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.


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.

  • 1
    $\begingroup$ Have you tried applying the universal coefficient theorem? $\endgroup$ – Matt Samuel Mar 4 '16 at 17:50

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 $R=k[x_1,...,x_n]$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.