Generalizations of Tucker–Fan–Shashkin Lemmas

Abstract

Tucker and Ky Fan’s lemma are combinatorial analogs of the Borsuk–Ulam theorem (BUT). In 1996, Yu. A. Shashkin proved a version of Fan’s lemma, which is a combinatorial analog of the odd mapping theorem (OMT). We consider generalizations of these lemmas for BUT–manifolds, i.e. for manifolds that satisfy BUT. Proofs rely on a generalization of the OMT and on a lemma about the doubling of manifolds with boundaries that are BUT–manifolds.

1 Tucker’s, Fan’s and Shashkin’s Lemmas

Throughout this paper the symbol ${\mathbb R}^d$ denotes the Euclidean space of dimension d. We denote by ${\mathbb B}^d$ the d-dimensional unit ball and by ${\mathbb S}^d$ the d-dimensional unit sphere. If we consider ${\mathbb S}^d$ as the set of unit vectors x in ${\mathbb R}^{d+1}$, then points x and $-x$ are called antipodal and the symmetry given by the mapping $x \rightarrow -x$ is called the antipodality on ${\mathbb S}^d$.

1.1 Tucker and Fan’s Lemma

Let T be a triangulation of the d-dimensional ball ${\mathbb B}^d$. We call T antipodally symmetric on the boundary if the set of simplices of T contained in the boundary of ${\mathbb B}^d = {\mathbb S}^{d-1}$ is an antipodally symmetric triangulation of ${\mathbb S}^{d-1}$; that is if $s\subset {\mathbb S}^{d-1}$ is a simplex of T, then $-s$ is also a simplex of T.

Tucker’s Lemma Tucker (1945) Let T be a triangulation of ${\mathbb B}^d$ that is antipodally symmetric on the boundary. Let

$$\begin{aligned} L:V(T)\rightarrow \{+1,-1,+2,-2,\ldots , +d,-d\} \end{aligned}$$

be a labelling of the vertices of T that is antipodal (i. e. $L(-v)=-L(v)$) for every vertex v on the boundary. Then there exists an edge in T that is complementary , i.e., its two vertices are labelled by opposite numbers.

(In Fig. 1 is given an illustration of Tucker’s lemma.)

Fig. 1

Illustration of Tucker’s lemma

There is also a version of Tucker’s lemma for spheres:

Spherical Tucker’s Lemma Let T be a centrally symmetric triangulation of the sphere ${\mathbb {S}}^d$. Let

$$\begin{aligned} L:V(T)\rightarrow \{+1,-1,+2,-2,\ldots , +d,-d\} \end{aligned}$$

be an antipodal labelling. Then there exists a complementary edge.

Tucker’s lemma was extended by Fan (1952):

Ky Fan’s Lemma Let T be a centrally symmetric triangulation of the sphere ${\mathbb {S}}^d$. Suppose that each vertex v of T is assigned a label L(v) from $\{\pm 1,\pm 2,\ldots ,\pm n\}$ in such a way that $L(-v)=-L(v)$. Suppose this labelling does not have complementary edges. Then there are an odd number of d-simplices of T whose labels are of the form $\{k_0,-k_1,k_2,\ldots ,(-1)^dk_d\}$, where $1\le k_0<k_1<\cdots <k_d\le n$. In particular, $n\ge d+1$.

1.2 Shashkin’s Lemma

In the 1990s, Yu. A. Shashkin published several works related to discrete versions of classic fixed point theorems (Shashkin 1991, 1994, 1996a, b, 1999). In Shashkin (1996b) he proved the following theorem:

Shashkin’s Lemma Let T be a triangulation of a planar polygon that is antipodally symmetric on the boundary. Let

$$\begin{aligned} L:V(T)\rightarrow \{+1,-1,+2,-2, +3,-3\} \end{aligned}$$

be a labelling of the vertices of T that satisfies $L(-v)=-L(v)$ for every vertex v on the boundary. Suppose that this labelling does not have complementary edges. Then for any numbers a, b, c, where $|a|=1,\; |b|=2, \; |c|=3$, the total number of triangles in T with labels (a, b, c) and $(-a,-b,-c)$ is odd.

