Show that $0 \to \operatorname{Ker} \partial \to C_n \to \operatorname{Im} \partial \to 0$ splits. This comes from Hatcher's exercise 43 of section 2.2:


  
*(a) Show that a chain complex of free abelian groups $C_n$ splits as a direct sum of subcomplexes $0 \to L_{n+1} \to K_n \to 0$ with at most two nonzero terms. [Show the short exact sequence $0 \to \operatorname{Ker}∂ \to C_n \to \operatorname{Im}∂ \to 0$ splits and take $K_n=\operatorname{Ker}∂$.]
  

I am trying to solve the hint. This is my first introduction to (splitting) exact sequences, so apologies if I'm missing something obvious.
My attempt: 
Because the $C_n$'s are free abelian, so are $\operatorname{Ker} ∂$ and $\operatorname{Im}∂$. Let $K$ be a basis of $Ker ∂$. Then, $i(K)$ generates the image of $\operatorname{Ker}∂$. Elements of $i(K)$ cannot contain powers of other elements, lest we create torsion in the quotient $\operatorname{Im}∂$, so we may extend $i(K)$ to a basis of $C_n$, (**) which I will denote $i(K) \cup K'$. Then, it's clear that $s(K')$ is a basis of $\operatorname{Im}∂$, and that $C_n \cong \operatorname{Ker}∂ \oplus \operatorname{Im}∂$.
(**) I am extremely uncertain about this step. Is it valid? What if $C_n$ isn't finitely generated?
Is this even the best way to approach this problem? 
 A: As a heads up, the fact that $C_n$ is free does not imply that $\partial(C_n)$ is free. For example, take $0\to \Bbb Z\to \Bbb Z\to \Bbb Z/2\to 0$.
In this case, $\operatorname{im} \partial$ is free because it is a submodule of $C_{n-1}$, which is free. A submodule of a free module over a PID is free, although even this is not entirely obvious (this fails over some rings).
Anyway, since $\operatorname{im} \partial$ is free, the sequence splits. Take a basis $\{b_i\}$ for $\operatorname{im}\partial$. (This exists for any free module, even infinitely generated.) For each $b_i$, pick any $c_i\in \partial^{-1}(b_i)\in C_n$. I suppose this requires the axiom of choice, but that is already needed to show that a submodule of a free $\Bbb Z$-module is free.
Since $\{b_i\}$ is a basis, this extends to a map $f\colon \operatorname{im}\partial\to C_n$ called a splitting map. In particular, $\partial\circ f=\operatorname{id}$. Finally, construct the isomorphism $C_n\cong \ker\partial\oplus \operatorname{im\partial}$ by writing $C_n=\ker\partial\oplus \operatorname{im}f$ as an internal direct sum and then identifying $\operatorname{f}$ with $\operatorname{\partial}$ (note that $f$ is injective since $\partial\circ f$ is an isomorphism).

Edit: to expand on the part where you write $C_n=\ker\partial\oplus \operatorname{im}f$, there is no need to explicitly extend $\{c_i\}$ to a basis of $C_n$.
Claim 1: $C_n=\ker\partial+\operatorname{im}f$. Proof: by the first isomorphism theorem $C_n/\ker \partial\cong \operatorname{im}\partial$ Thus $C_n$ is covered by the cosets $x+\ker\partial$, so any element in $C_n$ is a sum of an element of $\operatorname{im}f$ and an element of $\ker\partial$.
Claim 2: $\ker\partial\cap\operatorname{im}f=\{0\}$. Proof: exactness.
