following diagrams:

The existence of the dashed arrow ${\mathit{\varphi}}^{\prime}$ on the left for all possible maps $\mathit{\varphi}:C\to {X}^{\prime}$ shall imply the existence of the dashed arrow $h$ on the right (of course, the converse implication always holds true: if ${f}^{\prime}=fh$ for some morphism $h$, then ${f}^{\prime}\mathit{\varphi}=f(h\mathit{\varphi})$ for all morphisms $\mathit{\varphi}:C\to {X}^{\prime}$).

### **Proposition 3.1**

- (a)
Assume that $\mathrm{add}C=\mathrm{add}{C}^{\prime}$. Then $f$ is right $C$-determined if and only if $f$ is right ${C}^{\prime}$-determined.

- (b)
If $f$ is right $C$-determined, then $f$ is also right $(C\oplus {C}^{\prime})$-determined.

### *Proof*

Trivial verification. $\square $

We denote by ${}^{C}[\to Y\u3009$ the set of the right equivalence classes of the morphisms ending in $Y$ which are right $C$-determined. We will see below that also ${}^{C}[\to Y\u3009$ is a lattice, thus we call it the *right C-factorization lattice* for $Y$.

Note that ${}^{C}[\to Y\u3009$ is usually not closed under predecessors or successors inside $[\to Y\u3009$. But there is the following important property:

### **Proposition 3.2**

The subset ${}^{C}[\to Y\u3009$ of $[\to Y\u3009$ is closed under meets.

### *Proof*

Let ${f}_{1}:{X}_{1}\to Y$ and ${f}_{2}:{X}_{2}\to Y$ be right $C$-determined. As we know, the meet of $[{f}_{1}\u3009$ and $[{f}_{2}\u3009$ is given by forming the pullback of ${f}_{1}$ and ${f}_{2}$. Thus assume that $X$ is the pullback with maps ${g}_{1}:X\to {X}_{1}$ and ${g}_{2}:X\to {X}_{2}$ and let $f={f}_{1}{g}_{1}={f}_{2}{g}_{2}.$ We want to show that $f$ is right $C$-determined. Thus, assume that there is given ${f}^{\prime}:{X}^{\prime}\to Y$ such that for any $\mathit{\varphi}:C\to {X}^{\prime}$, there exists ${\mathit{\varphi}}^{\prime}:C\to X$ such that ${f}^{\prime}\mathit{\varphi}=f{\mathit{\varphi}}^{\prime}$. Then we see that for any $\mathit{\varphi}:C\to {X}^{\prime}$, we have ${f}^{\prime}\mathit{\varphi}=f\mathit{\varphi}={f}_{1}({g}_{1}\mathit{\varphi})$, thus ${f}^{\prime}\mathit{\varphi}$ factors through ${f}_{1}$. Since ${f}_{1}$ is right $C$-determined, it follows that ${f}^{\prime}$ factors through ${f}_{1}$, say ${f}^{\prime}={f}_{1}{h}_{1}$ for some ${h}_{1}:{X}^{\prime}\to {X}_{1}$. Similarly, for any $\mathit{\varphi}:C\to {X}^{\prime}$, the morphism ${f}^{\prime}\mathit{\varphi}$ factors through ${f}_{2}$ and therefore ${f}^{\prime}={f}_{2}{h}_{2}$ for some ${h}_{2}:{X}^{\prime}\to {X}_{2}$. Now ${f}_{1}{h}_{1}={f}^{\prime}={f}_{2}{h}_{2}$ implies that there is $h:{X}^{\prime}\to X$ such that ${g}_{1}h={h}_{1},$ and ${g}_{2}h={h}_{2}$. Thus ${f}^{\prime}={f}_{1}{h}_{1}={f}_{1}{g}_{1}h=fh$ shows that ${f}^{\prime}$ factors through $f$. $\square $

We should stress that ${}^{C}[\to Y\u3009$ usually is not closed under joins, see the examples at the end of the section. One of these examples is chosen in order to convince the reader that this is not at all a drawback, but an important feature if we want to work with lattices of finite height.

### **Theorem 3.3**

where $C$ runs through all the $\mathrm{\Lambda}$-modules (or just through representatives of all multiplicity-free $\mathrm{\Lambda}$-modules) and this is a filtered union of meet-semilattices.

By definition, the sets ${}^{C}[\to Y\u3009$ are subsets of $[\to Y\u3009$. By Proposition 3.1(a), we know that ${}^{C}[\to Y\u3009$ only depends on $\mathrm{add}C$, thus we may restrict to look at representatives of multiplicity-free $\mathrm{\Lambda}$-modules $C$. Proposition 3.1(b) asserts that both ${}^{C}[\to Y\u3009$ and ${}^{{C}^{\prime}}[\to Y\u3009$ are contained in ${}^{C\oplus {C}^{\prime}}[\to Y\u3009$, thus we deal with a filtered union. According to Proposition 3.2, we deal with embeddings of meet-semilattices. The essential assertion of Theorem 3.3 is that any morphism is right determined by some module, the usual formulation of Auslander’s First Theorem. A discussion of this assertion and its proof follows.

There is a precise formula which yields for $f$ the smallest possible module $C(f)$ which right determines $f$. We will call it the minimal right determiner of $f$, any other right determiner of $f$ will have $C(f)$ as a direct summand.

*almost factor through*$f$, provided there is a commutative diagram of the following form

*If the indecomposable projective module P almost factors through f, then*$P/\mathrm{rad}P$

*embeds into the cokernel*$\mathrm{Cok}(f).$ Namely, given a map $\mathit{\eta}:P\to Y$ such that the image of $\mathit{\eta}$ is not contained in the image of $f$, as well as the commutative diagram above, we may complete the diagram by adding the cokernels of the horizontal maps:

Since the image of $\mathit{\eta}$ is not contained in the image of $f$, we see that ${\mathit{\eta}}^{\prime}$ is non-zero, thus $P\phantom{\rule{-0.166667em}{0ex}}/\phantom{\rule{-0.166667em}{0ex}}\mathrm{rad}P$ is a submodule of $\mathrm{Cok}(f).$

### **Theorem 3.4**

(Determiner formula of Auslander–Reiten–Smalø) Let $f$ be a morphism ending in $Y$. Let $C(f)$ be the direct sum of the indecomposable modules of the form ${\mathit{\tau}}^{-}K$, where $K$ is an indecomposable direct summand of the intrinsic kernel of $f$ and of the indecomposable projective modules which almost factor through $f$, one from each isomorphism class. Then $f$ is right $C$-determined if and only if $C(f)\in \mathrm{add}C$.

The theorem suggests to call $C(f)$ the *minimal right determiner of f*. For the proof of Theorem 3.4, see [5] and also [38].

### **Corollary 3.5**

Any morphism $f$ is right $C$-determined by some $C$, for example by the module

### *Proof*

We have to show that $C(f)$ is a direct summand of ${\mathit{\tau}}^{-}\mathrm{Ker}(f)\oplus P(\mathrm{soc}\mathrm{Cok}(f))$. The intrinsic kernel of $f$ is a direct summand of $\mathrm{Ker}(f)$, thus if $K$ is an indecomposable direct summand of the intrinsic kernel of $f$, then ${\mathit{\tau}}^{-}K$ is a direct summand of ${\mathit{\tau}}^{-}\mathrm{Ker}(f).$ Now assume that $S$ is a simple module such that $P(S)$ almost factors through $f$. Then $S$ is a submodule of $\mathrm{Cok}(f)$, thus $P(S)$ is a direct summand of $P(\mathrm{soc}\mathrm{Cok}(f)).$$\square $

### **Corollary 3.6**

(Auslander) The module ${\mathit{\tau}}^{-}\mathrm{Ker}(f)\oplus \mathrm{\Lambda}$ right determines $f$.

### **Corollary 3.7**

Let $P$ be a projective module and $f:X\to Y$ a right minimal morphism. Then $f$ is right $P$-determined if and only if $f$ is a monomorphism and the socle of the cokernel of $f$ is generated by $P$.

### *Proof*

This is an immediate consequence of the determiner formula: First, assume that $f$ is right $P$-determined. Then the intrinsic kernel of $f$ has to be zero. Since we assume that $f$ is right minimal, $f$ must be a monomorphism. If $S$ is a simple submodule of the cokernel of $f$, then $P(S)$ almost factors through $f$, thus $P(S)$ is a direct summand of $P$. This shows that the socle of the cokernel of $f$ is generated by $P$. Conversely, assume that $f$ is a monomorphism and the socle of the cokernel of $f$ is generated by $P$. Since $f$ is a monomorphism, $C(f)$ is the direct sum of all indecomposable projective modules ${P}^{\prime}$ which almost factor through $f$. Such a module ${P}^{\prime}$ is the projective cover of a simple submodule of $\mathrm{Cok}(f)$. Since $P$ generates the socle of the cokernel of $f$, it follows that ${P}^{\prime}$ is a direct summand of $P$. Thus $C(f)$ is in $\mathrm{add}P$, therefore $f$ is right $P$-determined. $\square $

### **Corollary 3.8**

A right minimal morphism $f:X\to Y$ is a monomorphism if and only if it is right $\mathrm{\Lambda}$-determined.

### *Example 3*

*The subset*${}^{C}[\to Y\u3009$

*of*$[\to Y\u3009$

*is usually not closed under joins,*as the following example shows: Let $\mathrm{\Lambda}$ be the path algebra of the quiver of type ${\mathbb{A}}_{3}$ with two sources, namely of

Let ${f}_{i}:P({b}_{i})\to Q(a)$ be non-zero maps for $i=1,2$, these are monomorphisms, thus they are right $\mathrm{\Lambda}$-determined. The join of $[{f}_{1}\u3009$ and $[{f}_{2}\u3009$ in $[\to Y\u3009$ is given by the map $[{f}_{1},{f}_{2}]:P({b}_{1})\oplus P({b}_{2})\to Q(a)$. Clearly, this map is right minimal, but it is not injective. Thus $[{f}_{1},{f}_{2}]$ is not right $\mathrm{\Lambda}$-determined.