Fig. 2

Illustration of Shashkin’s lemma

Remark

In other words, Shashkin proved that if $(a,b,c)=(1,2,3), \, (1,-2,3), (1,2,-3)$ and $(1,-2,-3)$, then the number of triangles with labels (a, b, c) or $(-a,-b,-c)$ is odd. Denote this number by $\mathrm{SN}(a,b,c)$. Then in Fig. 2 we have

$$\begin{aligned} \mathrm{SN}(1,2,3)=3, \; \mathrm{SN}(1,-2,3)=1,\; \mathrm{SN}(1,2,-3)=3, \; \mathrm{SN}(1,-2,-3)=3. \end{aligned}$$

Note that this result was published only in Russian and only for two–dimensional case. Moreover, Shashkin attributes this theorem to Fan (1952).

Actually, Shashkin’s lemma can be derived from Ky Fan’s lemma for $n=d+1$. However. Shashkin’s proof is different and relies on the odd mapping theorem (OMT). In fact, this lemma is a discrete version of the OMT. That is why we distinguish this result as Shashkin’s lemma.

The following is a spherical version of Shashkin’s lemma.

Spherical Shashkin’s Lemma Let T be a centrally symmetric triangulation of   ${\mathbb {S}}^d$. Let

$$\begin{aligned} L:V(T)\rightarrow \Pi _{d+1}:=\{+1,-1,+2,-2,\ldots , +(d+1),-(d+1)\} \end{aligned}$$

be an antipodal labelling of T. Suppose that this labelling does not have complementary edges. Then for any set of labels $\Lambda :=\{\ell _1,\ell _2,\ldots ,\ell _{d+1}\}\subset \Pi _{d+1}$ with $|\ell _i|=i$ for all i, the number of d –simplices in T that are labelled by $\Lambda$ is odd.

1.3 Main Results

In Musin (2012) we invented BUT (Borsuk–Ulam Type) – manifolds. Theorems 3.1–3.4 in this paper extend Tucker’s and Shashkin’s lemmas for BUT–manifolds. Namely, Theorem 3.1 and Theorem 3.2 are extensions of the spherical Tucker and Shahskin lemmas, where ${\mathbb {S}}^d$ is substituted by a BUT–manifold. Theorems 3.3 and 3.4 are extensions of the original Tucker and Shashkin lemmas, where ${\mathbb {B}}^d$ is substituted by a manifold M with boundary $\partial M$ that is a BUT–manifold.

Our proof of Theorem 3.2 is relies on a generalization of the odd mapping theorem for BUT–manifolds:

Theorem 2.1

Let $(M_1,A_1)$ and $(M_2,A_2)$ be BUT–manifolds. Then any odd continuous mapping $f:M_1\rightarrow M_2$ has odd degree.

Theorems 3.3 and 3.4 follow from Theorems 3.1 and 3.2 by using Lemma 3.1, which is about the doubling of manifolds with boundaries that are BUT–manifolds.

In Section 4 we extend for BUT–manifolds Shaskin’s proof of two Tucker’s theorems about covering families from Tucker (1945). Actually, these theorems are corollaries of Theorem 3.2.

2 The Odd Mapping Theorem

We say that a mapping $f:{\mathbb {S}}^d\rightarrow {\mathbb {S}}^d$ is odd or antipodal if $f(-x)=-f(x)$ for all $x\in {\mathbb {S}}^d$. If f is a continuous mapping, then $\deg {f}$ (the degree of f) is well defined.

Let $f:M_1\rightarrow M_2$ be a continuous map between two closed manifolds $M_1$ and $M_2$ of the same dimension. The degree is a number that represents the amount of times that the domain manifold wraps around the range manifold under the mapping. Then ${{\mathrm{deg_2}}}(f)$ (the degree modulo 2) is 1 if this number is odd and 0 otherwise. It is well known that ${{\mathrm{deg_2}}}(f)$ of a continuous map f is a homotopy invariant [see (Milnor 1969)].

The classical odd mapping theorem states that

Every continuous odd mapping $f:{\mathbb {S}}^d\rightarrow {\mathbb {S}}^d$ has odd degree.

