Homology with coefficients for $S^n$.

Question

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?

Answer 1

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.