### *Example 4*

Here, the modules $R,{R}^{\prime},{R}^{\prime \prime},\cdots $ are the indecomposable representations of length 2, one from each isomorphism class and all the arrows are inclusion maps.

The join in ${}^{C}[\to Y\u3009$ of two different maps ${f}_{1},{f}_{2}$ in the height 2 layer is just the identity map $Y\to Y$, whereas the join of ${f}_{1},{f}_{2}$ in $[\to Y\u3009$ is the direct sum map $[{f}_{1},{f}_{2}]:{R}_{1}\oplus {R}_{2}\to Y.$ More generally, if there are given $n$ pairwise different regular modules ${R}_{1},\cdots ,{R}_{n}$ of length 2 with inclusion maps ${f}_{i}:{R}_{i}\to Y$, then the join in $[\to Y\u3009$ is the direct sum map $[{f}_{1},\cdots ,{f}_{n}]:{R}_{1}\oplus \cdots \oplus {R}_{n}\to Y$. Let us stress that all these direct sum maps are right minimal (thus here we deal with a cofork as defined in Sect. 13). Thus, if the base field $k$ is infinite, the smallest subposet of $[\to Y\u3009$ closed under meets and joins and containing the inclusion maps $R\to Y$ with $R$ regular of length 2 will have infinite height.

### **Proposition 3.9**

Let $f:X\to Y$ be a morphism. If ${C}^{\prime}$ is an indecomposable direct summand of $C(f)$, then $\mathrm{Hom}({C}^{\prime},Y)\ne 0.$

### *Proof*

By definition, there are two kinds of indecomposable direct summands of $C(f)$, the non-projective ones are of the form ${\mathit{\tau}}^{-}{K}^{\prime}$, where ${K}^{\prime}$ is an indecomposable direct summand of the intrinsic kernel of $f$, the remaining ones are the indecomposable projective modules which almost factor through $f$. Of course, if $P$ is an indecomposable projective module which almost factors through $f$, then $\mathrm{Hom}(P,Y)\ne 0$.

If we assume that ${\mathit{\varphi}}^{\prime}=0$, then $f\mathit{\varphi}=0$, thus $\mathit{\varphi}$ factors through the kernel of $f$, say $\mathit{\varphi}=u{\mathit{\varphi}}^{\prime \prime}.$ Consequently, $u{u}^{\prime}=\mathit{\varphi}\mathit{\mu}=u{\mathit{\varphi}}^{\prime \prime}\mathit{\mu}$. But $u$ is injective, thus ${u}^{\prime}={\mathit{\varphi}}^{\prime \prime}\mathit{\mu}$ and therefore ${1}_{{K}^{\prime}}=r{u}^{\prime}=r{\mathit{\varphi}}^{\prime \prime}\mathit{\mu}$. But this means that $\mathit{\mu}$ is split mono, a contradiction. It follows that ${\mathit{\varphi}}^{\prime}\ne 0$, thus $\mathrm{Hom}({C}^{\prime},Y)\ne 0.$$\square $

### *Remark*

Proposition 3.9 asserts that all the indecomposable direct summands ${C}^{\prime}$ of the minimal right determiner $C(f)$ of a map $f:X\to Y$ satisfy $\mathrm{Hom}({C}^{\prime},Y)\ne 0$. Actually, according to [5], Proposition XI.2.4 (see also [38]), such a module ${C}^{\prime}$ is equipped with a distinguished non-zero map ${C}^{\prime}\to Y$ which is said to “almost factor through” $f$. At the beginning of this section we gave a corresponding definition in the special case when ${C}^{\prime}$ is projective. See also the Remark 3 at the end of Sect. 4.

## 4 The Auslander bijection. Auslander’s Second Theorem

Let $C,Y$ be objects. Let $\mathrm{\Gamma}(C)=\mathrm{End}{(C)}^{\phantom{\rule{4.pt}{0ex}}\text{op}\phantom{\rule{4.pt}{0ex}}}$. We always will consider $\mathrm{Hom}(C,Y)$ as a $\mathrm{\Gamma}(C)$-module. For any module $M$, we denote by $\mathcal{S}M$ the set of all submodules (it is a lattice with respect to intersection and sum of submodules).

Here is a reformulation of the definition of ${\mathit{\eta}}_{CY}$.

### **Proposition 4.1**

Let $f:X\to Y$. Then ${\mathit{\eta}}_{CY}(f)$ is the set of all $h\in \mathrm{Hom}(C,Y)$ which factor through $f$. This subset of $\mathrm{Hom}(C,Y)$ is a $\mathrm{\Gamma}(C)$-submodule.

### *Proof*

We have mentioned already, that ${\mathit{\eta}}_{CY}(f)$ is a $\mathrm{\Gamma}(C)$-submodule of $\mathrm{Hom}(C,Y)$. Also, if $h\in {\mathit{\eta}}_{CY}(f)=f\mathrm{Hom}(C,X)$, then $h$ factors through $f$. And conversely, if $h$ factors through $f$, then $h$ belongs to $f\mathrm{Hom}(C,X)={\mathit{\eta}}_{CY}(f).$$\square $

### **Lemma 4.2**

If $X={X}_{0}\oplus {X}_{1}$ and $f({X}_{0})=0$, then ${\mathit{\eta}}_{CY}(f)={\mathit{\eta}}_{CY}(f|{X}_{1}).$

### *Proof*

*If*${f}_{1}$

*is a right minimal version of f, then*${\mathit{\eta}}_{CY}(f)={\mathit{\eta}}_{CY}({f}_{1})$. Thus, ${\mathit{\eta}}_{CY}$ is constant on right equivalence classes and we can define ${\mathit{\eta}}_{CY}([f\u3009)={\mathit{\eta}}_{CY}(f)$. We obtain in this way a map

### **Proposition 4.3**

### *Proof*

A trivial verification: First, let us show that ${\mathit{\eta}}_{CZ}$ is injective. Consider maps $f:X\to Y$ and ${f}^{\prime}:{X}^{\prime}\to Y$ such that $f\mathrm{Hom}(C,X)={f}^{\prime}\mathrm{Hom}(C,{X}^{\prime}).$ Since $f$ is right $C$-determined and ${f}^{\prime}\mathrm{Hom}(C,{X}^{\prime})\subseteq f\mathrm{Hom}(C,X)$, we see that ${f}^{\prime}\in f\mathrm{Hom}({X}^{\prime},X)$. Since ${f}^{\prime}$ is right $C$-determined and $f\mathrm{Hom}(C,X)\subseteq {f}^{\prime}\mathrm{Hom}(C,{X}^{\prime})$, we see that $f\in f\mathrm{Hom}(X,{X}^{\prime})$. But this means that ${f}^{\prime}\u2aaff\u2aaf{f}^{\prime}$, thus $[f\u3009=[{f}^{\prime}\u3009.$ $\square $

Auslander’s Second Theorem (as established in [2]) asserts:

### **Theorem 4.4**

*The map*${\mathit{\eta}}_{CY}$

*defined by*${\mathit{\eta}}_{CY}(f)=\mathrm{Im}\mathrm{Hom}(C,f)$

*yields a lattice isomorphism*

*The composition*

*of the inclusion map and the map*${\mathit{\eta}}_{CY}$ defined by ${\mathit{\eta}}_{CY}(f)=\mathrm{Im}\mathrm{Hom}(C,f)$*is a lattice isomorphism.*

**Convention.** In the following, several examples of Auslander bijections will be presented. When looking at the submodule lattice $\mathcal{S}M$ of a module $M$, we usually will mark (some of) the elements of $\mathcal{S}M$ by bullets $\bullet $ and connect comparable elements by a solid lines. Here, going upwards corresponds to the inclusion relation.

For the corresponding lattices ${}^{C}[\to Y\u3009$, we often will mark an element $[f:X\to Y\u3009$ (with $f$ a right minimal map) by just writing $X$ and we will connect neighboring pairs $[f:X\to Y\u3009\le [{f}^{\prime}:{X}^{\prime}\to Y\u3009$ by drawing an (upwards) arrow $X\to Y$. On the other hand, sometimes it seems to be more appropriate to refer to the right minimal map $f:X\to Y$ with kernel ${K}^{\prime}$ and image ${Y}^{\prime}$ by using the short exact sequence notation ${K}^{\prime}\to X\to {Y}^{\prime}$.

Note that the lattice $\mathcal{S}\mathrm{Hom}(C,Y)$ has two distinguished elements, namely $\mathrm{Hom}(C,Y)$ itself as well as its zero submodule. Under the bijection ${\mathit{\eta}}_{CY}$ the total submodule $\mathrm{Hom}(C,Y)$ corresponds to the identity map ${1}_{Y}$ of $Y$, this is not at all exciting. But of interest seem to be the maps in ${\mathit{\eta}}_{CY}^{-1}(0)$, we will discuss them in this will be discussed in Proposition 5.5

**The special case**$C={}_{\mathrm{\Lambda}}\mathrm{\Lambda}$. It is worthwhile to draw the attention on the special case when $C={}_{\mathrm{\Lambda}}\mathrm{\Lambda}$.

### **Proposition 4.5**

The special case of the Auslander bijection ${\mathit{\eta}}_{\mathrm{\Lambda}Y}$ is the obvious identification of both ${}^{\mathrm{\Lambda}}[\to Y\u3009$ and $\mathcal{S}\mathrm{Hom}(\mathrm{\Lambda},Y)$ with $\mathcal{S}Y$.