Shashkin (1996b) (see also [Matoušek (2003), Proposition 2.4.1]) gives a proof of this theorem for simplicial mappings $f:{\mathbb {S}}^d\rightarrow {\mathbb {S}}^d$. Conner and Floyd (1960) considered Theorem 2.1 for a wide class of spaces. Here we extend the odd mapping theorem for BUT–manifolds. In our paper Musin (2012), we extended the Borsuk–Ulam theorem for manifolds.

Let M be a connected compact PL (piece-wise linear) d-dimensional manifold without boundary with a free simplicial involution $A:M\rightarrow M$, i. e. $A^2(x)=A(A(x))=x$ and $A(x)\ne x$. We say that a pair (M, A) is a BUT (Borsuk-Ulam Type) manifold if for any continuous $g:M \rightarrow {\mathbb {R}}^d$ there is a point $x\in M$ such that $g(A(x))=g(x)$. Equivalently, if a continuous map $f:M \rightarrow {\mathbb {R}}^d$ is antipodal, i.e. $f(A(x))=-f(x)$, then the set of zeros $Z_f:=f^{-1}(0)$ is not empty.

In Musin (2012), we found several equivalent necessary and sufficient conditions for manifolds to be BUT. In particular,

M is a d–dimensional BUT–manifold if and only if M admits an antipodal continuous transversal to zeros mapping $h:M \rightarrow {\mathbb {R}}^d$ with $|Z_h|=2\pmod {4}$.

A continuous mapping $h:M \rightarrow {\mathbb {R}}^d$ is called transversal to zero if there is an open set U in ${\mathbb {R}}^d$ such that U contains 0, U is homeomorphic to the open d-ball and $h^{-1}(U)$ consists of a finite number open sets in M that are homeomorphic to open d-balls.

The class of BUT–manifolds is sufficiently large. It is clear that $({\mathbb {S}}^d,A)$ with $A(x)=-x$ is a BUT-manifold. Suppose that M can be represented as a connected sum $N\# N$, where N is a closed PL manifold. Then M admits a free involution. Indeed, M can be “centrally symmetrically” embedded to ${\mathbb {R}}^k$, for some k, and the antipodal symmetry $x\rightarrow -x$ in ${\mathbb {R}}^k$ implies a free involution $A:M\rightarrow M$ [Musin (2012), Corollary 1]. For instance, orientable two-dimensional manifolds $M^2_g$ with even genus g and non-orientable manifolds $P^2_m$ with even m, where m is the number of Möbius bands, are BUT-manifolds.

Let $M_i,\; i=1,2$, be a manifold with a free involution $A_i$. We say that a mapping $f:M_1\rightarrow M_2$ is antipodal (or odd, or equivariant) if $f(A_1(x))=A_2(f(x))$ for all $x\in M_1$.

Theorem 2.1

Let $(M_1,A_1)$ and $(M_2,A_2)$ be d-dimensional BUT–manifolds. Then any odd continuous mapping $f:M_1\rightarrow M_2$ has odd degree.

Proof

Since $(M_2,A_2)$ is BUT, there is a continuous antipodal transversal to zeros mapping $g:M_2 \rightarrow {\mathbb {R}}^d$ with $|Z_g|=4m_2+2$ [Musin (2012), Theorem 2].

Let $h:=g\circ f$. Then $h:M_1 \rightarrow {\mathbb {R}}^d$ is continuous and antipodal. Since the degree of a mapping is a homotopy invariant, without loss of generality we may assume that h is a transversal to zero mapping (see Musin (2012), [Lemma 3]). Therefore $|Z_h|=4m_1+2$. On the other hand,

$$\begin{aligned} |Z_h|=\sum \limits _{x\in Z_g}{|f^{-1}(x)|}. \end{aligned}$$

Then

$$\begin{aligned} 2m_1+1=(2m_2+1){{\mathrm{deg_2}}}{f}\pmod {2}. \end{aligned}$$

Thus, the degree of f is odd. $\square$

3 Tucker’s and Shashkin’s Lemmas for BUT–Manifolds

In our papers Musin (2012, (2015), Musin and Volovikov (2015) are considered extensions of Tucker’s lemma. Here we consider generalizations of Tucker’s and Shashkin’s lemmas for manifolds with and without boundaries.

