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Example of two complexes of chains $C_*$ and $D_*$ and a morphism of complexes of chains $f_*:C_*\to D_*$ in such a way that $f_n:C_n\to D_n$ is injective for all $n\in\mathbb{Z}$ but $f_*:H_n(C_*)\to H_n(D_*)$ is not injective for some $n\in \mathbb{Z}$.

I am trying to find an example where what is described in the statement of this question is fulfilled. I've thought about doing the following but I do not know if I'm on the right track:

Let's take $C_n=\mathbb Z/4\mathbb Z$, $D_n=\mathbb Z/8\mathbb Z$, $\partial_n:\mathbb Z/4\mathbb Z\to \mathbb Z/4\mathbb Z$, $\partial_n(x)=0 $, $\partial_n':\mathbb Z/8\mathbb Z\to\mathbb Z/8\mathbb Z$, $\partial_n'(x)=4x$, $f_n:\mathbb Z/4\mathbb Z\to \mathbb Z/8\mathbb Z$ the inclusion function for all $n\in\mathbb{Z}$, which clearly is injective. So we have the following:

$$\require{AMScd} \begin{CD} \dotsb@>\times0>>\mathbb Z/4\mathbb Z@>\times0>>\mathbb Z/4\mathbb Z@>\times0>>\mathbb Z/4\mathbb Z@>\times0>>\dotsb \\ @. @Vf_{n - 1}VV @Vf_nVV @Vf_{n + 1}VV @. \\ \dotsb@>\times4>>\mathbb Z/8\mathbb Z@>\times4>>\mathbb Z/8\mathbb Z@>\times4>>\mathbb Z/8\mathbb Z@>\times4>>\dotsb \end{CD} $$

Note that for all $n\in\mathbb{Z}$, it is true that $\ker(\partial_n)=\mathbb Z/4\mathbb Z$, $\operatorname{Im}(\partial_{n+1})=\{\bar{0}\}$ and so $H_n(C_*)=\mathbb Z/4\mathbb Z$. On the other hand $\ker(\partial_n')=\{\bar{0},\bar{2},\bar{4},\bar{6}\}$ and $\operatorname{Im}(\partial_{n+1}')=\{\bar{0},\bar{4}\}$ and so $H_n(D_*)=\frac{\{\bar{0},\bar{2},\bar{4},\bar{6}\}}{\{\bar{0},\bar{4}\}}$. My question is if $f_*:H_n(C_*)\to H_n(D_*)$ is injective? How can I prove this? Thank you.

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    $\begingroup$ There is no "inclusion function" $\mathbb Z/4\mathbb Z \to \mathbb Z/8\mathbb Z$. Do you mean the multiplication-by-2 map? Then your example seems to show that $f_*$, far from being injective, is identically $0$. $\endgroup$ – LSpice May 16 at 1:41
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    $\begingroup$ Aside from the wrong terminology, I think no improvement is needed; you've given an example of the behaviour you sought. Aside from the fact that you could replace your two-sided-infinite complex by $0 \to A \to A \to 0$ to make it even simpler, I think it's natural and accessible. $\endgroup$ – LSpice May 16 at 1:58
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    $\begingroup$ @Nash What group is $H_n(D_*)$ isomorphic to? $\endgroup$ – Balarka Sen May 16 at 2:15
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    $\begingroup$ Correct. Then you have a homomorphism $f_* : \Bbb Z_4 \to \Bbb Z_2$. Now can you answer why $f_*$ is not injective? $\endgroup$ – Balarka Sen May 16 at 2:20
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    $\begingroup$ For a topological example, consider $S^1 \subset D^2$. Then the inclusion of (singular or cellular) chains $C_*(S^1) \to C_*(D^2)$ is injective for all $*$, but $H_1(S^1) \to H_1(D^2)$ is the $0$ map $\mathbb{Z} \to 0$. $\endgroup$ – William May 16 at 2:24

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