at.algebraic topology - Equivariant maps inducing isomorphism in integral cohomology

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 $f:Xto Y$ can be replaced by an inclusion $iota:Xto M_f$, 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 $f:Xto Y$ 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 $f:Xto Y$ as in inclusion and there is a long exact sequence in cohomology
$cdotsto H^ast(X;mathbb{Z})to H^{ast+1}(Y,X;mathbb{Z})to H^{ast+1}(Y;mathbb{Z})to H^{ast+1}(X;mathbb{Z})to H^{ast+2}(X;mathbb{Z})tocdots$
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 $Xto S^{i_0}$ via $Xstackrel{f}{to} Yto S^{i_0}$ (all upto equivariant homotopy).

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