Let T be an antipodally symmetric (or antipodal) triangulation of a BUT–manifold (M, A). This means that $A:T\rightarrow T$ sends simplices to simplices. Denote by $\Pi _n$ the set of labels $\{+1,-1,+2,-2,\ldots , +n,-n\}$ and let $L:V(T)\rightarrow \Pi _n$ be a labeling of T. We say that this labelling is antipodal if $L(A(v))=-L(v)$. An edge uv in T is called complementary if $L(u)=-L(v)$.

Theorem 3.1

([Musin (2015), Theorem 4.1]) Let (M, A) be a d-dimensional BUT–manifold. Let T be an antipodal triangulation of M. Then for any antipodal labelling $L:V(T)\rightarrow \Pi _d$ there exists a complementary edge.

Any antipodal labelling $L:V(T)\rightarrow \Pi _n$ of an antipodally symmetric triangulation T of M defines a simplicial map $f_L:T\rightarrow {\mathbb {R}}^n$. Let $\{e_1,-e_1,e_2,-e_2,\ldots ,e_n,-e_n\}$ be the standard orthonormal basis in ${\mathbb {R}}^n$. For $v\in V(T)$, set $f_L(v):=e_i$ if $L(v)=i$ and $f_L(v):=-e_i$ if $L(v)=-i$. Since $f_L$ is defined on V(T), it defines a simplicial mapping $f_L:T\rightarrow {\mathbb {R}}^n$ (see details in [Matoušek (2003), Sect. 2.3].)

The following theorem is a version of Shashkin’s lemma for manifolds without boundary.

Theorem 3.2

Let (M, A) be a d-dimensional BUT–manifold. Let T be an antipodally symmetric triangulation of M. Let $L:V(T)\rightarrow \Pi _{d+1}$ be an antipodal labelling of T. Suppose that this labelling does not have complementary edges. Then for any set of labels $\Lambda :=\{\ell _1,\ell _2,\ldots ,\ell _{d+1}\}\subset \Pi _{d+1}$ with $|\ell _i|=i$ for all i, the number of d–simplices in T that are labelled by $\Lambda$ is odd.

Proof

Since L has no complimentary edges, $f_L:T\rightarrow {\mathbb {R}}^{d+1}$ is an antipodal mapping of M to the boundary of the crosspolytope $C^{d+1}$ that is the convex hull $\mathrm{conv}{\{e_1,-e_1,\ldots ,e_{d+1},-e_{d+1}\}}.$ Note that $\partial C^{d+1}$ is a simplicial sphere ${\mathbb {S}}^d$, which is a BUT-manifold. Therefore, Theorem 2.2 implies that the number of preimages of the simplex in $\partial C^{d+1}$ with indexes from $\Lambda$ is odd. It completes the proof. $\square$

Remark

Theorem 3.1 can be proved using the same arguments. Indeed, suppose that $L:V(T)\rightarrow \Pi _d$ has no complementary edges. Then $f_L$ sends M to $\partial C^d$. Since $\dim {\partial C^d}=d-1$, $\deg {f_L}=0$. This contradicts Theorem 2.1.

Now we extend Tucker’s and Shashkin’s lemmas for the case when M is a manifold with boundary that is a BUT–manifold. But first, prove that there exists a “double” of M that is a BUT-manifold.

Lemma 3.1

Let M be a compact PL manifold with boundary $\partial M$. Suppose $(\partial M,A)$ is a BUT–manifold. Then there is a BUT–manifold $(\tilde{M},\tilde{A})$ and a submanifold N in $\tilde{M}$ such that $N\simeq M$, $\tilde{A}|_{\partial N}\simeq A$, $(N{\setminus }\partial N) \cap \tilde{A}(N{\setminus }\partial N)=\emptyset$ and

$$\begin{aligned} \tilde{M}\simeq (N{\setminus }\partial N)\cup \partial N \cup \tilde{A}(N{\setminus }\partial N). \end{aligned}$$

Proof

• 1. First we prove the following statement: Let X be a finite simplicial complex. Let Y be a subcomplex of X with a free involution $A:Y\rightarrow Y$. Then there is a simplicial embedding F of X into ${\mathbb {R}}^q_+:=\{(x_1,\ldots ,x_q)\in {\mathbb {R}}^q: x_1\ge 0\}$, where q is sufficiently large, such that Y is centrally symmetrically embedded in ${\mathbb {R}}^q$, i.e. $F(A(y))=-F(y)$ for all $y\in Y$, and $X{\setminus } Y$ is mapped into the interior of ${\mathbb {R}}^q_+$. Indeed, let $v_1,v_{-1},\ldots ,v_m,v_{-m}$ denote vertices of Y such that $A(v_k)=v_{-k}$. Let $\{v_{m+1},\ldots , v_n\}$ be the set of vertices of $X{\setminus } Y$.