### *Proof*

First, consider ${}^{\mathrm{\Lambda}}[\to Y\u3009$: The determiner formula asserts: a right minimal morphism is right $\mathrm{\Lambda}$-determined if and only if it is a monomorphism. Thus ${}^{\mathrm{\Lambda}}[\to Y\u3009$ is just the set of right equivalence classes of monomorphisms ending in $Y$, and the map $f\mapsto \mathrm{Im}(f)$ yields an identification between the set of right equivalence classes of monomorphisms ending in $Y$ and the submodules of $\mathrm{\Lambda}$.

Next, we deal with $\mathcal{S}\mathrm{Hom}(\mathrm{\Lambda},Y)$. Note that $\mathrm{\Gamma}({}_{\mathrm{\Lambda}}\mathrm{\Lambda})=\mathrm{End}{({}_{\mathrm{\Lambda}}\mathrm{\Lambda})}^{\phantom{\rule{4.pt}{0ex}}\text{op}\phantom{\rule{4.pt}{0ex}}}=\mathrm{\Lambda}$ and there is a canonical identification $\mathit{\u03f5}:\mathrm{Hom}(\mathrm{\Lambda},Y)\simeq Y$ (given by $\mathit{\u03f5}(h)=h(1)$ for $h\in \mathrm{Hom}(\mathrm{\Lambda},Y)$), thus $\mathcal{S}\mathit{\u03f5}:\mathcal{S}\mathrm{Hom}(\mathrm{\Lambda},Y)\simeq \mathcal{S}Y$ (with $\mathcal{S}\mathit{\u03f5}(U)=\{h(1)\mid h\in U\}$ for $U$ a submodule of $\mathrm{Hom}(\mathrm{\Lambda},Y)$).

As a consequence, we see that *all possible submodule lattices*$\mathcal{S}Y$*occur as images under the Auslander bijections.* This assertion can be strengthened considerably, as we want to show now.

By definition, an artin algebra $\mathrm{\Lambda}$ is an artin $k$-algebra for some commutative artinian ring $k$ (this means that $\mathrm{\Lambda}$ is a $k$-algebra and that it is finitely generated as a $k$-module). Such an algebra is said to be *strictly wild* (or better *strictly*$k$*-wild*), provided for any artin $k$-algebra $\mathrm{\Gamma}$, there is a full exact embedding $\mathrm{mod}\mathrm{\Gamma}\to \mathrm{mod}\mathrm{\Lambda}$. If $M$ is a $\mathrm{\Lambda}$-module and ${M}^{\prime}$ is a ${\mathrm{\Lambda}}^{\prime}$-module, a *semilinear isomorphism* from $M$ to ${M}^{\prime}$ is a pair $(\mathit{\alpha},f)$, where $\mathit{\alpha}:\mathrm{\Lambda}\to {\mathrm{\Lambda}}^{\prime}$ is an algebra isomorphism, and $f:M\to {M}^{\prime}$ is an isomorphism of abelian groups such that $f(\mathit{\lambda}m)=\mathit{\alpha}(\mathit{\lambda})f(m)$ for all $\mathit{\lambda}\in \mathrm{\Lambda}$ and $m\in M$. It is clear that any semilinear isomorphism from $M$ to ${M}^{\prime}$ induces a lattice isomorphism $\mathcal{S}M\to \mathcal{S}{M}^{\prime}.$

### **Proposition 4.6**

Let $\mathrm{\Lambda}$ be an artin $k$-algebra which is strictly $k$-wild. Let $\mathrm{\Gamma}$ be an artin $k$-algebra and $M$ a $\mathrm{\Gamma}$-module. Then there are $\mathrm{\Lambda}$-modules $C,Y$ such that the $\mathrm{\Gamma}(C)$-module $\mathrm{Hom}(C,Y)$ is semilinearly isomorphic to $M$. Thus there is a lattice isomorphism ${}^{C}[\to Y\u3009\to \mathcal{S}M$.

### *Proof*

Let $F:\mathrm{mod}\mathrm{\Gamma}\to \mathrm{mod}\mathrm{\Lambda}$ be a full embedding (we do not need that it is exact). Let $C=F({}_{\mathrm{\Gamma}}\mathrm{\Gamma})$ and $Y=F(M)$. Let $\mathit{\alpha}:\mathrm{\Gamma}=\mathrm{End}{({}_{\mathrm{\Gamma}}\mathrm{\Gamma})}^{\phantom{\rule{4.pt}{0ex}}\text{op}\phantom{\rule{4.pt}{0ex}}}\to \mathrm{End}{(C)}^{\phantom{\rule{4.pt}{0ex}}\text{op}\phantom{\rule{4.pt}{0ex}}}=\mathrm{\Gamma}(C)$ as well as $f:M=\mathrm{Hom}({}_{\mathrm{\Gamma}}\mathrm{\Gamma},M)\to \mathrm{Hom}(C,M)$ both be given by applying the functor $F$. Since $F$ is a full embedding, $\mathit{\alpha}$ is an algebra isomorphism and $f$ is an isomorphism of abelian groups. The functoriality of $F$ asserts that we also have $f(\mathit{\gamma}m)=\mathit{\alpha}(\mathit{\gamma})f(m)$ for all $\mathit{\gamma}\in \mathrm{\Gamma}$ and $m\in M$. This shows that the pair $(\mathit{\alpha},f)$ is a semilinear isomorphism. $\square $

### *Remark*

but even if $F$ is exact, such a bijection will not be given by applying directly $F$. Namely, if $f:X\to Y$ is right minimal and right $C$-determined, then the kernel of $f$ belongs to $\mathrm{add}\mathit{\tau}C$, thus the kernel of $F(f)$ belongs to $\mathrm{add}F(\mathit{\tau}C)$, whereas the intrinsic kernel of any right $F(C)$-determined map has to belong to $\mathrm{add}\mathit{\tau}F(C)$ and the $\mathrm{\Lambda}$-modules $F(\mathit{\tau}C)$ and $\mathit{\tau}F(C)$ may be very different, as the obvious embeddings of the category of $n$-Kronecker modules into the category of $(n+1)$-Kronecker modules (using for one arrow the zero map) show.

Note that under a full exact embedding functor $F:\mathrm{mod}\mathrm{\Gamma}\to \mathrm{mod}\mathrm{\Lambda}$, submodule lattices are usually not preserved: given a $\mathrm{\Gamma}$-module $M$, the functor $F$ yields an embedding of $\mathcal{S}({}_{\mathrm{\Gamma}}M)$ into $\mathcal{S}({}_{\mathrm{\Lambda}}F(M))$, but usually this is a proper embedding. Actually, for any finite-dimensiona algebra $\mathrm{\Lambda}$, there are submodule lattices $\mathcal{S}({}_{\mathrm{\Gamma}}M)$ which cannot be realized as the submodule lattice of any $\mathrm{\Lambda}$-module. Namely, assume that the length of the indecomposable projective $\mathrm{\Lambda}$-modules is bounded by $t$ and take a finite-dimensional algebra $\mathrm{\Gamma}$ with a local $\mathrm{\Gamma}$-module $M$ of length $t+1$. Then $\mathcal{S}M$ is a modular lattice of height $t+1$ with a unique element of height $t$ (the radical of the module $M$). If $\mathcal{S}({}_{\mathrm{\Lambda}}Y)$ is of the form $\mathcal{S}M$, then $Y$ has to be a local $\mathrm{\Lambda}$-module of length $t+1$, thus a factor module of an indecomposable projective $\mathrm{\Lambda}$-module. But by assumption, the indecomposable projective $\mathrm{\Lambda}$-modules have length at most $t$.

### *Remark 1*

Let $Y=\u2a01{Y}_{i},$ then the subsets $\mathrm{Hom}(C,{Y}_{i})$ of $\mathrm{Hom}(C,Y)$ are actually $\mathrm{\Gamma}(C)$-submodules and there is an isomorphism of $\mathrm{\Gamma}(C)$-modules $\mathrm{Hom}(C,Y)\simeq {\u2a01}_{i}\mathrm{Hom}(C,{Y}_{i})$. Thus ${\mathit{\eta}}_{CY}$ maps the lattice ${}^{C}[\to Y\u3009$ bijectively onto the submodule lattice $\mathcal{S}({\u2a01}_{i}\mathrm{Hom}(C,{Y}_{i})$. The lattice $\mathcal{S}({\u2a01}_{i}\mathrm{Hom}(C,{Y}_{i})$ contains ${\prod}_{i}\mathcal{S}\mathrm{Hom}(C,{Y}_{i})$ as a sublattice and both have the same height. However, ${\prod}_{i}\mathcal{S}\mathrm{Hom}(C,{Y}_{i})$ may be a proper sublattice of $\mathcal{S}({\u2a01}_{i}\mathrm{Hom}(C,{Y}_{i})$, since isomorphisms of subfactors of the various modules $\mathrm{Hom}(C,{Y}_{i})$ yield diagonals in $\mathcal{S}({\u2a01}_{i}\mathrm{Hom}(C,{Y}_{i})$.

