Tucker’s lemma, Borsuk-Ulam, triangulating a ball in *truly* antipodally symmetric fashion

combinatorial-geometrydiscrete geometrygeneral-topologysimplicial-complex

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 MatouĊĦek'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.

Best Answer

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 $S^{n-1} \times [1-\epsilon, 1]$.

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