Why is $ 0 \to \ker h \xrightarrow {i_h} H^n(C,g) \xrightarrow {h} Hom(H^n(C),G) \to 0 $ (Hathcher's p192) split exact?

Question

I'm reading Hatcher's book of Algebraic Topology, and I got stuck at (page 192) concerning the following split exact sequence $$ 0 \to \ker h \xrightarrow {i_h} H^n(C,G) \xrightarrow {h} Hom(H_n(C),G) \to 0 $$ It is clear that it is short exact ($h$ is surjective), however I can't see why it splits. I guess if we could show that $Hom(H^n(C),G)$ or $H^n(C,G)$ are free abelian it would do the trick.

Answer 1

So this is the idea I got, as E.KOW suggested using the splitting lemma.

We have that the following sequence $$ 0 \to Z_n \xrightarrow {i_Z} C_n \xrightarrow {\partial_n} B_{n-1} \to 0 $$ is split exact. Thus, by the splitting lemma, there exists $t \colon C_n \to Z_n$ such that the restriction to $Z_n$ yields the identity over $Z_n$ ($id_{Z_n}$)

To show that $h$ is surjective, take $g \colon H_n(C) \to G$ and consider $\varphi = (g \circ q) \circ t \colon C_n \to G$ (where $q \colon Z_n \to Z_n/B_n$ is the quotient map). We can easily show that $[\varphi] \in H^n(C,G)$ and that $h([\varphi])=g$

Now consider \begin{align} q \colon Hom(H_n(C),G) &\to H^n(C,G) \\ g &\mapsto [(g \circ q) \circ t] \end{align} Showing that q is a morphism, we can conclude using the splitting lemma again (since $h \circ q = id$ over $Hom(H_n(C),G))$.