Conley index theory without index pairs. I: The point-set level theory

Abstract

We propose a new framework for Conley index theory. The main feature of our approach is that we do not use the notion of index pairs. We introduce, instead, the notions of compactifiable subsets and index neighbourhoods and formulate and prove basic results in Conley index theory using these notions. We treat both the discrete time case and the continuous time case.

Appendices

Part 3. Appendix

Appendix A. Proper maps of compactly generated weak Hausdorff spaces

In this appendix, we prove some results on proper maps that are needed in this paper. As in the other parts of the paper, we assume every topological space to be compactly generated weak Hausdorff.

We define proper maps as follows:

Definition A.1

Let $f :X \rightarrow Y$ be a continuous map of topological spaces. We say that f is proper if, for any compact Hausdorff subset L of Y, the inverse image $f^{-1}(L)$ is compact Hausdorff.

Lemma A.2

Let $f :X \rightarrow Y$ and $g :Y \rightarrow Z$ be two continuous maps.

• (1) If f and g are proper, gf is also proper.

• (2) If gf is proper, f is also proper.

Proof

• (1) Obvious.

• (2) Let L be a compact Hausdorff subset of Y. We have

  $$\begin{aligned} f^{-1}(L) \subset (gf)^{-1}(g(L)). \end{aligned}$$

  Since Z is weak Hausdorff, g(L) is a compact Hausdorff subset of Z. Since gf is proper, the inverse image $(gf)^{-1}(g(L))$ is a compact Hausdorff subset of X. On the other hand, since Y is weak Hausdorff, L is closed in Y. Hence, $f^{-1}(L)$ is closed in X (and therefore, in $(gf)^{-1}(g(L))$). Thus, $f^{-1}(L)$ is compact Hausdorff. $\square$

Lemma A.3

Proper maps are stable under base change.

Proof

Let $f :X \rightarrow Y$ be a proper map and $g :Y' \rightarrow Y$ be a continuous map. Consider the following pullback diagram:

Take any compact Hausdorff subset $L'$ of $Y'$. Since Y is weak Hausdorff, $g(L')$ is a compact Hausdorff subset of Y. Since f is proper, the inverse image $f^{-1}(g(L')) \ (= g'(f'^{-1}(L')))$ is a compact Hausdorff subset of X. The following is a pullback diagram:

We see that $f'^{-1}(L)$ is compact Hausdorff because it is a fibre product of two compact Hausdorff spaces. Thus, $f'$ is proper. $\square$

Lemma A.4

Let $f :X \rightarrow Y$ be a continuous map of topological spaces. Then, the following two conditions are equivalent:

• (i) f is proper.

• (ii) For any continuous map $g :L \rightarrow Y$ with L compact Hausdorff, the fibre product $L \times _Y X$ is compact Hausdorff.

Proof

(ii) $\Rightarrow$ (i): Consider the case when g is an inclusion.

(i) $\Rightarrow$ (ii): By Lemma A.3, the base change $f' :L \times _Y X \rightarrow L$ is proper. Since L is compact Hausdorff, it follows that $L \times _Y X \ (= f'^{-1}(L))$ is also compact Hausdorff. $\square$

Lemma A.5

Proper maps are universally closed.

Proof

By Lemma A.3, it suffices to verify that proper maps are closed. Let $f :X \rightarrow Y$ be a proper map. Let A be a closed subset of X. Since Y is compactly generated, it suffices to verify that, for any continuous map $g :L \rightarrow Y$ with L compact Hausdorff, $g^{-1}(f(A))$ is closed in L. Consider the following pullback diagram:

By Lemma A.4, $L \times _Y X$ is compact Hausdorff. We see that $f' :L \times _Y X \rightarrow L$ is a continuous map between compact Hausdorff spaces, hence closed. Thus, $g^{-1}(f(A)) \ (= f'(g'^{-1}(A)))$ is closed in L. $\square$

Lemma A.6

An inclusion is proper if and only if it is closed.

Proof

The ‘if’ part holds because closed subsets of compact Hausdorff spaces are compact Hausdorff. The ‘only if’ part follows from Lemma A.5. $\square$

Lemma A.7

Let $f :K \rightarrow X$ be a continuous map with K compact Hausdorff. Then, f is proper.

Proof

Take any compact Hausdorff subset L of X. Since X is weak Hausdorff, L is closed in X, and the inverse image $f^{-1}(L)$ is closed in K. Since K is compact Hausdorff, $f^{-1}(L)$ is also compact Hausdorff. $\square$

Lemma A.8

Let $f :X \rightarrow Y$ be a continuous map of topological spaces. Suppose that there exists a finite closed cover $(A_i)_{i \in I}$ of X such that $f|_{A_i} :A_i \rightarrow Y$ is proper for each $i \in I$. Then, f is proper.

