Equivariant maps inducing isomorphism in integral cohomology

Question

Consider the following statement.

Suppose $X$, $Y$ are finite CW-complexes with free involution and $\mu:X\to Y$ is an equivariant map. If $\mu^*:H^i(Y;\mathbb{Z})\to H^i(X;\mathbb{Z})$ is an isomorphism for $i>i_0$ and is onto for $i=i_0$, then $\mu^{\sharp}:\pi_{\mathrm{eq}}^i(Y)\to\pi_{\mathrm{eq}}^i(X)$ is a 1-1 correspondence for $i>i_0$ and is onto for $i=i_0$.

Moreover, the preimage of each element of $\pi_{\mathrm{eq}}^{i_0}(X)$ is in 1-1 correspondence with the elements of the kernel of $\mu^*:H^{i_0}(Y;\mathbb{Z})\to H^{i_0}(X;\mathbb{Z})$.

Here $\pi_{\mathrm{eq}}^i(Y)$ denotes the set of all equivariant maps $Y\to S^i$ up to equivariant homotopy. This statement, without the 'moreover' part, appears to be known. Becker and Glover in Note on the embedding of manifolds in euclidean space (1971) state it (without the 'moreover' part) and claim that it is well known from obstruction theory. But what about the `moreover' part: is it true? if yes, can you give a reference?

Answer 1

I don't know of a reference off-hand but here's one way to think about it. First, one can think of $H^i(X;\mathbb{Z})$ as $[X,K(\mathbb{Z},n)]$, the set of homotopy classes of maps. Notice that a cellular model for $K(\mathbb{Z},n)$ can be taken to be $S^n$ union higher cells that kill off the higher homotopy groups. Second, any map $f:X\to Y$ can be replaced by an inclusion $\iota:X\to M_f$, where $M_f$ is the mapping cylinder and it has the same homotopy type as that of $Y$. This works in the equivariant setting also. The third fact is that if any equivariant map $f:X\to Y$ induces an isomorphism in cellular cohomology and $f$ acts freely on both $X$ and $Y$ then $f$ induces an isomorphism on equivariant cohomology as well. The equivariant cohomology can be thought of as maps from spaces to $K(\mathbb{Z},n)$ up to equivariant homotopy.

Now think of $f:X\to Y$ as in inclusion and there is a long exact sequence in cohomology $\cdots\to H^\ast(X;\mathbb{Z})\to H^{\ast+1}(Y,X;\mathbb{Z})\to H^{\ast+1}(Y;\mathbb{Z})\to H^{\ast+1}(X;\mathbb{Z})\to H^{\ast+2}(X;\mathbb{Z})\to\cdots$ which tells you in your case that $H^{i}(Y,X;\mathbb{Z})=0$ if $i>i_0$. The kernel of $H^{i_0}(Y;\mathbb{Z})\to H^{i_0}(X;\mathbb{Z})$ is just the image of $H^{i_0}(Y,X;\mathbb{Z})$ in $H^{i_0}(Y;\mathbb{Z})$. Thinking of $H^{i_0}(Y,X;\mathbb{Z})$ as relative homotopy classes of maps from $(Y,X)$ to $K(\mathbb{Z},i_0)$. These maps only probe the $i_0$ skeleton of $(Y,X)$ because $H^i(Y,X;\mathbb{Z})=0$ if $i>i_0$. And since any map can be made homotopic to a cellular map we need only study homotopy classes of maps from the $i_0$-skeleton of $(Y,X)$ to the $i_0$-skeleton of $K(\mathbb{Z},i_0)$ which is $S^{i_0}$. These are precisely the different ways of factoring a given equivariant map $X\to S^{i_0}$ via $X\stackrel{f}{\to} Y\to S^{i_0}$ (all upto equivariant homotopy).