### *Remark 2*

When dealing with the Auslander bijections ${\mathit{\eta}}_{CY}:{}^{C}[\to Y\u3009\to \mathcal{S}\mathrm{Hom}(C,Y)$, we always can assume that $C$ is multiplicity-free and supporting, here *supporting* means that $\mathrm{Hom}({C}_{i},Y)\ne 0$ for any indecomposable direct summand ${C}_{i}$ of $C$. Namely, let ${C}^{\prime}$ be the direct sum of all indecomposable direct summands ${C}_{i}$ of $C$ with $\mathrm{Hom}({C}_{i},Y)\ne 0$, one from each isomorphism class. Then, on the one hand, ${}^{C}[\to Y\u3009={}^{{C}^{\prime}}[\to Y\u3009$ (since a map $f$ ending in $Y$ is right $C$-determined if and only if it is right ${C}^{\prime}$-determined. On the other hand, there is an idempotent $e\in \mathrm{\Gamma}(C)$ such that $e\mathrm{\Gamma}(C)e=\mathrm{\Gamma}({C}^{\prime})$ and $e\mathrm{Hom}(C,Y)e=\mathrm{Hom}({C}^{\prime},Y)$, and there is a lattice isomorphism $\mathcal{S}\mathrm{Hom}(C,Y)\to \mathcal{S}\mathrm{Hom}({C}^{\prime},Y),$ given by $U\mapsto eU$, where $U$ is a submodule of $\mathrm{Hom}(C,Y)$.

### *Remark 3*

Both objects ${}^{C}[\to Y\u3009$ and $\mathcal{S}\mathrm{Hom}(C,Y)$ related by the Auslander bijection ${\mathit{\eta}}_{CY}$ concern morphisms ending in $Y$. Of course, in Proposition 3.9 we have seen already that all the indecomposable direct summands ${C}^{\prime}$ of the minimal right determiner $C(f)$ of a map $f:X\to Y$ satisfy $\mathrm{Hom}({C}^{\prime},Y)\ne 0.$

**not**start at $C$, thus the relationship between the elements of ${}^{C}[\to Y\u3009$ and the submodules of $\mathrm{Hom}(C,Y)$ is really of interest! Note however that in case we deal with a map $f:C\to Y$ which is right $C$-determined (and starts in $C$), then

We use the next two sections in order to transfer well-known properties of the lattice of submodules of a finite length module to the right $C$-factorization lattices, in particular the Jordan–Hölder theorem. In Sect. 5, we introduce the right $C$-length of a right $C$-determined map $f$ ending in $Y$, it corresponds to the the length of the factor module $\mathrm{Hom}(C,Y)/{\mathit{\eta}}_{CY}(f).$ In Sect. 6 we will define the $C$-type of $f$ as the dimension vector of $\mathrm{Hom}(C,Y)/{\mathit{\eta}}_{CY}(f).$

## 5 Right $C$-factorizations and right $C$-length

The Auslander bijection asserts that the lattice ${}^{C}[\to Y\u3009$ is a modular lattice of finite height, thus there is a Jordan–Hölder Theorem for ${}^{C}[\to Y\u3009$; it can be obtained from the corresponding Jordan–Hölder Theorem for the submodule lattice $\mathcal{S}\mathrm{Hom}(C,Y)$. In Sects. 5 and 6, we are going to formulate the assertions for ${}^{C}[\to Y\u3009$ explicitly. Here we consider composition series of submodules and factor modules of ${}^{C}[\to Y\u3009$.

Let ${h}_{i}:{X}_{i}\to {X}_{i-1}$ be maps, where $1\le i\le t,$ with composition $f={h}_{1}\cdots {h}_{t}$. The sequence $({h}_{1},{h}_{2},\cdots ,{h}_{t})$ is called a *right C-factorization of f of length t* provided the maps ${h}_{i}$ are non-invertible and the compositions ${f}_{i}={h}_{1}\cdots {h}_{i}$ are right minimal and right $C$-determined, for $1\le i\le t.$ It sometimes may be helpful to deal also with right $C$-factorizations of length $0$; by definition these are just the identity maps (or, if you prefer, the isomorphisms).

If $({h}_{1},\cdots ,{h}_{t})$ is a right $C$-factorization of a map $f$, then any integer sequence $0=i(0)<i(1)<\cdots <i(s)=t$ defines a sequence of maps $({h}_{1}^{\prime},{h}_{2}^{\prime},\cdots ,{h}_{s}^{\prime})$ with ${h}_{j}^{\prime}={h}_{i(j-1)+1}\cdots {h}_{i(j)}$ for $1\le j\le s$. We use the following lemma inductively, in order to show that $({h}_{1}^{\prime},{h}_{2}^{\prime},\cdots ,{h}_{s}^{\prime})$ is again a right $C$-factorization of $f$ and we say that $({h}_{1},{h}_{2},\cdots ,{h}_{t})$ is a *refinement* of $({h}_{1}^{\prime},{h}_{2}^{\prime},\cdots ,{h}_{s}^{\prime})$. In particular, any right $C$-factorization $({h}_{1},\cdots ,{h}_{t})$ of $f$ is a refinement of $f$.

### **Lemma 5.1**

### *Proof*

We only have to check that ${h}_{i}{h}_{i+1}$ cannot be invertible. Assume ${h}_{i}{h}_{i+1}$ is invertible. Then ${h}_{i}$ is a split epimorphism. Since ${f}_{i}={h}_{1}\cdots {h}_{i}$ is right minimal, it follows that ${h}_{i}$ is invertible, a contradiction. $\square $

We say that a right $C$-factorization $({h}_{1},{h}_{2},\cdots ,{h}_{t})$ is *maximal* provided it does not have a refinement of length $t+1$.

### **Proposition 5.2**

### *Proof*

This is a direct consequence of Auslander’s Second Theorem. $\square $

### **Corollary 5.3**

Any right $C$-factorization $({h}_{1},\cdots ,{h}_{t})$ has a refinement which is a maximal right $C$-factorization and all maximal right $C$-factorizations of $({h}_{1},\cdots ,{h}_{t})$ have the same length.

### *Proof*

This follows from Proposition 5.2 and the Jordan–Hölder theorem. $\square $

In particular, any right minimal right $C$-determined map $f$ has a refinement which is a maximal right $C$-factorization, say $({h}_{1},\cdots ,{h}_{t})$ and its length $t$ will be called the *right*$C$*-length* of $f$, we write ${|f|}_{C}$ for the right $C$-length of $f$. There is the following formula:

### **Proposition 5.4**

where $|\mathrm{Hom}(C,Y)|$ denotes the length of the $\mathrm{\Gamma}(C)$-module $\mathrm{Hom}(C,Y)$ and $|{\mathit{\eta}}_{CY}(f)|$ the length of its $\mathrm{\Gamma}(C)$-submodule ${\mathit{\eta}}_{CY}(f)$.

**The right equivalence class**${\mathit{\eta}}_{CY}^{-1}(0).$ As we have mentioned in Sect. 4, it is of interest to determine the maps in the right equivalence class ${\mathit{\eta}}_{CY}^{-1}(0).$

### **Proposition 5.5**

Let $C,Y$ be modules. Up to right equivalence, there is a unique right $C$-determined map $f$ ending in $Y$ with ${|f|}_{C}$ maximal. The submodule ${\mathit{\eta}}_{CY}(f)$ of $\mathrm{Hom}(C,Y)$ is the zero module. If ${f}^{\prime}$ is any right $C$-determined map ending in $Y$, then $f={f}^{\prime}h$ for some $h$.

### *Proof*

The lattice ${}^{C}[\to Y\u3009$ has a unique zero element, namely ${\mathit{\eta}}_{CY}^{-1}(0)$. Let ${\mathit{\eta}}_{CY}^{-1}(0)=[f\u3009$ for some right minimal map $f$. Then, the right $C$-length of $f$ has to be maximal and $[f\u3009\le [{f}^{\prime}\u3009$ for any right $C$-determined map ${f}^{\prime}$ ending in $Y$. $\square $

In general it seems to be quite difficult to describe the maps $f$ such that $[f\u3009={\mathit{\eta}}_{CY}^{-1}(0).$ But one should be aware that such a map $f$ always does exist: any pair $C,Y$ of $\mathrm{\Lambda}$-modules determines uniquely up to right equivalence a map $f$ ending in $Y$, namely the right minimal, right $C$-determined map $f$ with ${\mathit{\eta}}_{CY}(f)=0.$

### **Proposition 5.6**

Let $C,Y$ be modules. The set ${\mathit{\eta}}_{CY}^{-1}(0)$ is the right equivalence class of the zero map $0\to Y$ if and only if $P(\mathrm{soc}Y)$ belongs to $\mathrm{add}C$.

### *Proof*

This is a direct consequence of Corollary 3.7. $\square $

**The special case of**$C$**being projective.** For an arbitrary projective module $C$, there is the following description of the right $C$-length of a right minimal, right $C$-determined morphism $f$. Here, we denote by $[M:S]$ the Jordan–Hölder multiplicity of the simple module $S$ in the module $M$, this is the number of factors in a composition series of $M$ which are isomorphic to $S$.

### **Proposition 5.7**

Let $C$ be projective. The right minimal, right $C$-determined maps $f:X\to Y$ are up to right equivalence just the inclusion maps of submodules $X$ of $Y$ such that the socle of $Y/X$ is generated by $C$.

The minimal element ${\mathit{\eta}}_{CY}^{-1}(0)$ of ${}^{C}[\to Y\u3009$ is the inclusion map $X\to Y$, where $X$ is the intersection of the kernels of all maps $Y\to Q(S)$, where $S$ is a simple module with $P(S)$ a direct summand of $C$.

