A $\mathbb{Z}_2$-space is a pair $(T, \nu)$, where $T$ is a topological space and $\nu: T \rightarrow T$, called the $\mathbb{Z}_2$-action, is a homeomorphism such that $\nu \circ \nu= id_{T}$ . If $(T_1,\nu_1)$ and $(T_2, \nu_2)$ are $\mathbb{Z}_2$-spaces, a $\mathbb{Z}_2$-map between them is a continuous mapping $f : T_1 \rightarrow T_2$ such that $f \circ \nu_1 = \nu_2 \circ f$. The sphere $S^n$ is considered as a $\mathbb{Z}_2$-space with the antipodal homeomorphism $x \rightarrow -x$.
A simplicial $\mathbb{Z}_2$-complex is a simplicial complex $K$ with a simplicial map $\nu$ of $K$ into itself such that $\nu$ is a $\mathbb{Z}_2$-action on $K$.
I am working on a huge 35-dimensional simplicial $\mathbb{Z}_2$-complex that is very complicated. Let $B$ be this simplicial $\mathbb{Z}_2$-complex with the $\mathbb{Z}_2$-action $\nu$. I want to show there is no (continuous) $\mathbb{Z}_2$-map from $B$ to $S^3$. In this text when we talk about an $S^4$, we mean a "hollow" $S^4$ such that $\nu(S^4)=S^4$. If a copy of $S^4$ (or any structure homeomorphic to an $S^4$) were included in $B$, then by Borsuk-Ulam Theorem, there would not be any $\mathbb{Z}_2$-map from $B$ to $S^3$. I applied a discrete Morse function on $B$ and got $B'$ which is simpler than $B$, and homotopy equivalent to $B$.
Since $B$ and $B'$ are homotopy equivalent, their homology groups $H_k$ are isomorphic and their Betti numbers $b_k$ are equal: $$ H_k(B') \simeq H_k(B) \quad b_k(B')=b_k(B).$$
I am thinking to compute the homology group $H_4(B')$ and the Betti number $b_4(B')$ to get $H_4(B)$ and $b_4(B)$, and prove there exists a copy of $S^4$ included in $B$. I am not sure what homology group or Betti number show there exists an $S^4$ in $B$.
I guess that if I get $b_4(B)\neq 0$, I could conclude there is an $S^4$ included in $B$; or if $H_{4}(B)\neq 0$ and $b_4(B)= 0$, I could conclude there is no $S^4$ included in $B$ but still there is no $\mathbb{Z}_2$-map from $B$ to $S^3$.
Am I right? Are there any other ideas? My question is mostly about the homology groups. How do I realize that there is an $S^4$ in $B$ from its homology group (or Bettie number)?
I do not believe that you can determine the lack of existence of a $\mathbb{Z}/2$-equivariant map $B \to S^3$ by knowing whether $B$ contains a "copy of $S^4$." For example, let $B$ consist of $S^3$ with its standard antipodal $\mathbb{Z}/2$-action, along with two copies of $S^4$, one glued to each "pole" $(0, 0, 0, \pm 1)$. Define a $\mathbb{Z}/2$-action on $B$ by, as I said, the antipodal action on $S^3$ and where $\nu$ interchanges the two copies of $S^4$. Then there is a $\mathbb{Z}/2$-equivariant map $B \to S^3$ which is the identity on $S^3$ and which sends each copy of $S^4$ to the pole where it's glued.