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Suppose that $f:X\rightarrow Y$ is a continuous mapping between connected topological spaces such that for any field $k$, $H_{\ast}(X,k)\rightarrow H_{\ast}(Y,k)$ is an isomorphism. Does it follow that $H_{\ast}(X,\mathbf{Z})\rightarrow H_{\ast}(Y,\mathbf{Z})$ is an isomorphism in integral homolgy ? Do we need to assume that $X$ and $Y$ are simply connected ?

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  • $\begingroup$ I think the universal coefficient theorem implies it has to be surjective. $\endgroup$ – ronno Jan 14 at 15:58
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In fact, you only need it for $\mathbb{Q}$ and $\mathbb{Z}/p$. There is a very useful short exact sequence $0 \rightarrow \mathbb{Z}\rightarrow \mathbb{Q} \rightarrow \oplus _p \mathbb{Z}/p^\infty \rightarrow 0$. In this case, it gives us a long exact sequence in homology $ \dots \rightarrow H_{n+1} (X;\oplus _p \mathbb{Z}/p^\infty) \rightarrow H_n(X; \mathbb{Z}) \rightarrow H_n(X; \mathbb{Q}) \rightarrow H_n(X;\oplus _p \mathbb{Z}/p^\infty) \rightarrow \dots$ . If we knew that an isomorphism with coefficients in $\mathbb{Z}/p$ gave rise to an isomorphism with coefficients in $\mathbb{Z}/p ^\infty$ we would be done by the five lemma. So let's prove that it does.

For all $k$ we have a short exact sequence $0 \rightarrow \mathbb{Z}/p^{k-1} \rightarrow \mathbb{Z}/p^k \rightarrow \mathbb{Z}/p \rightarrow 0$, so by induction, the long exact sequence on homology coming from this short exact sequence, and the five lemma we have that the induced map on homology with coefficients in $\mathbb{Z}/p^m$ is an isomorphism for all $m$. Now we want to show that the homology with coefficients in $\lim\limits_{k \rightarrow \infty}\mathbb{Z}/p^k=\mathbb{Z}/p^\infty$ is the same as the colimit of the homology with coefficients in $\mathbb{Z}/p^k$. This follows directly from the fact that on the chain complex level the colimit of the $C_n(X ; \mathbb{Z}/p^k)$ is $C_n(X ; \lim\limits_{k \rightarrow \infty}\mathbb{Z}/p^k)$ and that taking homology commutes with direct limits. So we have an isomorphism on homology with coefficients in $\mathbb{Z}/p^\infty$. Direct summing over all $p$ we can then apply the five lemma to the initial long exact sequence, and we are done.

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