  Denote by $C^n$ the n–dimensional crosspolytope that is the boundary of convex hull

  $$\begin{aligned} \mathrm{conv}{\{e_1,-e_1,\ldots ,e_n,-e_n\}} \end{aligned}$$

  of the vectors of the standard orthonormal basis and their negatives.

  Now define an embedding $F:X\rightarrow C^n$. Let $F(v_k):=e_k$, $F(v_{-k}):=e_{-k}$, where $1\le k\le m$, $F(v_k):=e_k$, and $k=m+1,\ldots ,n$. Since F is defined for all of the vertices of X, it uniquely defines a simplicial (piecewise linear) mapping $F:X\rightarrow C^n\subset {\mathbb {R}}^n$. Then

  $$\begin{aligned} F(Y)\subset C^m\subset {\mathbb {R}}^m=\{(x_1,\ldots ,x_n)\in {\mathbb {R}}^n: x_i=0, \; i=m+1,\ldots ,n \}, \end{aligned}$$

  $F(A(y))=-F(y)$ for all $y\in Y$ and

  $$\begin{aligned} F(X{\setminus } Y)\subset {\mathbb {R}}^{n-1}_+:=\{(x_1,\ldots ,x_n)\in {\mathbb {R}}^n: x_{m+1}+\cdots +x_n>0\}, \end{aligned}$$

  as required.

• 2. Let $X=M$ and $Y=\partial M$. Then it follows from 1 that there is an embedding $F:M\rightarrow {\mathbb {R}}^q_+$ with $F(\partial M)\subset {\mathbb {R}}^q$ and $F(A(y))=-F(y)$ for all $y\in \partial M$, where $q=n-1$. Let

  $$\begin{aligned} \tilde{M}:= & {} F(M)\cup (-F(M))\subset {\mathbb {R}}^{q+1}={\mathbb {R}}^q_+\cup (-{\mathbb {R}}^q_+) \; \text{ and } \\ \tilde{A}(x):= & {} -x \text{ for } \text{ all } x\in \tilde{M}. \end{aligned}$$

  It is clear that $\tilde{M}\simeq (N{\setminus }\partial N)\cup \partial N \cup \tilde{A}(N{\setminus }\partial N),$ where $N:=F(M).$

• 3. Let us prove that $(\tilde{M},\tilde{A})$ is BUT. Indeed, since $(\partial M,A)$ is BUT, there is a continuous antipodal transversal to zeros mapping $g:\partial M\simeq \partial N \rightarrow {\mathbb {R}}^{d-1}$ with $|Z_g|=4m+2$, where $d:=\dim {M}.$ We extend this mapping to $h:\tilde{M} \rightarrow {\mathbb {R}}^{d}$ with $h|_{\partial N}=g$ and $|Z_h|=|Z_g|=4m+2$.

Let $v=(x_1,\ldots ,x_{n})\in {\mathbb {R}}^{n}$ be a vertex of $\tilde{M}$. If $v\in \partial N$, then put

$$\begin{aligned} h(v):=(g(v),0)\in {\mathbb {R}}^{d}. \end{aligned}$$

For $v\in \tilde{M}{\setminus }\partial N$ define

$$\begin{aligned} h(v):=(0,\ldots ,0,x_{m+1}+\cdots +x_n)\in {\mathbb {R}}^{d}. \end{aligned}$$

Then $h:\tilde{M}\rightarrow {\mathbb {R}}^{d}$ is an antipodal transversal to zeros mapping and $h^{-1}(0)=g^{-1}(0)$. $\square$

Theorem 3.3

Let M be a d–dimensional compact PL manifold with boundary $\partial M$. Suppose $(\partial M,A)$ is a BUT–manifold. Let T be a triangulation of M that antipodally symmetric on $\partial M$. Let $L:V(T)\rightarrow \Pi _d$ be a labelling of T that is antipodal on the boundary. Then there is a complementary edge in T.

