Given two groups $G,H$ and a group homomorphism $\varphi:G\to H$ its well-known that
$\varphi$ is injective if and only if the kernel $\ker\varphi$ is trivial.
In order to derive a similiar criterion for $\varphi$ being surjective (I think) I was able to prove the following proposition
$\varphi$ is surjective if and only if the quotient group $H/{\rm im}~\varphi$ exists.
Proof
If $\varphi$ is surjective, then ${\rm im}~\varphi=H$. Therefore the quotient group of our interest is given by $H/{\rm im}~\varphi=H/H\cong\{e\}$, in particular the quotient exists.
Now assume that $H/{\rm im}~\varphi$ exists. Then ${\rm im}~\varphi$ is a normal subgroup and as such kernel of a homomorphism with domain $H$. Let $G'$ be group and $\psi:H\to G'$ such that $\ker\psi={\rm im}~\varphi$. Composition yields the homomorphism $\psi\circ\varphi:G\to G'$ and by definition $\ker(\psi\circ\varphi)=G$. By the First Isomorphism Theorem we have $G/\ker(\psi\circ\varphi)\cong\{e\}\cong G'$. But then $\ker\psi=H$, hence by construction $\ker\psi=H={\rm im}~\varphi$. The result follows.
Is my argumentation sound; if so: why do I fail to locate a source actually stating this (occasionally) useful proposition? If not, where did I went wrong?
Thanks in advance!
EDIT
From the comments I realised that I have overlooked a crucial part: if $H$ is abelian, then $H/{\rm im}~\varphi$ always admits a group structure; regardless of $\varphi$ being surjective as in an abelian group every subgroup is normal. So I would like to rephrase the claimed proposition.
Let $G,H$ be groups and consider $H$ to be non-abelian. A group homomorphism $\varphi:G\to H$ is surjective if and only if the coset $H/{\rm im}~\varphi$ is a group.