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

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 $$f_\sharp \partial=\partial f_\sharp$$.

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.

• Yes, that is the definition of $f_*$. – Pedro Tamaroff Feb 26 '19 at 17:16
• 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. – 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 $$\alpha+Im \partial^X_{n+1} \mapsto f_\#(\alpha) + Im \partial^Y_{n+1}$$.