Equivariant maps from simplicial complexes to spheres

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?

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 $$\operatorname{cup-length}\ker(p^*:H^*(\overline{X})\to H^*(X))\le \operatorname{ind}(X)\le \dim(\overline{X}).$$
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.
• Could you please give a little more detail about the involution $\nu$? Am I right that (forgetting the triangulation) it can be defined as $(x,y,z) \mapsto (-x,-y,-z)$, where the torus is radially symmetric about the $z$-axis? But then the index of $K$ would be $1$ and not $2$. – user1272680 Jun 15 '16 at 10:05
• @user1272680: I'm not entirely sure if the orientation-reversing involution you describe has as its quotient the Klein bottle. I prefer to think of the torus as $S^1\times S^1\subset \mathbb{C}^2$ and then my involution is given by $\nu(z,w)=(-z,\bar{w})$. – Mark Grant Jun 15 '16 at 10:45
• I think the cohomology calculation I had in mind went as follows: the 1st Stiefel-Whitney class $w_1\in H^1(\mathrm{Klein};\mathbb{Z}/2)$ pulls back to zero in the cohomology of the torus. I think I thought that $w_1^2$ was nonzero, but now I'm not sure. I'll have another think about this when I get time. – Mark Grant Jun 15 '16 at 11:05