### *Proof*

Let $\mathcal{Q}$ be the set of modules $Q(S)$, where $S$ is a simple module with $P(S)$ a direct summand of $C$. Let $X$ be the intersection of the kernels of all maps $Y\to Q$ with $Q\in \mathcal{Q}.$ Since $Y$ is of finite length, there are finitely many maps ${g}_{i}:Y\to Q({S}_{i})$ with $Q({S}_{i})\in \mathcal{Q}$, say $1\le i\le m$, such that $X={\bigcap}_{i=1}^{m}\mathrm{Ker}({g}_{i}).$ Then $Y/X$ embeds into ${\u2a01}_{i=1}^{m}Q({S}_{i})$, thus its socle is generated by $C$. It follows that the inclusion map $X\to Y$ is right $C$-determined. On the other hand, if ${X}^{\prime}\to Y$ is right minimal and right $C$-determined, then it is a monomorphism, thus we can assume that it is an inclusion map. In addition, we know that the socle of $Y/{X}^{\prime}$ is generated by $C$, thus $Y/{X}^{\prime}$ embeds into a finite direct sum of modules in $\mathcal{Q}$. It follows that ${X}^{\prime}$ is the intersection of some maps $Y\to Q$, where $Q\in \mathcal{Q}$, thus $X\subseteq {X}^{\prime}.$$\square $

There is the following consequence: *The*${}_{\mathrm{\Lambda}}\mathrm{\Lambda}$*-length of any inclusion map*$X\to Y$ (such a map is obviously right minimal and right $\mathrm{\Lambda}$-determined) *is precisely the length of*$Y/X.$

For further results concerning the right $C$-length of maps, see Sect. 9.

## 6 The right $C$-type of a right $C$-determined map

Recall that we consider $\mathrm{Hom}(C,Y)$ as a $\mathrm{\Gamma}(C)$-module, where $\mathrm{\Gamma}(C)=\mathrm{End}{(C)}^{\phantom{\rule{4.pt}{0ex}}\text{op}\phantom{\rule{4.pt}{0ex}}}.$ The indecomposable projective $\mathrm{\Gamma}(C)$-modules are of the form $\mathrm{Hom}(C,{C}_{i})$, where ${C}_{i}$ is an indecomposable direct summand of $C$, thus the simple $\mathrm{\Gamma}(C)$-modules are of the form $S({C}_{0})=\mathrm{top}\mathrm{Hom}(C,{C}_{0}).$

Given an artin algebra $\mathrm{\Gamma}$, we denote by ${K}_{0}(\mathrm{\Gamma})$ its Grothendieck group (of all $\mathrm{\Gamma}$-modules modulo all exact sequences), it is the free abelian group with basis the set of isomorphism classes $[S]$ of the simple $\mathrm{\Gamma}$-modules $S$. Given a $\mathrm{\Gamma}$-module $M$, we denote by $\mathbf{dim}M$ the corresponding element in ${K}_{0}(\mathrm{\Gamma})$, called the *dimension vector* of $M$. Of course, $\mathbf{dim}M$ can be written as an integral linear combination $\mathbf{dim}M={\sum}_{[S]}[M:S][S],$ where the coefficient of $[S]$ is just the Jordan–Hölder multiplicity $[M:S]$ of $S$ in $M$. The elements of ${K}_{0}(\mathrm{\Gamma})$ with non-negative coefficients will be said to be the $\mathrm{\Gamma}$-dimension vectors. If $\mathbf{e}$ is a $\mathrm{\Gamma}$-dimension vector and $M$ is a $\mathrm{\Gamma}$-module, we denote by ${\mathcal{S}}_{\mathbf{e}}M$ the subset of $\mathcal{S}M$ consisting of all submodules of $M$ with dimension vector $\mathbf{e}.$

Let us return to the artin algebra $\mathrm{\Gamma}(C)$, where $C$ is a $\mathrm{\Lambda}$-module. Thhe Grothendieck group ${K}_{0}(\mathrm{\Gamma}(C))$ is the free abelian group with basis the set of modules $S({C}_{i})$, where ${C}_{i}$ runs through a set of representatives of the isomorphism classes of the indecomposable direct summands ${C}_{i}$ of $C$. We are interested here in the dimension vectors of $\mathrm{Hom}(C,Y)$ and of its factor modules. Actually, we want to attach to each right $C$-determined map ending in $Y$ its right $C$-type $\mathbf{t}yp{e}_{C}(f)$ so that $\mathbf{t}yp{e}_{C}(f)=\mathbf{dim}\mathrm{Hom}(C,Y)/{\mathit{\eta}}_{CY}(f).$ We start with pairs of neighbors in the right $C$-factorization lattice ${}^{C}[\to Y\u3009$, since they correspond under ${\mathit{\eta}}_{CY}$ to the composition factors of $\mathrm{Hom}(C,Y)$.

Let $f=h{f}^{\prime}$ and ${f}^{\prime}$ be right minimal, right $C$-determined maps. We say that the pair $(f,{f}^{\prime})$ is a pair of *C-neighbors* provided ${|f|}_{C}={|{f}^{\prime}|}_{C}+1.$ Note that the pair $(f,{f}^{\prime})$ in ${}^{C}[\to Y\u3009$ is a pair of $C$-neighbors provided $[f\u3009<[{f}^{\prime}\u3009$ and there is no ${f}^{\prime \prime}$ with $[f\u3009<[{f}^{\prime \prime}\u3009<[{f}^{\prime}\u3009$ (of course, it is the condition $[f\u3009<[{f}^{\prime}\u3009$ which implies that there is a map $h$ with $f={f}^{\prime}h$).

### *Remark*

Let us consider a composition $f={f}^{\prime}h$, where ${f}^{\prime}$ both are right minimal and right $C$-determined. It can happen that $h$ is also right minimal and right $C$-determined, but $f={f}^{\prime}h$ is not right $C$-determined. Also it can happen that both maps ${f}^{\prime}$ and $f={f}^{\prime}h$ are right minimal and right $C$-determined, whereas $h$ is not right $C$-determined. Here are corresponding examples.

### *Example 5*

First, let $C=S(b)\oplus S(c)$, thus $\mathit{\tau}C=S(a)\oplus S(b)$ and both ${f}^{\prime}$ and $h$ are right $C$-determined, whereas $f$ is not right $C$-determined.

Second, let $C=Q(b)\oplus S(c),$ thus $\mathit{\tau}C=P(b)\oplus S(b)$. Then both $f$ and ${f}^{\prime}$ are right $C$-determined, whereas $h$ is not right $C$-determined.

Let $f:X\to Y$ and ${f}^{\prime}:{X}^{\prime}\to Y$ such that $(f,{f}^{\prime})$ is a pair of neighbors. We say that $(f,{f}^{\prime})$ is *of type*${C}_{0}$ (or better of type $[S({C}_{0})]$) where ${C}_{0}$ is an indecomposable direct summand of $C$, provided there is a map $\mathit{\varphi}:{C}_{0}\to {X}^{\prime}$ such that ${f}^{\prime}\mathit{\varphi}$ does not factor through $f$. Such a summand ${C}_{0}$ must exist, since otherwise ${f}^{\prime}$ would factor through $f$, due to the fact that $f$ is right ${C}_{0}$-determined. The following proposition shows that ${C}_{0}$ is uniquely determined.

### **Proposition 6.1**

If $(f,{f}^{\prime})$ is a pair of $C$-neighbors of type ${C}_{0}$, then ${\mathit{\eta}}_{CY}({f}^{\prime})/{\mathit{\eta}}_{CY}(f)$ is isomorphic to the simple $\mathrm{\Gamma}(C)$-module $S({C}_{0})=\mathrm{top}\mathrm{Hom}(C,{C}_{0}).$ Thus, the type of a pair of $C$-neighbors is well-defined.

Thus, if $(f,{f}^{\prime})$ is a pair of $C$-neighbors of type ${C}_{0}$, we may write $\mathbf{t}yp{e}_{C}(f,{f}^{\prime})=[S({C}_{0})]\in {K}_{0}(\mathrm{\Gamma}(C)).$

### *Proof*

We note the following: *If*$(f,{f}^{\prime})$*is a pair of*$C$*-neighbors and*$f={f}^{\prime}h$, *then*$h$*may be neither injective nor surjective.* Let us exhibit examples with ${f}^{\prime}={1}_{Y}$.

### *Example 6*

As in the examples 5, let $\mathrm{\Delta}$ be the linearly directed quiver of type ${\mathbb{A}}_{3}$ and take now as $\mathrm{\Lambda}$ the path algebra of $\mathrm{\Delta}$ modulo the zero relation $\mathit{\alpha}\mathit{\beta}$.

Now consider a right minimal right $C$-determined map $f$ ending in $Y$. As we have mentioned, we want to attach to $f$ an element $\mathbf{t}yp{e}_{C}(f)\in {K}_{0}(\mathrm{\Gamma}(C)).$

### **Proposition 6.2**

with $\mathbf{d}=\mathbf{dim}\mathrm{Hom}(C,Y)$.

### *Proof*

into a finite number of disjoint subsets.

### **Proposition 6.3**

for every $\mathrm{\Gamma}(C)$-dimension vector $\mathbf{e}$.

Assume now that $k$ is an algebraically closed field and that $\mathrm{\Lambda}$ and $\mathrm{\Gamma}$ are $k$-algebra. If $M$ is a $\mathrm{\Gamma}$-module and $\mathbf{e}$ a dimension vector for $\mathrm{\Gamma}$, we write ${\mathbb{G}}_{\mathbf{e}}M$ instead of ${\mathcal{S}}_{\mathbf{e}}M$. Note that ${\mathbb{G}}_{\mathbf{e}}M$ is in a natural way an algebraic variety, it is called a *quiver Grassmannian.* Namely, all the $\mathrm{\Gamma}$-modules with dimension vector $\mathbf{e}$ have the same $k$-dimension, say $e$ (if $\mathbf{e}={\sum}_{[S]}{e}_{S}[S]$, then $e={\sum}_{[S]}{e}_{S}{dim}_{k}S$). Denote by ${\mathbb{G}}_{e}({}_{k}M)$ the usual Grassmannian of all $e$-dimensional subspaces $U$ of the vector space ${}_{k}M.$ Using Plücker coordinates one knowns that ${\mathbb{G}}_{e}({}_{k}M)$ is a closed subset of a projective space, thus ${\mathbb{G}}_{e}({}_{k}M)$ is a projective variety. Now ${\mathbb{G}}_{\mathbf{e}}M$ is a subset of ${\mathbb{G}}_{e}({}_{k}M)$ defined by the vanishing of some polynomials (which express the fact that we consider submodules $U$ with a fixed dimension vector), thus also ${\mathbb{G}}_{\mathbf{e}}M$ is an algebraic variety and indeed a projective variety (but usually not even connected).

