If $f: X\to Y$ is a map, then the chain map $f_\sharp: C_n(X) \to C_n(Y)$ is defined by composing a simplex with $f$, that is $f_\sharp(\sigma)=f\sigma$. We can then extend this linearly on all of $C_n(X)$.

The chain maps also satisfy $f_\sharp \partial=\partial f_\sharp$.

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How do the chain maps $f_\sharp: C_n(X) \to C_n(Y)$ induce a homomorphism $f_*: H_n(X) \to H_n(Y)$?

What is the induced map?

Is it $g+Im\partial_{n+1}^X \mapsto f_\sharp(g) +Im\partial_{n+1}^Y$?

This map is well-defined since if $g+Im\partial_{n+1}^X=h+Im\partial_{n+1}^X$, then $g-h \in Im\partial_{n+1}^X$. Since $f_\sharp$ takes boundaries to boundaries, then $f_\sharp(g-h) \in Im\partial_{n+1}^Y$ and so $f_\sharp(g)+Im\partial_{n+1}^Y=f_\sharp(h)+Im\partial_{n+1}^Y$.

However, I'm not sure if this is what the induced map should be.

  • $\begingroup$ Yes, that is the definition of $f_*$. $\endgroup$ – Pedro Tamaroff Feb 26 '19 at 17:16
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    $\begingroup$ Just a general comment- often at this level of abstraction, there's really only one obvious simple definition. In fact that's is why it's defined at this level of abstraction; all the other details in all the different places $f_*$ shows up end up being irrelevant so mathematicians of the past ended up searching for the most general definition possible. $\endgroup$ – Neal Feb 26 '19 at 17:21

It is easy to check that $f_\#$ maps $\ker \partial^X_n$ to $\ker \partial^Y_n$, and maps $Im \partial^X_{n+1}$ to $Im \partial^Y_{n+1}$.

So, we get a map $\ker \partial^X_n \to \ker \partial^Y_n \to \ker \partial^Y_n/ Im \partial^Y_{n+1}= H_n(Y)$ where $\alpha \mapsto f_\#(\alpha) \mapsto f_\#(\alpha) + Im \partial^Y_{n+1}$. Furthermore, the kernel of this map includes $Im \partial^X_{n+1}$.

Therefore, we can factor this map to obtain $$H_n(X)=\ker \partial^X_n/Im \partial^X_{n+1} \to \ker \partial^Y_n/ Im \partial^Y_{n+1}= H_n(Y)$$ where $\alpha+Im \partial^X_{n+1} \mapsto f_\#(\alpha) + Im \partial^Y_{n+1}$.


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