I'm trying to prove that if the sequence
$$ M \xrightarrow{\varphi} W \rightarrow 0$$
is exact with $ W $ being a free module, then $ M \simeq \ker{\varphi} \oplus W $
What I got is that since $ W $ is free, then there exists a $ \psi: W \rightarrow M $ such that $ \varphi(\psi(w)) = w $ for all $ w \in W $.
With this I can proof that $ \psi $ is injective. But I can't finish the proof, meaning that if $ m \notin \psi(W) $, then $ m \in \ker\varphi $.
Am I getting this right? I would appreciate some help
You're misinterpreting the meaning of direct sum: it's not a disjoint union. What you need to show is that each $m\in M$ can be written as the sum of an element in $\ker\varphi$ and an element in $\psi(W)$, and that the latter two submodules intersect in zero. That the intersection is zero is clear from the definition of $\psi$, and for the other half of the claim I'd consider for each $m$ the element $m-\psi(\varphi(m))$.