Tucker's lemma, Borsuk-Ulam, triangulating a ball in *truly* antipodally symmetric fashion I'm attempting to prove Tucker's Lemma from the Borsuk-Ulam theorem by means of the proof sketched as "immediate" on page 36 of Matoušek's Using the Borsuk-Ulam Theorem. In order to do this, I need an auxiliary result.
Let $\Delta$ denote a simplicial complex that triangulates the closed ball $B^n$ by means of a homeomorphism $h: ||\Delta||\rightarrow B^n$, where $||\Delta||$ is the topological space corresponding to this simplicial complex. Let this triangulation be antipodal on the boundary in the following sense: that for a set $U\subseteq S^{n-1}=\partial B^n$, $U$ is the image of a simplex of $\Delta$ iff $-U$ is the image of a simplex of $\Delta$. ($-U$ denotes the set of antipodes of the points of $U$.)
We want to say that there exists an $h'$ that improves on this "antipodality on the boundary" slightly. We want $h'$ (and its inverse) to preserve antipodality for all pairs of points on the boundary. The boundary of $B^n$ and antipodality has been defined. For $||\Delta||$, we define antipodality for points on the boundary as follows. A point of $||\Delta||$ is "on the boundary" iff it is in $h^{-1}(\partial B^n)$. For a vertex on the boundary $||\Delta||$, its antipode is the vertex that winds up antipodal to it on $\partial B^n$. Every non-vertex point of $||\Delta||$ is uniquely identified as some affine combination of certain vertices. For a non-vertex point of $||\Delta||$ that's on the boundary, we identify its antipode as the same affine combination of the antipodes of the vertices.
More explicitly, any point of $||\Delta||$ in $h^{-1}(\partial B^n)$ is in a simplex of $||\Delta||$ whose vertices are also in $\partial B^n$, so we can define the antipode of an arbitrary point $x$ in $h^{-1}(\partial B^n)$ as $\sum_{i=0}^k a_iv_i\mapsto\sum_{i=0}^k a_i(-v_i)$.
To repeat ourselves, we want to show the existence of a homeomorphism $h': ||\Delta||\rightarrow B^n$ that carries antipodes to antipodes. Anybody got ideas? Thanks so much.
 A: Sketch: Show that $h$ restricted to the boundary is homotopic to an antipodal map using the Alexander trick and inducting on the dimensions of the simplices. Now find an $\epsilon$ such that either $h(v)$ is on $S^{n-1}$ or $||h(v)|| < 1 - \epsilon$ for each vertex $v$ of $\Delta$ and glue on the homotopy along $S^{n-1} \times [1-\epsilon, 1]$.
A: As promised above, here is the elaboration of Ronno's sketch.
We alter $h: ||\Delta|| \rightarrow B^n$, which is weakly antipodal on the boundary, to a homeomorphism $h': ||\Delta|| \rightarrow B^n$ that is antipodal on the boundary.
In fact, it is sufficient to construct an antipodal homeomorphism $h'|_E: E\rightarrow \partial B^n$. That's because we know that $E\cong \partial B^n$ and $||\Delta||\cong B^n$, and any homeomorphism $S^{n-1}\rightarrow S^{n-1}$ extends to a homeomorphism $B^n\rightarrow B^n$.\footnote{For instance, for $f: S^{n-1}\rightarrow S^{n-1}$, $x\in S^{n-1}$ a vector, and $r\in[0,1]$ a scalar, a point in $B^n$ is identified by some $rx$. So define $g: B^n\rightarrow B^n$ by $F(rx)=rf(x)$.} So a triangulation of $S^{n-1}$ by $E$ extends to a triangulation of $B^n$ by $||\Delta||$, and if the former map is antipodal, the latter map will be antipodal on the boundary.
We show the existence of such an antipodal $h'|_{E}$ by a list of homeomorphisms $h_i: E\rightarrow S^{n-1}$ for $0\leq i\leq n-1$, with $h_0:=h|_E$ and $h'|_{E}:=h_{n-1}$. The \textit{first condition} on this list is that $h_i$ is antipodal on all points of $E$ that are in a $i$-simplex. The \textit{second condition} to keep track of is that for $U\subseteq S^{n-1}$, $h^{-1}_i(U)$ is a simplex if and only if $h^{-1}_0(U)$ is a simplex, and if so, they're the same simplex. We show the list's existence by induction.
The base case is easy: because a $0$-simplex consists of a single point, $h_0$ is definitionally antipodal on all points in $0$-simplices, and the second condition is tautological.
For the inductive step, assume that $h_i$ (with $i<n-1$) satisfies the condition. Now consider the $i+1$-simplices of $E$. Now consider specific antipodal simplices\footnote{We're abusing notation/phrasing a little bit by allowing ``simplex'' to refer not only to a formal entity, but also to the appropriate subspaces of $||\Delta||$ that a given simplex induces. We ought to write $||\sigma||$, but that's bulky.} $\sigma, -\sigma$. Because all $i+1$-simplices are homeomorphic to $B^{i+1}$ and because $h_i$ is a homeomorphism, $\sigma \cong -\sigma \cong h_i(\sigma) \cong h_i(-\sigma) \cong B^{i+1}$. Note further that the points corresponding to the $S^i\subseteq B^{i+1}$ in this homeomorphism are those in the $i+1$-simplices $\sigma,-\sigma$ that are in $i$-simplices. So we consider the maps $h_i|_{\sigma}$ and $- \circ h_i|_{-\sigma} \circ -$. The rightmost $-$ is in $\Delta$; the leftmost is in $\partial B^n$. Thanks to the second condition on $h_i$ and the definition of Tucker triangulation, these maps have the same domain and range. Composed with homeomorphisms $B^{i+1}\rightarrow \sigma$ and $h_i(\sigma)\rightarrow B^{i+1}$, they are homeomorphisms $B^{i+1}\rightarrow B^{i+1}$ that are equal on the boundary $S^i\subseteq B^{i+1}$. It is a fact of topology, referred to as \textit{Alexander's Trick},\footnote{This term is also used for the earlier fact that a homeomorphism of spheres induces a homeomorphism of balls.} that homeomorphisms $B^{i+1}\rightarrow B^{i+1}$ are isotopic when they agree on $S^{i}$. You don't need to know what isotopic means except that it implies homotopic. The homotopy holds the boundary fixed, so it induces a homotopy from $h_i$ to an $g$ such that $g=h_i$ except on $\sigma$ and such that $g$ is antipodal on $\sigma$ (and by impliciation, on $-\sigma$). Choosing an $i+1$-simplex from each antipodal pair and repeat this process. Since the homotopies don't disturb any simplices other than whichever $i+1$-simplex they arose from, we can compose them. This composition takes $h_i$ to a function $f$ that agrees with $h_i$ on all $i$-simplices and is antipodal on all $i+1$-simplices. The function $f$ satisfies the conditions for $h_{i+1}$, so this completes the inductive step.
The list exists, so $h_{n-1}$ is an antipodal homeomorphism $h'|_E: E\rightarrow S^{n-1}$. As stated above, this is sufficient for a homeomorphism $h': ||\Delta||\rightarrow B^n$ that preserves antipodes.