Proof

Take any compact Hausdorff subset L of Y. For each $i \in I$, we see that $f^{-1}(L) \cap A_i$ is compact Hausdorff. We have a continuous map

$$\begin{aligned} g :\coprod _{i \in I} (f^{-1}(L) \cap A_i) \rightarrow X \end{aligned}$$

induced from the inclusions. Since X is weak Hausdorff, $f^{-1}(L)$, which is the image of g, is compact Hausdorff. $\square$

Appendix B. Shift equivalences and the Szymczak category

In this appendix, we introduce a class of morphisms, called shift equivalences, and explain its relation to the Szymczak category, following the idea of Franks–Richeson [3]. The results in this appendix are not used in this paper. We include them, however, because they clarify a categorical meaning of the Szymczak category.

We use the language of localization of categories. For basics on this subject, see, e.g. [12, §04VB].

Definition B.1

Let $\textsf{C}$ be a category.

• (1) Let $f :X \rightarrow X$ be an endomorphism in $\textsf{C}$. Then, we write $\widehat{f}$ for f seen as a morphism from f to itself in $\textsf{End}(\textsf{C})$.

• (2) Let $f :X \rightarrow X$ and $g :Y \rightarrow Y$ be two endomorphisms in $\textsf{C}$. We say that a morphism $\varphi :f \rightarrow g$ in $\textsf{End}(\textsf{C})$ is a shift equivalence if there exist a morphism $\psi :g \rightarrow f$ in $\textsf{End}(\textsf{C})$ and $n \in \mathbb {N}$ such that $\psi \varphi = \widehat{f}^n$ and $\varphi \psi = \widehat{g}^n$.

We write $\textsf{ShiftEq}(\textsf{C})$ for the class of all shift equivalences in the category $\textsf{End}(\textsf{C})$.

Lemma B.2

Let $\textsf{C}$ be a category. Then, $\textsf{ShiftEq}(\textsf{C})$ is a saturated multiplicative system in $\textsf{End}(\textsf{C})$ (in the sense of [12, Defn. 05Q8]).

Proof

Obviously, $\textsf{ShiftEq}(\textsf{C})$ is closed under composition and contains the isomorphisms. Suppose that we are given a diagram

in $\textsf{End}(\textsf{C})$ such that $\varphi \in \textsf{ShiftEq}(\textsf{C})$. Let us take $\psi :g \rightarrow f$ and $n \in \mathbb {N}$ such that $\psi \varphi = \widehat{f}^n$ and $\varphi \psi = \widehat{g}^n$. Then, the following diagram is commutative, and the right map $\widehat{h}^n$ is a morphism in $\textsf{ShiftEq}(\textsf{C})$:

Suppose that $\varphi , \psi :f \rightrightarrows g$ are two morphisms in $\textsf{End}(\textsf{C})$ and $\chi :h \rightarrow f$ is a morphism in $\textsf{ShiftEq}(\textsf{C})$ such that $\varphi \chi = \psi \chi$. Let us take $\omega :f \rightarrow h$ and $n \in \mathbb {N}$ such that $\omega \chi = \widehat{h}^n$ and $\chi \omega = \widehat{f}^n$. Then, $\widehat{g}^n \varphi = \widehat{g}^n \psi$, and $\widehat{g}^n$ is a morphism in $\textsf{ShiftEq}(\textsf{C})$. Thus, $\textsf{ShiftEq}(\textsf{C})$ is a left multiplicative system. One can see that $\textsf{ShiftEq}(\textsf{C})$ is a right multiplicative system in the same way.

Let us prove that $\textsf{ShiftEq}(\textsf{C})$ is saturated. Suppose that we are given a sequence

$$\begin{aligned} f \xrightarrow {\varphi } g \xrightarrow {\psi } h \xrightarrow {\chi } i \end{aligned}$$

in $\textsf{End}(\textsf{C})$ such that $\psi \varphi$ and $\chi \psi$ are morphisms in $\textsf{ShiftEq}(\textsf{C})$. We want to prove that $\psi$ is in $\textsf{ShiftEq}(\textsf{C})$. Take $\omega :h \rightarrow f$, $\tau :i \rightarrow g$, and $m,n \in \mathbb {N}$ such that

$$\begin{aligned} \omega \psi \varphi = \widehat{f}^m, \qquad \psi \varphi \omega = \widehat{h}^m, \qquad \tau \chi \psi = \widehat{g}^n, \qquad \chi \psi \tau = \widehat{i}^n. \end{aligned}$$

Put $\sigma = \varphi \omega \widehat{h}^n$. Then, we have $\sigma \psi = \widehat{g}^{m+n}$ and $\psi \sigma = \widehat{h}^{m+n}$. $\square$

Definition B.3

Let $\textsf{C}$ be a category. We define a functor $Q :\textsf{End}(\textsf{C}) \rightarrow \textsf{Sz}(\textsf{C})$ as follows:

• Q is identity on the objects.

• $Q\varphi = \overline{(\varphi , 0)}$ for each morphism $\varphi$ in $\textsf{End}(\textsf{C})$.

Proposition B.4

Let $\textsf{C}$ be a category.

• (1) $(\textsf{Sz}(\textsf{C}), Q)$ is a localization of the category $\textsf{End}(\textsf{C})$ at the class of morphisms $\textsf{ShiftEq}(\textsf{C})$.

