TITLE: The formula of the induced homomorphism of chain maps
QUESTION [7 upvotes]: It is known that a chain map $f_\bullet$ between chain complexes $(A_\bullet, d_{A,\bullet})$ and $(B_\bullet, d_{B,\bullet})$ induces a homomorphism $$(f_\bullet)_*: H_\bullet(A_\bullet)\to H_\bullet(B_\bullet).$$
(https://en.wikipedia.org/wiki/Chain_complex#Chain_maps)
The sources (Hatcher / wikipedia) I read do not explicitly mention how the homomorphism is induced, so I would like to confirm if my idea is correct?
For $\alpha+\text{Im}\,\partial_{A,n+1}\in H_n(A_\bullet)$, 
$(f_n)_*(\alpha+\text{Im}\,\partial_{A,n+1})=f_n(\alpha)+\text{Im}\,\partial_{B,n+1}$?
And this works since $f_n$ maps cycles to cycles and boundaries to boundaries?
Thanks for any help.

REPLY [8 votes]: You are exactly right. To elaborate, the fact that $f_n$ sends cycles to cycles means that we have a map (abusing notation slightly)
$$f_n:\ker(d_n^A)\to\ker(d_n^B)$$
and the fact that $f_n$ sends boundaries to boundaries means that $f_n(\operatorname{Im}(d_{n+1}^A))\subseteq \operatorname{Im}(d_{n+1}^B)$. Hence the map
$$\pi_n^B\circ f_n:\ker(d_n^A)\to H_n(B_{\bullet})$$
(where $\pi_n^B:\ker(d_n^B)\to H_n(B_{\bullet})$ is the projection) sends every element of $\operatorname{Im}(d_{n+1}^A)\subseteq \ker(d_n^A)$ to the identity in $H_n(B_{\bullet})$. Therefore, by the universal property of quotient groups, $\pi_n^B\circ f_n$ factors uniquely through $H_n(A_{\bullet})$:
$$\pi_n^B\circ f_n:\quad \ker(d_n^A)\xrightarrow{\ \pi_n^A \ }H_n(A_{\bullet})\xrightarrow{\ f_* \ }H_n(B_{\bullet}).$$
Since $\pi_n^B\circ f_n = f_*\circ\pi_n^A$, we see that
$$f_*(\alpha+\operatorname{Im}(d_{n+1}^A)) = f_n(\alpha)+\operatorname{Im}(d_{n+1}^B),$$
just as you claimed.