I'm reading about a homomorphism diagram that is commute in Dummit's Abstract Algebra page 100. The diagram is as following:
I understand that for $\varphi$ to be well defined, it must be independent of the choice of $g \in G$, thus we must show that if $g \in bN$ then:
$\Phi(g) = \varphi(g) = \varphi(bnN) = \varphi(bN) = \Phi(b)$
But before that, he says that the independence of $g$ is equivalent to requiring that $\Phi$ be trivial on $N$, so that:
$\varphi$ is well defined on $G/N$ if and only if $N \le ker \: \Phi$
How do we prove that statement?
Assume $\phi$ is well-defined. That is, $\Phi(g)=\Phi(b)$ whenever $g\in bN$. In particular, $\Phi(g)=\Phi(1)$, whenever $g\in N$. As $\Phi(1)=1$, this means $\Phi(g)=1$ whenever $g\in N$, i.e., $N\subseteq \ker\Phi$.
Assume $N\subseteq \ker \Phi$. Then for $g,b$ with $g\in b N$ (say, $g=bn$ with $n\in N$) we have $\Phi(g)=\Phi(gn)\Phi(b)\Phi(n)=\Phi(b)1=\Phi(b)$. That is, $\phi$ is well-defined.