Proposition 6.3 can be reformulated as follows:

### **Proposition 6.4**

for every $\mathrm{\Gamma}(C)$-dimension vector $\mathbf{e}$.

In particular, we see that the set ${}^{C}{[\to Y\u3009}^{\mathbf{e}}$ is a projective variety: these *Auslander varieties* (as they should be called) furnish an important tool for studying the right equivalence classes of maps ending in a given module. As we have mentioned at the end of Sect. 2, the study of the set of right minimal maps ending in a fixed module $Y$ can be separated nicely into that of the local symmetries described by the right automorphism groups and that of the global directedness given by the right factorization lattice. Auslander’s first theorem describes the right factorization lattice as the filtered union of the right $C$-factorization lattices, and, as we now see, these right $C$-factorization lattices are finite disjoint unions of (transversal) subsets which are projective varieties, the Auslander varieties.

### *Remark*

It seems that quiver Grassmannians first have been studied by Schofield [42] and Crawley-Boevey [10] in order to deal with generic properties of quiver representations. In 2006, Caldero and Chapoton [9] observed that quiver Grassmannians can be used effectively in order to analyse the structure of cluster algebras as introduced by Fomin and Zelevinsky. Namely, it turns out that cluster variables can be described using the Euler characteristic of quiver Grassmannians. In this way quiver Grassmannians are now an indispensable tool for studying cluster algebras and quantum cluster algebras. We should add that quiver Grassmannians were also used (at least implicitly) in the study of quantum groups, see for example the calculation of Hall polynomials in [34]. A large number of papers is presently devoted to special properties of quiver Grassmannians.

There is the famous assertion that any projective variety is a quiver Grassmannian, see the paper [29] by Reineke (answering in this way a question by Keller) as well as blogs by Le Brujn [26] (with a contribution by Van den Bergh) and by Baez [6]. Actually, the construction as proposed by Van den Bergh in Le Bruyn’s blog is much older, it has been mentioned explicitly already in 1996 by Hille [21] dealing with moduli spaces of thin representations (see the example at the end of that paper), and it can be traced back to earlier considerations of Huisgen-Zimmermann dealing with moduli spaces of serial modules, even if they were published only later (see [22], Theorem G, but also [8], Corollary B, and [12], Example 5.4). It follows from Proposition 4.6 above that given a strictly wild algebra $\mathrm{\Lambda}$ and any projective variety $V$, there are $\mathrm{\Lambda}$-modules $C,Y$ and a dimension vector $\mathbf{e}$ such that ${}^{C}{[\to Y\u3009}^{\mathbf{e}}$ is isomorphic to $V$. We will show in [39] that this holds true for all controlled wild algebras.

## 7 Maps of right $C$-length 1

By definition, a right minimal right $C$-determined map $f$ has right $C$-length 1 provided $f$ is not invertible and given any factorization $f={f}^{\prime}h$ with ${f}^{\prime}$ right minimal right $C$-determined, then one of the maps ${f}^{\prime},h$ is invertible. Let us denote by ${}^{C}{[\to Y\u3009}^{1}$ the set of right equivalence classes of the maps ending in $Y$ which have right $C$-length 1.

Warning: an irreducible map $f$ is of course right minimal, but if $f$ is irreducible and right $C$-determined, we may have ${|f|}_{C}>1.$ For example, consider the Kronecker quiver, take $C={}_{\mathrm{\Lambda}}\mathrm{\Lambda}$. The irreducible map $f:{P}_{0}\to {P}_{1}$ has ${|f|}_{C}=2$ (note the factorizazion ${P}_{0}\subset \mathrm{rad}{P}_{1}\subset {P}_{1}$).

Here is an immediate consequence of Proposition 5.4.

### **Corollary 7.1**

Let $f:X\to Y$ be right minimal and right $C$-determined. Then ${|f|}_{C}=1$ if and only if ${\mathit{\eta}}_{CY}(f)$ is a maximal $\mathrm{\Gamma}(C)$-submodule of $\mathrm{Hom}(C,Y)$.

In order to analyze maps of right $C$-length 1, we will need the following lemma.

### **Lemma 7.2**

If $f$ is right minimal and ${h}^{\prime}$ is a split epimorphism, then also ${f}^{\prime}$ is right minimal.

### *Remark*

Observe that it is not enough to assume that ${h}^{\prime}$ is an epimorphism. As an example, take the indecomposable injective Kronecker module $X={Q}_{1}$ of length $3$, let $K$ be a submodule of length 2, and ${K}^{\prime}=K/\mathrm{soc}.$ Then ${Q}_{1}\to {Q}_{1}/K$ is right minimal. But the induced sequence is just the short exact sequence $K/\mathrm{soc}\to {Q}_{1}/\mathrm{soc}\to Y$ which splits.

### *Proof*

Denote the kernel of ${h}^{\prime}$ by ${K}^{\prime \prime}$, thus we can assume that $K={K}^{\prime}\oplus {K}^{\prime \prime}$ such that ${h}^{\prime}$ is the canonical projection $K\to {K}^{\prime}$ with kernel ${K}^{\prime \prime}$. Assume that ${X}^{\prime}=U\oplus V$, where $U$ is contained in the kernel of ${f}^{\prime}$, thus $U\subseteq {K}^{\prime}.$ Since ${X}^{\prime}=X/{K}^{\prime \prime}$, there are submodules ${U}^{\prime},{V}^{\prime}$ of ${X}^{\prime}$ both containing ${K}^{\prime \prime}$ such that ${U}^{\prime}+{V}^{\prime}={X}^{\prime}$ and ${U}^{\prime}\cap {V}^{\prime}={K}^{\prime \prime}$, with $U={U}^{\prime}/{K}^{\prime \prime}$ and $V={V}^{\prime}/{K}^{\prime \prime}.$

Consider ${U}^{\prime \prime}={U}^{\prime}\cap {K}^{\prime}$, this is a submodule of the kernel $K$ of $f$. Also, ${U}^{\prime}={K}^{\prime \prime}+{U}^{\prime \prime}$ (using the modular law). Thus we have ${U}^{\prime \prime}\cap {V}^{\prime}={U}^{\prime}\cap {K}^{\prime}\cap {V}^{\prime}\subseteq {K}^{\prime}\cap {K}^{\prime \prime}=0$ and ${U}^{\prime \prime}+{V}^{\prime}={U}^{\prime \prime}+{V}^{\prime}+{K}^{\prime \prime}={U}^{\prime}+{V}^{\prime}=X$. This shows that ${U}^{\prime \prime}$ is a direct summand of $X$ which is contained in the kernel of $f$. Since $f$ is right minimal, we see that ${U}^{\prime \prime}=0.$ Since ${U}^{\prime}={K}^{\prime \prime}+{U}^{\prime \prime}={K}^{\prime \prime}$, it follows that $U=0.$$\square $

### **Corollary 7.3**

Let $C$ be a module. Let $f:X\to Y$ be a right minimal right $C$-determined epimorphism with ${|f|}_{C}=1$. Then the kernel of $f$ is indecomposable.

### *Proof*

Since $f$ is right minimal and ${p}_{1}$ is a split epimorphism, lemma 7.2 asserts that ${f}^{\prime}$ is right minimal. Since ${f}^{\prime}$ is also right $C$-determined, we see that ${|f|}_{C}\ge 2,$ a contradiction. $\square $

### **Proposition 7.4**

### *Proof*

with the upper row being the sequence $\mathit{\u03f5}$. Since the lower sequence does not split, the map ${f}^{\prime}$ is right minimal. The kernel shows that ${f}^{\prime}$ is also $C$-determined. Since ${\mathit{\varphi}}^{\prime}$ is not invertible, the factorization $f={f}^{\prime}{\mathit{\varphi}}^{\prime}$ shows that $({\mathit{\varphi}}^{\prime},{f}^{\prime})$ is a $C$-factorization of $f$ of length at least $2$, thus ${|f|}_{C}\ge 2,$ a contradiction. This shows that $[\mathit{\u03f5}]$ belongs to the $\mathrm{\Gamma}(K)$-socle of ${\mathrm{Ext}}^{1}(Y,K).$

where again the upper row is $\mathit{\u03f5}$. Write ${h}_{1}^{\prime}=({\mathit{\varphi}}_{1},\cdots ,{\mathit{\varphi}}_{t})$ with endomorphisms ${\mathit{\varphi}}_{i}:K\to K$. If all ${\mathit{\varphi}}_{i}$ belong to the radical of $\mathrm{\Gamma}(K)$, then all the sequences induced from $\mathit{\u03f5}$ by the maps ${\mathit{\varphi}}_{i}$ split, thus also the lower sequence splits, since it is induced from $\mathit{\u03f5}$ by ${h}_{1}^{\prime}.$ Thus, at least one of the maps ${\mathit{\varphi}}_{i}$ has to be invertible and therefore ${h}_{1}^{\prime}$ is a split monomorphism.

