# Question about the proof of Second Isomorphism Theorem

The Second Isomorphism Theorem: Let $$H$$ be a subgroup of a group $$G$$ and $$N$$ a normal subgroup of $$G$$. Then $$H/(H\cap N)\cong(HN)/N$$

There is the proof of Abstract Algebra Thomas by W. Judson:

Define a map $$\phi$$ from $$H$$ to $$HN/N$$ by $$H\mapsto hN$$. The map $$\phi$$ is onto, since any coset $$hnN=hN$$ is the image of $$h$$ in $$H$$. We also know that $$\phi$$ is a homomorphism because $$\phi(hh')=hh'N=hNh'N=\phi(h)\phi(h')$$ By the First Isomorphism Theorem, the image of $$\phi$$ is isomorphic to $$H/\ker\phi$$, that is $$HN/N=\phi(H)\cong H/\ker\phi$$ Since $$\ker\phi=\{h\in H:h\in N\}=H\cap N$$ $$HN/N=\phi(H)\cong H/H\cap N$$

My question:

Is it necessary to prove that the map $$\phi$$ is onto? Can we only prove that $$\phi$$ is well defined and the image of $$\phi$$ is a subset of $$HN/N$$? And then we can use the First Isomorphism Theorem and continue the proof.

Thank you.

The First Isomorphism Theorem states that if $$\varphi: G \to G'$$, then $$\mathrm{im}(\varphi) \cong G/\mathrm{ker}(\varphi)$$. If we do not know that your $$\phi$$ is surjective, then the First Isomorphism Theorem only shows us that $$H/H \cap N \cong \mathrm{im}(\phi) \subseteq HN/N$$, which does not finish the job.