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Let $F$ be a finite field (for example I could take $\mathbb{Z}_2$) and $f:X\longrightarrow Y$ a continuous map between compact, orientable and connected manifolds of dimension $n$. Suppose I have an isomorphism $H^n(f,F):H^n(Y,F)\longrightarrow H^n(X,F)$. How do I prove that $H_n(f,F):H_n(X,F)\longrightarrow H_n(Y,F)$ or $H_n(f,\mathbb{Z}):H_n(X,\mathbb{Z})\longrightarrow H_n(Y,\mathbb{Z})$ is an isomorphism?

So far I have tried to UCT for cohomology the following way:

$H^n(f,F)$ isomorphism implies I have and isomorphism between $\text{Hom}(H_n(Y,\mathbb{Z}),F)$ and $\text{Hom}(H_n(X,\mathbb{Z}),F)$. I can't follow from here.

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Let $M$ be an orientable (connected, closed) $n$-manifold. Then classically $H_n(M;F) \cong F$. But moreover, $H_{n-1}(M;\mathbb{Z})$ is torsion-free. This is Corollary 3.28 in Hatcher's book. The proof basically goes like this: if $H_{n-1}(M;\mathbb{Z})$ weren't torsion free, by the UCT then $H_n(M; \mathbb{F}_p)$ would be bigger than just $\mathbb{F}_p = H_n(M;\mathbb{Z}) \otimes \mathbb{F}_p$, which contradicts orientability.

It follows then that $H^n(M;F) \cong \hom_\mathbb{Z}(H_n(M;\mathbb{Z}), F)$, and moreover the isomorphism is functorial, i.e. if $f : X \to Y$ as in your question, then $f^* : H^n(Y;F) \to H^n(X;F)$ is the dual map of $f_* : H_n(X;\mathbb{Z}) \to H_n(Y;\mathbb{Z})$. This is because the short exact sequence in the UCT is functorial, and both $\operatorname{Ext}$-terms vanish.

Now $H_n(X;\mathbb{Z}) \cong H_n(Y;\mathbb{Z}) \cong \mathbb{Z}$, so the question boils down to this:

Let $F$ be a finite field. Given some group morphism $\phi : \mathbb{Z} \to \mathbb{Z}$, if $\phi^* : \hom_\mathbb{Z}(\mathbb{Z}, F) \to \hom_\mathbb{Z}(\mathbb{Z}, F)$ is an isomorphism, is $\phi$, respectively $\phi \otimes F$, an isomorphism?

(To apply this to your situation, let $\phi = f_* : H_n(X;\mathbb{Z}) \to H_n(Y;\mathbb{Z})$.)

The answer to the first question is actually "no". Indeed, suppose $\phi(k) = ak$, where $a$ is some number prime to $p$ (where $p$ is the characteristic of $p$). Then $\phi^* : F \to F$ is multiplication by $a$, which is an isomorphism! But $\phi$ is clearly not an isomorphism. So just choose a map of degree prime to the characteristic of your field to get a counterexample.

However $\phi \otimes F : \mathbb{Z} \otimes F \to \mathbb{Z} \otimes F$ will be an isomorphism, because multiplication by $a$ is an isomorphism.

tl;dr If $H^n(f,F)$ is an isomorphism, then $H_n(f,F)$ is an isomorphism, but $H_n(f,\mathbb{Z})$ may not be.

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  • $\begingroup$ If you happen to know it for all finite fields then you can say it is an isomorphism $\endgroup$ – Spotty May 16 '16 at 9:35

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