# $f_n:C_n\to D_n$ is injective for all $n\in\mathbb{Z}$ but $f_*:H_n(C_*)\to H_n(D_*)$ is not injective for some $n\in \mathbb{Z}$.

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.

• 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 $0$. – LSpice May 16 at 1:41
• 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. – LSpice May 16 at 1:58
• @Nash What group is $H_n(D_*)$ isomorphic to? – Balarka Sen May 16 at 2:15
• Correct. Then you have a homomorphism $f_* : \Bbb Z_4 \to \Bbb Z_2$. Now can you answer why $f_*$ is not injective? – Balarka Sen May 16 at 2:20
• 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 $\mathbb{Z} \to 0$. – William May 16 at 2:24