Theorem 3.4

Let M be a d–dimensional compact PL manifold with boundary $\partial M$. Suppose $(\partial M,A)$ is a BUT–manifold. Let T be a triangulation of M that antipodally symmetric on $\partial M$. Let $L:V(T)\rightarrow \Pi _{d+1}$ be a labelling of T that is antipodal on the boundary and has no complementary edges. Then for any set of labels $\Lambda :=\{\ell _1,\ell _2,\ldots ,\ell _{d+1}\}\subset \Pi _{d+1}$ with $|\ell _i|=i$ for all i, the number of d–simplices in T that are labelled by $\Lambda$ or $(-\Lambda )$ is odd.

Proof

By Lemma 3.1 there is a BUT–manifold $(\tilde{M},\tilde{A})$ that is the double of M. We can extend T and L from M to an antipodal triangulation $\tilde{T}:=T\cup \tilde{A}(T)$ of $\tilde{M}$ and an antipodal labelling $\tilde{L}:V(\tilde{T})\rightarrow \Pi _n$, where $n=d$ in Theorem 3.3 and $n=d+1$ in Theorem 3.4, such that $\tilde{L}|_T=L$.

Thus, for the case $n=d$ Theorem 3.3 follows from Theorem 3.1 and for $n=d+1$ Theorem 3.2 yields Theorem 3.4. $\square$

4 Shashkin’s Proof of Tucker’s Theorems

In this section we consider two Tucker’s theorems about covering families. Note that Tucker (1945) obtained these theorem only for   ${\mathbb {S}}^2$. Bacon (1966) proved that statements in Theorems 4.1 and 4.2 are equivalent to the Borsuk–Ulam theorem for normal topological spaces X with free continuous involutions $A:X\rightarrow X$. [See also Theorem 2.1 in our paper (Musin and Volovikov 2015)]. Actually, these theorems can be proved from properties of Schwarz’s genus (Svarc 1966) or Yang’s cohomological index (Karasev 2009; Musin and Volovikov 2015).

For the two–dimensional case in the book Shashkin (1999) Shashkin derives Tucker’s theorems from his lemma. Here we extend his proof for BUT–manifolds of all dimensions.

Theorem 4.1

Let (M, A) be a d-dimensional BUT–manifold. Consider a family of closed sets $\{B_i, B_{-i}\},\, i=1,\ldots , d+1$, where $B_{-i}:=A(B_i)$, is such that $B_i\cap B_{-i}=\emptyset$ for all i. If this family covers M, then for any set of indices $\{k_1,k_2,\ldots ,k_{d+1}\}\subset \Pi _{d+1}$ with $|k_i|=i$ for all i, the intersection of all $B_{k_i}$ is not empty.

Proof

Note that any PL manifold admits a metric. For a triangulation T of M, the norm of T, denoted by |T|, is the diameter of the largest simplex in T.

Let $T_1,T_2,\ldots$ be a sequence of antipodal triangulations of M such that $|T_i|\rightarrow 0$. Now for all i define an antipodal labelling $L_i:V(T_i)\rightarrow \Pi _{d+1}$. For every $v\in V(T_i)\subset M$ set

$$\begin{aligned} L_i(v):=\ell , \quad \text{ where }\, v\in B_\ell \text{ and } |\ell |=\min {\{|k|: v\in B_k\}}. \end{aligned}$$

Then $L_i$ satisfies the assumptions in Theorem 3.2 and $T_i$ contains a simplex $s_i$ with labels $\{k_1,k_2,\ldots ,k_{d+1}\}\subset \Pi _{d+1}$.

Since M is compact and $|s_i|\rightarrow 0$, the sequence $\{s_i\}$ contains a converging subsequence P with limit $w\in M$. Then for $s_i\in P$ we have $V(s_i)\rightarrow w$.

