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Let $f: X \to Y$ be a map of topological spaces which induces isomorphisms $H_*(f;\mathbb{Z})$ on singular homology with $\mathbb{Z}$-coefficients. Show that $f$ induces isomorphisms $H_*(f;\mathbb{Z}/k\mathbb{Z})$ on singular homology with $\mathbb{Z}/k\mathbb{Z}$-coefficients for every $k \in \mathbb{Z}$.

What's the general approach to proofing something like that? I don't know any algebraic tools to change the ring of singular homology.

Thanks in advance.

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    $\begingroup$ Consider the universal coefficient theorem. $\endgroup$
    – anomaly
    Jun 26, 2015 at 18:10
  • $\begingroup$ I know no algebraic topology. But, it sounds like you want to show that if $f$ induces an isomorphism of two long exact sequences of $\mathbb{Z}$-modules, then $f$ induces an isomorphism of the sequences as $\mathbb{Z}/k\mathbb{Z}$-modules. If so, then I would think to tensor with $(--)\otimes_\mathbb{Z} \mathbb{Z}/k\mathbb{Z}$. A google search gave me this. $\endgroup$
    – Eoin
    Jun 26, 2015 at 18:12
  • $\begingroup$ We didn't cover the universal coefficient theorem yet, there has to be another way. What do I have to tensor to get the desired result? $\endgroup$
    – Cosmare
    Jun 26, 2015 at 18:22
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    $\begingroup$ You don't need tensor products or the universal coefficient theorem for this. Hint: $f$ induces an iso in $H_*(\_;R)$ for a commutative ring $R$ iff $\tilde{H}(cone(f),R)$ vanishes. Consider the self chain map on the singular chain complex of $cone(f)$ which is given by multiplying the coefficients with $k$. As soon as $k\neq 0$ (this case is trivial) this is injective and it's cokernel computes the $\mathbb{Z}/k\mathbb{Z}$ homology. Now take a look at the LES of this SES. The same can be done with reduced homology, too. $\endgroup$
    – archipelago
    Jun 26, 2015 at 22:17
  • $\begingroup$ Thanks, that helped a lot. $\endgroup$
    – Cosmare
    Jun 27, 2015 at 11:25

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