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Let $f:(X,A)\to (Y,B)$ be a continuous map of pairs $(X,A), (Y,B)$ of topological spaces. $f$ induces a chain map on singular chain complexes $$f_*:C_*(X,A;R)\to C_*(Y,B;R),\; \sum_{\sigma}r_{\sigma}\sigma\mapsto \sum_{\sigma}r_{\sigma}f\circ \sigma,$$where $R$ is commutative ring with unit $1_R$. Suppose that $f_*$ induces an isomorphism in singular homology. My question is: Does $f_*$ imply an isomorphism in singular cohomology?

My question arises from a step in the proof of excision in singular cohomology.

Or more generaly, I would be happy if you have an example of a chain map $f_*:C_*\to D_*$ (of chain complexes of $R$-modules $C_*$ and $D_*$), such that $f_*$ induces an isomorphism in homology of chain complexes, (but if you dualize $f_*$ with the hom-functor $hom(-,R)$ such that the cochain map $f^*:D^*\to C^*$ does not induce an isomorphism in cohomology.

Regards

Edit: Ok I know the answer of the first question. The answer of my first qestion is yes, because the modules $C_n(X,A;R)$ and $C_n(Y,B;R)$ are projective for all $N\in\mathbb{N}$.

The answer of the second question should be no, and I think this goes wrong if you take an example whith $\mathbb{Z}/n\mathbb{Z}$-modules, where $n$ is not prime. But I'm still interested in an example.

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    $\begingroup$ Have you tried applying the universal coefficient theorem? $\endgroup$ – Matt Samuel Mar 4 '16 at 17:50
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Examples for (2) can be constructed as follows:

Let $C$ be exact and take $f = 0: C \to C$ the zero chain map. This trivially induces an isomorphism in homology. If the cohomology of the dual complex is non-trivial, it works because $f^\ast = 0$ is then no isomorphism.

For an explicit example let $R=k[x]$ be a polynomial ring over a field $k$. Take $$C:\qquad 0 \to k[x] \xrightarrow{x} k[x] \to k \to 0$$ ($k$ in degree $0)$. Dualizing gives $$C^\ast:\qquad 0 \to 0\to k[x] \xrightarrow{x} k[x] \to 0$$ Hence $H^2(C^\ast)=k[x]/(x)=k$.

Tensoring this complex with itself gives further examples over $R=k[x_1,...,x_n]$.

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