The induced map on homology from a chain map is well defined.

Question

I'm having a hard time understanding a proof that should be a simple "check".

Background info: Let $f:A_\ast\to B_\ast$ be a chain map of abelian groups. The induced map on homology $f_\ast:H_n(A)\to H_n(B)$ is defined by setting $f_\ast([\sigma]):=[f_n(\sigma)]$.

Question: Check that it is well defined: Let $\sigma,\sigma'\in A_n$.

$$[\sigma]=[\sigma']\iff \exists \tau\in A_{n+1}:\sigma-\sigma'=\partial^A_{n+1}(\tau) \implies f_\ast([\sigma])=[f_n(\sigma)]=[f_n(\partial^A_{n+1}(\tau)+\sigma')] $$

Now what is left to show is that $f_n\circ\partial^A_{n+1}(\tau)=0$. We can use the identity $f_n\circ\partial^A_{n+1}(\tau)=\partial^B_{n+1}\circ f_{n+1}(\tau)$, but I don't know why this identity implies the latter term is $0$.

Answer 1

You don't need to show that $f_n\circ\partial^A_{n+1}(\tau)=0$, consider a chain $(A_n,d_n)$ with $d_{n}\circ d_{n-1}=0$, we denote by $Z_n(A)=\{s:d_n(s)=0\}$ the homology $H_n(A)$ is the quotient of $Z_n(A)$ by the image of $d_{n+1}$.

So $[s]=[s']$ is equivalent tgat $s=s'+d_{n+1}(r)$, it results that $f_n(s)=f_n(s')+f_n(d_{n+1}(r))=f_n(s')+d_{n+1}(f_{n+1})(r)$, you deduce that $f_n(s)-f_n(s')$ is a boundary and $[f_n(s)]=[f_n(s')]$.