• (2) A morphism $\varphi$ in $\textsf{End}(\textsf{C})$ is a shift equivalence if and only if $Q \varphi$ is an isomorphism.

Proof

(1) Let f and g be two endomorphisms in $\textsf{C}$. Let $\varphi :f \rightarrow g$ be a shift equivalence. Take $\psi :g \rightarrow f$ and $n \in \mathbb {N}$ such that $\psi \varphi = \widehat{f}^n$ and $\varphi \psi = \widehat{g}^n$. We see that $Q\varphi \ (= \overline{(\varphi , 0)})$ is an isomorphism in $\textsf{Sz}(\textsf{C})$ with inverse $\overline{(\psi , n)}$.

Let $\Phi :\textsf{End}(\textsf{C}) \rightarrow \textsf{D}$ be a functor that sends shift equivalences to isomorphisms in $\textsf{D}$. We want to show that there exists a unique functor $\overline{\Phi } :\textsf{Sz}(\textsf{C}) \rightarrow \textsf{D}$ such that $\overline{\Phi }Q = \Phi$. Obviously, such $\overline{\Phi }$ must be equal to $\Phi$ on the objects. Notice that, for any morphism $\varphi :f \rightarrow g$ in $\textsf{End}(\textsf{C})$ and any $n \in \mathbb {N}$, we have $\overline{(\varphi , n)} = Q\varphi \circ (Q \widehat{f})^{-n}$. This means that $\overline{\Phi }$ must satisfy $\overline{\Phi }\,\overline{(\varphi , n)} = \Phi \varphi \circ (\Phi \widehat{f})^{-n}$. To the contrary, if we define $\overline{\Phi }$ in this way, it is indeed a well-defined functor satisfying $\overline{\Phi }Q = \Phi$.

(2) This follows from (1), Lemma B.2, and [12, Lem. 05Q9]. $\square$

Next, let us consider the continuous time case.

Definition B.5

Let $\textsf{C}$ be either $\textsf{Set}_*$ or $\textsf{CHaus}_*$.

• (1) Let F be an object of $\textsf{SFlow}(\textsf{C})$. Then, for each $t \in \mathbb {R}_{\geqslant 0}$, we write $\widehat{f}^t$ for $f^t$ seen as a morphism from F to itself in $\textsf{SFlow}(\textsf{C})$.

• (2) Let F and G be two objects in $\textsf{SFlow}(\textsf{C})$. We say that a morphism $\varphi :F \rightarrow G$ in $\textsf{SFlow}(\textsf{C})$ is a shift equivalence if there exist a morphism $\psi :G \rightarrow F$ in $\textsf{SFlow}(\textsf{C})$ and $s \in \mathbb {R}_{\geqslant 0}$ such that $\psi \varphi = \widehat{f}^s$ and $\varphi \psi = \widehat{g}^s$.

We write $\textsf{ShiftEq}_\textsf{cont}(\textsf{C})$ for the class of all shift equivalences in the category $\textsf{SFlow}(\textsf{C})$.

Lemma B.6

Let $\textsf{C}$ be either $\textsf{Set}_*$ or $\textsf{CHaus}_*$. Then, $\textsf{ShiftEq}_\textsf{cont}(\textsf{C})$ is a saturated multiplicative system in $\textsf{SFlow}(\textsf{C})$.

Proof

The same as the proof of Lemma B.2. $\square$

Definition B.7

Let $\textsf{C}$ be either $\textsf{Set}_*$ or $\textsf{CHaus}_*$. We define a functor $Q :\textsf{SFlow}(\textsf{C}) \rightarrow \textsf{Sz}_\textsf{cont}(\textsf{C})$ as follows:

• Q is identity on the objects.

• $Q\varphi = \overline{(\varphi , 0)}$ for each morphism $\varphi$ in $\textsf{SFlow}(\textsf{C})$.

Proposition B.8

Let $\textsf{C}$ be either $\textsf{Set}_*$ or $\textsf{CHaus}_*$.

• (1) $(\textsf{Sz}_\textsf{cont}(\textsf{C}), Q)$ is a localization of the category $\textsf{SFlow}(\textsf{C})$ at the class of morphisms $\textsf{ShiftEq}_\textsf{cont}(\textsf{C})$.

• (2) A morphism $\varphi$ in $\textsf{SFlow}(\textsf{C})$ is a shift equivalence if and only if $Q \varphi$ is an isomorphism.

Proof

The same as the proof of Proposition B.4. $\square$

Remark B.9

The relation between the Szymczak category and shift equivalences are explained in Franks–Richeson [3]. However, in [3], localization is not used; the Conley index is thus defined as a shift equivalence class in the endomorphism category. In other words, it is defined as an isomorphism class of the objects in the Szymczak category. The Conley index defined in Szymczak [13] carries more information: it is a connected simple system in the Szymczak category, i.e. an object of the Szymczak category defined up to unique isomorphism, which retains the information of automorphisms.