Generalizations of Tucker–Fan–Shashkin Lemmas

Since $(M_{2},A_{2})$ is BUT, there is a continuous antipodal transversal to zeros mapping $g:M_{2}\to{\mathbb{R}}^{d}$ with $|Z_{g}|=4m_{2}+2$ [[Musin2012], Theorem 2].

Let $h:=g\circ f$. Then $h:M_{1}\to{\mathbb{R}}^{d}$ 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 [Musin2012], [Lemma 3]). Therefore $|Z_{h}|=4m_{1}+2$. On the other hand,

Then

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

Since $L$ has no complimentary edges, $f_{L}:T\to{\mathbb{R}}^{d+1}$ is an antipodal mapping of $M$ to the boundary of the crosspolytope $C^{d+1}$ that is the convex hull ${\rm conv}{\{e_{1},-e_{1},\ldots,e_{d+1},-e_{d+1}\}}.$ 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. $\square$

1. 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 $A:Y\to Y$ . Then there is a simplicial embedding $F$ of $X$ into ${\mathbb{R}}^{q}_{+}:=\{(x_{1},\ldots,x_{q})\in{\mathbb{R}}^{q}:x_{1}\geq 0\}$ , 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

   $\displaystyle{\rm conv}{\{e_{1},-e_{1},\ldots,e_{n},-e_{n}\}}$

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

   Now define an embedding $F:X\to C^{n}$ . Let $F(v_{k}):=e_{k}$ , $F(v_{-k}):=e_{-k}$ , where $1\leq k\leq m$ , $F(v_{k}):=e_{k}$ , 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 $F:X\to C^{n}\subset{\mathbb{R}}^{n}$ . Then

   $\displaystyle F(Y)\subset C^{m}\subset{\mathbb{R}}^{m}=\{(x_{1},\ldots,x_{n}) \in{\mathbb{R}}^{n}:x_{i}=0,\;i=m+1,\ldots,n\},$

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

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

   as required.

2. 2.

   Let $X=M$ and $Y=\partial M$ . Then it follows from 1 that there is an embedding $F:M\to{\mathbb{R}}^{q}_{+}$ 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

   	$\displaystyle\tilde{M}$	$\displaystyle:=$	$\displaystyle F(M)\cup(-F(M))\subset{\mathbb{R}}^{q+1}={\mathbb{R}}^{q}_{+} \cup(-{\mathbb{R}}^{q}_{+})\;\mbox{ and }$
   	$\displaystyle\tilde{A}(x)$	$\displaystyle:=$	$\displaystyle-x\mbox{ for all }x\in\tilde{M}.$

   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. 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 $g:\partial M\simeq\partial N\to{\mathbb{R}}^{d-1}$ with $|Z_{g}|=4m+2$ , where $d:=\dim{M}.$ We extend this mapping to $h:\tilde{M}\to{\mathbb{R}}^{d}$ 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

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

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

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 $\tilde{L}:V(\tilde{T})\to\Pi_{n}$, 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. $\square$

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 $|T_{i}|\to 0$. Now for all $i$ define an antipodal labelling $L_{i}:V(T_{i})\to\Pi_{d+1}$. For every $v\in V(T_{i})\subset M$ set

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 $|s_{i}|\to 0$, the sequence $\{s_{i}\}$ contains a converging subsequence $P$ with limit $w\in M$. Then for $s_{i}\in P$ we have $V(s_{i})\to w$.

By assumption, all $B_{k}$ are closed sets. Therefore $w\in B_{k_{j}}$ for all $j=1,\ldots,d+1$, and thus $w\in\cap_{j}{B_{k_{j}}}$. $\square$

Without loss of generality, we can assume that $k\geq(d+2)/2$ 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

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

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

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

Let $x\in Q$. Then

Since $k\geq m+1$ 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. $A(x)\in C_{d+2}$. $\square$

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

a contradiction. $\square$