# Homology with coefficients for $S^n$.

It is claimed in page 154 Lemma 2.49 that if there is a group homomoprhism $$\varphi:\Bbb Z \rightarrow G$$, then we have the following commutativity $$\require{AMScd}\begin{CD}\Bbb Z @> {\varphi}>> G \\@V{\simeq}VV@VV{\simeq}V\\ \tilde{H}_k(S^k;\Bbb Z) @>{\varphi_*}>>\tilde{H}_k(S^k;G)\end{CD}$$ by induction. How is this done?

So Hatcher makes two claims few lines earlier that are important:

1. $$\varphi_*$$ commutes with homology long exact sequence
2. $$\varphi_*$$ commutes with induced homomorphisms $$f_*$$ for any $$f:(X,A)\to(Y,B)$$

Both are easy to show (hopefuly). What is harder is that I think that 1. implies that $$\varphi_*$$ also commutes with the Mayer-Vietoris sequence. Which will be important.

Now consider the sphere $$S^k$$. Take $$A,B$$ to be the upper and lower hemisphere and apply the Mayer-Vietoris sequence to it. Since $$A,B$$ are contractible then there are lots of $$0$$s in the sequence and so we get an isomorphism $$\tilde{H}_i(S^k)\simeq \tilde{H}_{i-1}(A\cap B)$$ from it. And $$\varphi_*$$ commutes with it.

On the other hand $$A\cap B$$ is homotopy equivalent to $$S^{k-1}$$ and so $$\varphi_*$$ commutes with some $$\tilde{H}_{i-1}(A\cap B)\simeq\tilde{H}_{i-1}(S^{k-1})$$ isomorphism by property 2.

Both these facts are coefficient independent and together they show that there is a commutative diagram of the form:

$$\require{AMScd}\begin{CD}\tilde{H}_{k-1}(S^{k-1};\Bbb Z) @> {\varphi_*}>> \tilde{H}_{k-1}(S^{k-1}; G) \\@V{\simeq}VV@VV{\simeq}V\\ \tilde{H}_{k}(S^{k};\Bbb Z) @> {\varphi_*}>> \tilde{H}_{k}(S^{k}; G)\end{CD}$$

Finally by induction we reduce the problem to $$k=0$$. This case has to be calculated manually I'm afraid. And I'm not sure why it holds to be honest.