TITLE: Equivariant maps from simplicial complexes to spheres
QUESTION [5 upvotes]: Given a topological space $X$ with  involution $\nu$, the $\mathbb Z_2$-index $\text{ind}(X)$ is the minimum integer $n$ such that there exists a map $f:X \to S^n$ which is equivariant with respect to the antipodal map on the sphere $S^n$.
Let $K$ be a (finite) simplicial complex with a fixed-point-free involution such that $\text{ind}(|K|)=d$ and $|K|$ is not homotopy equivalent to $S^d$. Can we always find a (maximal) simplex $\sigma$ such that deleting $\sigma$ and $\nu(\sigma)$ does not decrease the $\mathbb Z_2$-index?

REPLY [3 votes]: I believe the following should work as a counter-example. Let $K$ be a simplicial torus $T^2$, obtained as the orientation double cover of a triangulated Klein bottle $\overline{K}$. Then $K$ comes equipped with an (orientation-reversing) involution $\nu$. I believe $\operatorname{ind}(|K|)=2$, and that removing any simplex $\sigma$ and its involute $\nu(\sigma)$ decreases the index.
The key observation (for me at least) is this: if $X$ is a (reasonably nice) free $\mathbb{Z}_2$-space and $p:X\to \overline{X}$ is the resulting quotient double cover, then
$$
\operatorname{ind}(X) = \operatorname{secat}(p:X\to \overline{X}).
$$
Here $\operatorname{secat}(p)$ is the sectional category (or normalized Schwarz genus) of the double cover, which by definition is one less than the smallest number of open sets needed to cover $\overline{X}$, on each of which $p$ admits a section. It follows that
$$
\operatorname{cup-length}\ker(p^*:H^*(\overline{X})\to H^*(X))\le \operatorname{ind}(X)\le \dim(\overline{X}).
$$   
Returning to the example of the torus covering the Klein bottle, the calculation of the index follows from mod 2 cohomology calculations, and the fact that the index decreases on removing any top-dimensional cell follows from homotopy invariance of the sectional category together with the fact that the Klein bottle minus a cell deformation retracts onto its 1-skeleton.  
In fact, this should generalize: whenever $X$ is a closed $d$-dimensional free $\mathbb{Z}_2$-manifold with $\operatorname{ind}(X)=d$ we should see this behaviour.