By assumption, all $B_k$ are closed sets. Therefore $w\in B_{k_j}$ for all $j=1,\ldots , d+1$, and thus $w\in \cap _j{B_{k_j}}$. $\square$

Theorem 4.2

Let (M, A) be a d-dimensional BUT–manifold. Suppose that M is covered by a family $\mathcal F$ of $d+2$ closed subsets $C_1,\ldots ,C_{d+2}$. Suppose that all $C_i$ have no antipodal pairs (x, A(x)), in other words, $C_i\cap A(C_i)=\emptyset$. Let $0<k<d+2$. Then any k subsets from $\mathcal F$ intersect and there is a point x in this intersection such that A(x) belongs to the intersection of the remaining $(d+2-k)$ subsets in $\mathcal F$.

Proof

Without loss of generality, we can assume that $k\ge (d+2)/2$ and that the k subsets from $\mathcal F$ are $C_1,\ldots ,C_k$. Therefore, we have to prove that there is $x\in M$ such that

$$\begin{aligned} x\in \bigcap \limits _{i=1}^k{C_i} \; \text{ and } \; A(x)\in \bigcap \limits _{i=k+1}^{d+2}{C_i} \end{aligned}$$

Set $C_{-i}:=A(C_i)$. Let $m:=\lceil {d/2}\rceil ,$

$$\begin{aligned} B_1:=C_1\cap (C_{-2}\cup \cdots \cup C_{-(m+1)}\cup C_{-(d+2)}), \end{aligned}$$

$$\begin{aligned} B_2:=C_2\cap (C_{-3}\cup \cdots \cup C_{-(m+2)}\cup C_{-(d+2)}), \end{aligned}$$

$$\begin{aligned} \vdots \end{aligned}$$

$$\begin{aligned} B_d:=C_d\cap (C_{-(d+1)}\cup C_{-1}\cup \cdots \cup C_{-(m-1)}\cup C_{-(d+2)}), \end{aligned}$$

$$\begin{aligned} B_{d+1}:=C_{d+1}\cap (C_{-1}\cup \cdots \cup C_{-m}\cup C_{-(d+2)}). \end{aligned}$$

If $B_{-i}:=A(B_i)$, then

$$\begin{aligned} \bigcup \limits _{i=1}^{d+1}{B_i\cup B_{-i}}=\bigcup \limits _{i=1}^{d+2}{C_i\cap (C_1\cup \cdots \cup C_{d+2})}=\bigcup \limits _{i=1}^{d+2}{C_i\cap M}=\bigcup \limits _{i=1}^{d+2}{C_i}=M. \end{aligned}$$

On the other hand, $B_i\subset C_i$ and $B_{-i}\subset C_{-i}$, hence $B_i\cap B_{-i}=\emptyset$. Therefore, the family of subsets $\{B_i\}$ satisfies the assumptions of Theorem 4.1. It follows that

$$\begin{aligned} Q:=B_1\cap \cdots \cap B_k\cap B_{-(k+1)}\cap \cdots \cap B_{-(d+1)}\ne \emptyset . \end{aligned}$$

Let $x\in Q$. Then

$$\begin{aligned} x\in C_1\cap \ldots \cap C_k \, \text{ and } \, A(x)\in C_{-(k+1)}\cap \cdots \cap C_{-(d+1)}. \end{aligned}$$

Since $k\ge m+1$ and $x\in B_1=C_1\cap (C_{-2}\cup \ldots \cup C_{-(m+1)}\cup C_{-(d+2)})$, we have $x\in C_{-(d+2)}$, i.e. $A(x)\in C_{d+2}$. $\square$

Corollary 4.1

Let (M, A) be a d-dimensional BUT–manifold. Then M cannot be covered by $d+1$ closed sets, none containing a pair (x, A(x)) of antipodal points.

Note that the case $M={\mathbb {S}}^d$ was first considered by Lusternik and Schnirelmann in 1930.

Proof

Suppose the converse, so M can be covered by closed subsets $C_1,\ldots ,C_{d+1}$. Let $C_{d+2}:=C_1$. Then this covering satisfies the assumptions of Theorem 4.2. So there is x such that

$$\begin{aligned} x\in \bigcap \limits _{i=1}^{d+1}{C_i} \; \text{ and } \; A(x)\in C_{d+2}, \quad \text{ i.e. } \; (x,A(x)) \in C_1, \end{aligned}$$

a contradiction. $\square$