# The formula of the induced homomorphism of chain maps

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 $$(f_\bullet)_*: H_\bullet(A_\bullet)\to H_\bullet(B_\bullet).$$

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.

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.