I am trying to understand the proof of the following statement
Let $\varphi: M\to N$ be an $R$-module homomorphism. Then it has a left-inverse if and only if the sequence $$ 0\rightarrow M\xrightarrow{\varphi}N\rightarrow \mathrm{coker}\:\varphi\rightarrow 0 $$ splits.
By 'splits' I mean that there is a commutative diagram
$$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllll} 0 & \rightarrow & M & \xrightarrow{\varphi} & N &\rightarrow& \mathrm{coker}\:\varphi &\rightarrow & 0 \\ & & \da{\sim} & & \da{\psi} & & \da{\sim} & & & & \\ 0 & \rightarrow & M'_1 & \rightarrow & M'_1\oplus M'_2 & \rightarrow & M'_2 & \rightarrow & 0 & \\ \end{array} $$ where vertial maps are isomorphisms.
The part I am having problem with is the one where we assume that $\varphi$ has a left-inverse $\psi$ and we claim that $N$ is isomorphic to $M\oplus \ker\varphi$. Does not this imply that $\ker\varphi\cong\mathrm{coker}\:\varphi$?
Assume that the injective homomorphism $\varphi: M\to N$ has a left inverse $\psi$. Let $K = \rm{ker}(\psi)$ and $Q = \rm{im}(\varphi)$
The claim is then that $N = K \oplus Q$.
It is clear that $K$ and $Q$ and submodules of $N$. If $v\in K\cap Q$ then $v = \varphi(x)$ for some $x$ and $\psi(v) = 0$. But we have $x = \psi(\varphi(x)) = \psi(v) = 0$ so $v = \varphi(0) = 0$ and we get that $K\cap Q = 0$.
Finally, we need to check that $N = K + Q$, so let $v\in N$. Let $w = \varphi(\psi(v))\in Q$ and note that $\psi(v - w) = \psi(v) - \psi(w) = \psi(v) - \psi(\varphi(\psi(v))) = \psi(v) - (\psi\circ\varphi)(\psi(v)) = \psi(v) - \psi(v) = 0$ so $v - w \in K$ and we get that $v = (v - w) + w \in K + Q$ which completes the proof.