Theorem: Let $\phi: G \rightarrow H$ be a group homomorphism. Then $\operatorname{Ker}(\phi)$ is a normal subgroup of G. $\phi(G)$ is a subgroup of H and the map $\phi'$: $G/\operatorname{Ker}(\phi) \rightarrow \phi(G)$, $a\operatorname{Ker}(\phi)\mapsto \phi(a)$ is a group isomorphism.
My question relates to the part where it is shown that $\phi'$ is surjective. From the proof: Because of $\phi'(G/N)=\phi(G)$, $\phi'$ is obviously surjective. So why do we know that $\phi'(G/N)=\phi(G)$ holds?
Anything in $\phi(G)$ takes the form $\phi(a)$ for $a \in G$, and the coset $aN$ will map to $\phi(a)$. Check the definition of $\phi'$.