If $t>1$, then the lower sequence splits off a sequence $0\to K\stackrel{1}{\to}K\to 0\to 0$, but this means that ${h}_{2}$ is not right minimal. Thus $t=1.$ But then ${h}_{1}^{\prime}$ is an automorphism, thus ${h}_{1}$ is invertible, a contradiction. This shows that ${|f|}_{C}=1.$$\square $

We say that an epimorphism $f$ is *epi-irreducible,* provided for any factorization $f={f}^{\prime}{f}^{\prime \prime}$ with ${f}^{\prime \prime}$ a proper epimorphism, the map ${f}^{\prime}$ is a split epimorphism (the dual concept of mono-irreducible maps has been considered in [35]).

### **Proposition 7.5**

Let $C$ be indecomposable, non-projective and let $K=\mathit{\tau}C$. If $f:X\to Y$ is an epi-irreducible epimorphism with kernel $K=\mathit{\tau}C$, then $f$ belongs to ${}^{C}{[\to Y\u3009}^{1}$.

### *Proof*

Since ${\mathit{\varphi}}^{\prime \prime}$ is a proper epimorphism, also ${f}^{\prime \prime}$ is a proper epimorphism, thus ${f}^{\prime}$ is a split epimorphism. But this implies that also the exact sequence induced from $\mathit{\u03f5}$ by $\mathit{\varphi}$ splits. $\square $

We will need some basic facts concerning the Gabriel–Roiter measure of finite length modules, see [35]. The Gabriel–Roiter measure of a module $M$ will be denoted by $\mathit{\gamma}(M).$ We recall that any indecomposable module $M$ which is not simple has a Gabriel–Roiter submodule ${M}^{\prime}$, this is a certain indecomposable submodule of $M$ and the embedding ${M}^{\prime}\to M$ is called a *Gabriel–Roiter inclusion.* Recall that a Gabriel–Roiter inclusion ${M}^{\prime}\to M$ is mono-irreducible: this means that for any proper submodule ${M}^{\prime \prime}$ of $M$ with ${M}^{\prime}\subseteq {M}^{\prime \prime}$, the inclusion ${M}^{\prime}\subseteq {M}^{\prime \prime}$ splits. As a consequence, given any nilpotent endomorphism $f$ of $M/{M}^{\prime}$, the sequence induced from $0\to {M}^{\prime}\to M\to M/{M}^{\prime}\to 0$ using $f$ splits. Also, it follows that the cokernel $M/{M}^{\prime}$ of a Gabriel–Roiter inclusion is indecomposable (and not projective).

Of course, we may use duality and consider a Gabriel–Roiter submodule $U$ of $DM$, the corresponding projection $M={D}^{2}M\to DU$ will be called a *co-Gabriel–Roiter projection.* By duality, a co-Gabriel–Roiter projection is an epi-irreducible epimorphism with (non-injective) indecomposable kernel.

### **Corollary 7.6**

Let $M$ be an indecomposable module which is not simple and let $f:M\to Y$ be a co-Gabriel–Roiter projection, say with kernel $K$. Let $C={\mathit{\tau}}^{-}K.$ Then $f$ is right minimal, right $C$-determined and ${|f|}_{C}=1,$ thus $[f\u3009$ belongs to ${}^{C}{[\to Y\u3009}^{1}$.

### *Remark*

If $M$ is indecomposable and not simple, we also may consider a Gabriel–Roiter submodule $U$ of $M$, say with projection $p:M\to M/U={Y}^{\prime}$ and consider ${C}^{\prime}={\mathit{\tau}}^{-}U.$ Then $p$ is right minimal and right ${C}^{\prime}$-determined, however in general there is not a fixed number $t$ such that $[p\u3009$ belongs to ${}^{{C}^{\prime}}{[\to {Y}^{\prime}\u3009}^{t}$ or to ${}^{{C}^{\prime}}{[\to {Y}^{\prime}\u3009}_{t}$. A typical example is example 8 presented in the next section. The two modules $P(b)$ and ${\mathit{\tau}}^{-}S(a)$ both have $P(a)$ as a Gabriel–Roiter submodule with factor module $S(b)$. Let ${C}^{\prime}={\mathit{\tau}}^{-}P(a).$ The projection $P(b)\to S(b)$ belongs to ${}^{{C}^{\prime}}{[\to S(b)\u3009}^{2}={}^{{C}^{\prime}}{[\to S(b)\u3009}_{0}$ whereas the projection ${\mathit{\tau}}^{-}S(a)\to S(b)$ belongs to ${}^{{C}^{\prime}}{[\to S(b)\u3009}^{1}={}^{{C}^{\prime}}{[\to S(b)\u3009}_{1}.$

## 8 Epimorphisms in $[\to Y\u3009$

The set of right equivalence classes $[f\u3009$ where $f$ is an epimorphism is obviously a coideal of the lattice $[\to Y\u3009,$ we denote it by ${[\to Y\u3009}_{\phantom{\rule{4.pt}{0ex}}\text{epi}\phantom{\rule{4.pt}{0ex}}}$. Since the pullback of an epimorphism is again an epimorphism, we see that ${[\to Y\u3009}_{\phantom{\rule{4.pt}{0ex}}\text{epi}\phantom{\rule{4.pt}{0ex}}}$ is closed under meets. Also, for any module $C$, the subset ${}^{C}{[\to Y\u3009}_{\phantom{\rule{4.pt}{0ex}}\text{epi}\phantom{\rule{4.pt}{0ex}}}$ of ${}^{C}[\to Y\u3009$ consisting of the right equivalence classes of all right $C$-determined epimorphisms ending in $Y$ is a coideal which is closed under meets. Since ${}^{C}[\to Y\u3009$ is a lattice of finite height, we see that ${}^{C}{[\to Y\u3009}_{\phantom{\rule{4.pt}{0ex}}\text{epi}\phantom{\rule{4.pt}{0ex}}}$ has a unique minimal element, say $[{f}_{0}\u3009$ and our first aim will be to describe ${\mathit{\eta}}_{CY}({f}_{0}).$

Before we deal with this question, let us point out in which way the projectivity or non-projectivity of indecomposable direct summands of $C$ are related to the fact that right minimal right $C$-determined morphisms $f$ are mono or epi. If $f$ is a monomorphism, then $f$ is right minimal and right $\mathrm{\Lambda}$-determined (see Corollary 3.8), thus right $C$-determined for some projective module $C$. Conversely, if $C$ is projective, then any right minimal, right $C$-determined morphism is a monomorphism (see Corollary 3.7). Namely, if $K$ is an indecomposable direct summand of the kernel of $f$, where $f$ is right minimal, then $K$ is not injective and ${\mathit{\tau}}^{-}K$ is a direct summand of any module $C$ such that $f$ is right $C$-determined. Of course, since $K$ is not injective, ${\mathit{\tau}}^{-}K$ is an indecomposable non-projective module. One should be aware that a morphism may be right $C$-determined for some module $C$ without any indecomposable projective direct summand, without being surjective.

### *Example 7*

As in example 6, we take as $\mathrm{\Lambda}$ the path algebra of the linearly directed quiver $\mathrm{\Delta}$ of type ${\mathbb{A}}_{3}$, modulo the zero relation $\mathit{\alpha}\mathit{\beta}$. Again, let $Y=P(c)$ and $C=S(b)$. As we have mentioned already, the non-zero maps $f:P(b)\to P(c)$ are not surjective, but right $S(b)$-determined, and, of course, $S(b)$ is not projective. (As we will see below, it is essential for this feature that the kernel of $f$ has injective dimension at least 2; for a general discussion of maps which are not surjective, but right $C$-determined by a module $C$ without any indecomposable projective direct summands, we refer to [38]).

**The submodule**$\mathrm{Hom}(C,\mathcal{P},Y)$**of**$\mathrm{Hom}(C,Y).$ We denote by $\mathrm{Hom}(C,\mathcal{P},Y)$ the set of morphisms $C\to Y$ which factor through a projective module. Note that $\mathrm{Hom}(C,\mathcal{P},Y)$ is a $\mathrm{\Gamma}(C)$-submodule of $\mathrm{Hom}(C,Y)$.

### **Proposition 8.1**

Assume that $f:X\to Y$ is right $C$-determined. Then $f$ is surjective if and only if ${\mathit{\eta}}_{CY}(f)\supseteq \mathrm{Hom}(C,\mathcal{P},Y)$.

### *Proof*

One direction is a trivial verification: First assume that $f$ is surjective. Let $h$ belong to $\mathrm{Hom}(C,\mathcal{P},Y)$, thus $h={h}_{2}{h}_{1}$ where ${h}_{1}:C\to P)$ and ${h}_{2}:P\to Y$ with $P$ projective. Since $f$ is surjective and $P$ is projective, there is ${h}_{2}^{\prime}:P\to X$ such that ${h}_{2}=f{h}_{2}^{\prime}.$ Thus shows that $h={h}_{2}{h}_{1}=f{h}_{2}^{\prime}{h}_{1}$ belongs to $f\mathrm{Hom}(C,X)={\mathit{\eta}}_{CY}(f).$

The converse is more interesting, here we have to use that $f$ is right $C$-determined. We assume that ${\mathit{\eta}}_{CY}(f)\supseteq \mathrm{Hom}(C,\mathcal{P},Y)$. Let $p:P(Y)\to Y$ be a projective cover of $Y$. Consider an arbitrary morphism $\mathit{\varphi}:C\to P(Y)$. The composition $p\mathit{\varphi}$ belongs to $\mathrm{Hom}(C,\mathcal{P},Y)$, thus to ${\mathit{\eta}}_{CY}(f)=f\mathrm{Hom}(C,X)$. Since $f$ is right $C$-determined, it follows that $p$ itself factors through $f$, say $p=f{p}^{\prime}$ for some ${p}^{\prime}:P(Y)\to X$. Now the composition $f{p}^{\prime}=p$ is surjective, there $f$ has to be surjective. $\square $

Let us denote by ${}^{C}{[\to Y\u3009}_{\phantom{\rule{4.pt}{0ex}}\text{epi}\phantom{\rule{4.pt}{0ex}}}$ the subset of ${}^{C}[\to Y\u3009$ given by all elements $[f\u3009$ with $f$ an epimorphism.

