$G, H, K$ are groups and let $\phi: G\to H, \psi: G\to K$ be homomorphisms such that $\phi$ is onto and $\ker\phi \subset \ker\psi$. Prove that there is a unique homomorphism $\alpha: H\to K$ such that $\alpha\circ\phi = \psi$.
My attempt
My idea is that $\phi$ is injective on $G$ minus $\ker\phi$ so it is injective on $G$ minus $\ker\psi$, then define $\alpha$ to be $\psi\circ\phi^{-1}$, where the inverse of $\phi$ is for the $\phi$ as a map from $G - \ker\psi$ to $H$.
I'm having trouble writing this down rigorously. Also, I can't see how the uniqueness can be shown. Maybe using the fact that the inverse and $\psi$ are both uniquely determined?
Any help would be greatly appreciated, thanks.
For every $y\in H$, there exists $x\in G$ such that $\phi(x)=y$ since $\phi$ is surjective, write $\alpha(y)=\psi(x)$. $\alpha(y)$ does not depend of $x$ such that $\phi(x)=y$. Suppose that $\phi(x)=\phi(x')=y$. This implies that $\phi(x^{-1}x')=1$ and $x^{-1}x'\in ker\phi\subset ker\psi$, we deduce that $\psi(x^{-1}x')=\psi(x)^{-1}\psi(x')=1$ and $\psi(x)=\psi(x')$.