We recently introduced chain complexes in the algebraic topology course I'm taking; I'm currently trying to show that the map induced on homology by an injective chain map does not have to be injective itself. To do this, I constructed the following two bounded chain complexes, together with the chain map described by the vertical arrows: $$\require{AMScd} \begin{CD} 0 @>{}>> \mathbb{Z} @>{\times 2}>> \mathbb{Z} @>{}>> 0\\ @. @V{\times 2}VV @V{id}VV @. \\ 0 @>{}>> \mathbb{Z} @>{id}>> \mathbb{Z} @>{}>> 0 \end{CD}. $$ The middle square commutes, so the vertical arrows really describe a chain homomorphism. It is also clear that each vertical arrow is injective. The homology of the first chain complex is $H_0^1 = \mathbb{Z}/2\mathbb{Z}$ and $H_1^1 = 0$, and the homology of the second chain complex is $H_0^2 = H_1^2 = 0$. Clearly, the induced function $f_* : H_0^1 \to H_0^2$ can't be injective.
I still very new to chain complexes and homology, so a proof check would be appreciated!
Thanks