### **Proposition 8.2**

Here, the vertical maps are the canonical inclusions.

### *Proof*

It is well-known that given a module $M$ and a submodule ${M}^{\prime}$, then the lattice of submodules of the factor module $M/{M}^{\prime}$ is canonically isomorphic to the lattice of the submodules $U$ of $M$ satisfying ${M}^{\prime}\subseteq U.$ This is the vertical map on the right. $\square $

More generally, dealing with a morphism $f$ which is right $C$-determined, we can recover the image of $f$ as follows:

### **Proposition 8.3**

Let $f:X\to Y$ be right $C$-determined. Then one recovers the image of $f$ as the largest submodule ${Y}^{\prime}$ of $Y$ (with inclusion map $u:{Y}^{\prime}\to Y$) such that $u\mathrm{Hom}(C,\mathcal{P},{Y}^{\prime})\subseteq f\mathrm{Hom}(C,X).$

### *Proof*

Let ${Y}^{\prime}$ be the image of $f$ with inclusion map $u$ and $u{f}^{\prime}=f$ (with ${f}^{\prime}$ surjective). First of all, we show that $u\mathrm{Hom}(C,\mathcal{P},{Y}^{\prime})\subseteq f\mathrm{Hom}(C,X).$ Let ${\mathit{\varphi}}^{\prime}:C\to \mathrm{\Lambda}$ and ${\mathit{\varphi}}^{\prime \prime}:\mathrm{\Lambda}\to {Y}^{\prime}$ (the maps ${\mathit{\varphi}}^{\prime \prime}{\mathit{\varphi}}^{\prime}$ obtained in this way generate $\mathrm{Hom}(C,\mathcal{P},{Y}^{\prime})$ additively). We want to show that $u{\mathit{\varphi}}^{\prime \prime}\mathit{\varphi}$ factors through $f$. Since ${f}^{\prime}:X\to {Y}^{\prime}$ is surjective, there is $\mathit{\psi}:\mathrm{\Lambda}\to X$ such that ${\mathit{\varphi}}^{\prime \prime}={f}^{\prime}\mathit{\psi}$ (since $\mathrm{\Lambda}$ is projective). Thus $u{\mathit{\varphi}}^{\prime \prime}{\mathit{\varphi}}^{\prime}=u{f}^{\prime}\mathit{\psi}{\mathit{\varphi}}^{\prime}=f\mathit{\psi}{\mathit{\varphi}}^{\prime}.$ Thus $u{\mathit{\varphi}}^{\prime \prime}{\mathit{\varphi}}^{\prime}$ factors through $f.$

On the other hand, let ${u}^{\prime \prime}:{Y}^{\prime \prime}\to Y$ be a submodule of $Y$ such that ${u}^{\prime \prime}\mathrm{Hom}(C,\mathcal{P},{Y}^{\prime \prime})\subseteq f\mathrm{Hom}(C,X)$. Let $p:P({Y}^{\prime \prime})\to {Y}^{\prime \prime}$ be a projective cover. Consider the map ${f}^{\prime}={u}^{\prime \prime}p:P({Y}^{\prime \prime})\to Y.$ It has the property that for all maps $\mathit{\varphi}:C\to P({Y}^{\prime \prime})$ the composition ${f}^{\prime}\mathit{\varphi}$ factors through $f$ (namely ${f}^{\prime}\mathit{\varphi}={u}^{\prime \prime}p\mathit{\varphi}$ belongs to ${u}^{\prime \prime}\mathrm{Hom}(C,\mathcal{P},{Y}^{\prime \prime})\subseteq f\mathrm{Hom}(C,X)$). But $f$ is right $C$-determined, thus we conclude that ${f}^{\prime}$ factors through $\mathit{\alpha},$ say ${f}^{\prime}=f{\mathit{\varphi}}^{\prime}$ for some ${\mathit{\varphi}}^{\prime}:C\to P({Y}^{\prime \prime}).$ Thus the image ${Y}^{\prime \prime}$ of ${f}^{\prime}$ is contained in the image ${Y}^{\prime}$ of $f$. This is what we wanted to prove. $\square $

We recover in this way Proposition 8.1. Namely, if $f$ is surjective, then $Y$ is the image of $f$, thus $Y$ is one of the submodule ${Y}^{\prime}$ with $u\mathrm{Hom}(C,\mathcal{P},{Y}^{\prime})\subseteq f\mathrm{Hom}(C,X)$, thus $\mathrm{Hom}(C,\mathcal{P},Y)\subseteq f\mathrm{Hom}(C,Y).$

Conversely, if $\mathrm{Hom}(C,\mathcal{P},Y)\subseteq f\mathrm{Hom}(C,Y),$ then $Y$ is one of the submodules ${Y}^{\prime}$ with $u\mathrm{Hom}(C,\mathcal{P},{Y}^{\prime})\subseteq f\mathrm{Hom}(C,X)$ and therefore the image of $f$ contains $Y$, thus is equal to $Y$. This shows: *If f is right C-determined, then f is surjective if and only if*$\mathrm{Hom}(C,\mathcal{P},Y)\subseteq f\mathrm{Hom}(C,X).$

### **Corollary 8.4**

- (a)
All maps in ${}^{C}[\to Y\u3009$ are epimorphisms if and only of $\mathrm{Hom}(C,\mathcal{P},Y)=0.$

**Kernels with injective dimension at most**$1$.

### **Proposition 8.5**

- (i)
The injective dimension of $K$ is at most $1$.

- (ii)
If $Y$ is any module, then all maps in ${}^{C}[\to Y\u3009$ are epimorphisms.

- (iii)
We have $\mathrm{Hom}(C,\mathcal{P},Y)=0$ for all modules $Y$.

### *Proof*

Recall from [33], 2.4 that $K$ has injective dimension at most $1$ if and only if $\mathrm{Hom}(C,\mathrm{\Lambda})=0.$ Thus, if $K$ has injective dimension at most $1$ and $Y$ is an arbitrary module, then $\mathrm{Hom}(C,\mathcal{P},Y)=0,$ this shows that (i) implies (iii). Conversely, assume the condition (iii), thus $\mathrm{Hom}(C,\mathcal{P},Y)=0$ for all modules $Y$. If the injective dimension of $K$ would be at least $2$, then $\mathrm{Hom}(C,\mathrm{\Lambda})\ne 0.$ But $\mathrm{Hom}(C,\mathcal{P},\mathrm{\Lambda})=\mathrm{Hom}(C,\mathrm{\Lambda})$. This contradiction shows that (iii) implies (i). For the equivalence of (ii) and (iii) see Corollary 8.4(a). $\square $

### **Corollary 8.6**

Let $\mathrm{\Lambda}$ be hereditary and $C$ a module without any indecomposable projective direct summand. Then any right $C$-determined morphism $f:X\to Y$ is an epimorphism.

### *Proof*

Let $K=\mathit{\tau}C$. Since $C$ has no indecomposable projective direct summand, it follows that $C={\mathit{\tau}}^{-}K.$ Since $\mathrm{\Lambda}$ is hereditary, the injective dimension of any module is at most $1$. Since the injective dimension of $K$ is at most $1$, it follows from the proposition that all right $C$-determined maps are epimorphisms. $\square $

Of course, we also can show directly that $\mathrm{Hom}(C,\mathcal{P},Y)=0.$ Namely, let $g:C\to Y$ be in $\mathrm{Hom}(C,\mathcal{P},Y)$. Then $g={g}_{2}{g}_{1}$ with ${g}_{1}:C\to P$, where $P$ is a projective module. The image ${P}^{\prime}$ of ${g}_{1}$ is a submodule of $P$, thus, since $\mathrm{\Lambda}$ is hereditary, the module ${P}^{\prime}$ is also projective. Thus, we have a surjective map $C\to {P}^{\prime}$ with ${P}^{\prime}$ projective. Such a map splits. This shows that ${P}^{\prime}$ is isomorphic to a direct summand of $C$. It follows that ${P}^{\prime}=0$ and therefore $g=0.$

**Riedtmann–Zwara degenerations.**Recall that ${M}^{\prime}$ is a Riedtmann–Zwara degeneration of $M$ if and only if there is an exact sequence of the form

In terms of the Auslander bijection, we may deal with these data in several different ways: namely, we may look at the right equivalence classes of both $[M\to {M}^{\prime}\u3009$ and $[K\oplus M\to {M}^{\prime}\u3009$ in $[\to {M}^{\prime}\u3009$ as well as at the right equivalence class $[{M}^{\prime}\to M\u3009$ in $[\to M\u3009.$ In case we deal with $[K\oplus M\to {M}^{\prime}\u3009$, one should be aware that this map $[K\oplus M\to {M}^{\prime}\u3009$ is the join of the two maps $[K\to {M}^{\prime}\u3009$ and $[M\to {M}^{\prime}\u3009$ in $[\to {M}^{\prime}\u3009$.

In addition, we also may concentrate on the possible maps $K\to K$ and $L\to L$ (sometimes called steering maps).

When dealing with epimorphisms in ${}^{C}[\to Y\u3009$, Riedtmann–Zwara degenerations play a decisive role, as the following proposition shows:

### **Proposition 8.7**

Let $$