Greetings,

I have problem understanding something I read of Dummit & Foote. I'm not sure I got it correctly, so I need to be checked:

If $\displaystyle \pi: G \rightarrow G/N$ is the natural projection homomorphism of $\displaystyle G$ on $\displaystyle G/N$ and $\displaystyle \Phi: G \rightarrow H$ is a homomorphism, then let's prove that $\displaystyle \phi$, given by $\displaystyle gN \mapsto \Phi(g)$ is a function and homomorphism if and only if $\displaystyle N \subseteq ker \Phi$.

If I understand correctly, the issue here is proving that $\displaystyle \phi$ is a function. If we know that $\displaystyle \phi$ is a function, then proving that

$\displaystyle \phi(g_1N)\phi(g_2N)=\phi(g_1g_2N)$ is easy: $\displaystyle \phi(g_1N)=\Phi(g_1)$,$\displaystyle \phi(g_2N)=\Phi(g_2)$,$\displaystyle \phi(g_1g_2N)=\Phi(g_1g_2)$ and $\displaystyle \Phi(g_1)\Phi(g_2)=\Phi(g_1g_2)$ because $\displaystyle \Phi$ is homomorphism.

So, assume $\displaystyle N \subseteq ker \Phi$ and let $\displaystyle g_1N=g_2N$ (we want to prove that this implies $\displaystyle \phi(g_1N)=\phi(g_2N)$). Then $\displaystyle g_1$ and $\displaystyle g_2$ are in the same coset of N and this coset is contained in one coset of $\displaystyle ker\Phi$, because $\displaystyle gN \subseteq g ker\Phi$ for any $\displaystyle g$. So $\displaystyle g_1$ and $\displaystyle g_2$ must be in the same coset of $\displaystyle ker\Phi$, so $\displaystyle \Phi(g_1)=\Phi(g_2)$ which means that $\displaystyle \phi(g_1N)=\phi(g_2N)$.

And I'm so far. I would like to know if the above checks up ok and how to prove the reverse statement.

Thanks in advance....