$\varphi$ is well defined on $G/N$ if and only if $N\le \ker \Phi$.

These are concluding remarks from Dummit and Foote Abstract Algebra, Ch 3, sec 3:

We have, in the course of the proof of the isomorphism theorems, encountered situations where a homomorphism $$\varphi$$ on the quotient group $$G/N$$ is specified by giving the value of $$\varphi$$ on the coset $$gN$$ in terms of the representative $$g$$ alone. In each instance we then had to prove $$\varphi$$ was well defined, i.e., was independent of the choice of $$g$$. In effect we are defining a homomorphism, $$\Phi$$, on $$G$$ itself by specifying the value of $$\varphi$$ at $$g$$. Then 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 $$N\le \ker \Phi$$.

And then after some more commentary, they include this image:

Although I don't quite understand what is going on in this part of the section, I would like to see a proof of bullet assertion. Here is my attempt:

$$\implies$$: Suppose $$\varphi$$ well defined then, and let $$n\in N$$, then $$\Phi(n)=\varphi\circ\pi(n)=\varphi(nN)=\varphi(N)=e_H$$.

But, I can't show reverse implication. I think that since we have not shown yet that $$\varphi$$ is indeed well defined, using $$\Phi=\varphi\circ\pi$$ will be illegal.

Also, if you can attach other reading or lecture note link, that discusses aforementioned topic in more detail, I will be obliged.

The idea is that if $$N \leqslant \ker \Phi$$ and we have two representatives $$x, x'$$ of the same coset $$xN$$ then $$x' = xn$$ for some $$n \in N$$ and $$\Phi(n) = 1_H$$ so
$$\Phi(x) = \Phi(x)1_H = \Phi(x)\Phi(n) = \Phi(xn) = \Phi(x')$$
Therefore the definition $$\varphi(xN) = \Phi(x)$$ is well defined since the value $$\Phi(x)$$ we get does not depend on the representative we choose.