I don't understand why this statement requires N to be subgroup of kernel $(N \leq \ker\Phi)$, not just kernel itself $(N = \ker\Phi)$
"φ is well defined on G/N if and only if N ≤ ker Φ"
where N≤G, Φ:G→H, φ:G/N→H, probably. (The information given in the textbook is a bit vague that I'm not quite sure that is even right–or simply I'm a fool! So I decided to bring...the entire context around it.)
I think,
If such N were a proper subgroup of $\ker\Phi$ i.e. $\exists g\in \ker\Phi-N$, then $gN\neq N$.
However, for all $g'\in gN$ which is not $g$, $\Phi(g)=1_H\neq\Phi(g')$, hence φ obviously not well-defined.
Context:
The notion here that if $N \leqslant \ker \Phi$ an, say, you have $x, x'$ of the same coset $xN$ then $x' = xn$ for some $n \in N$ and $\Phi(n) = 1_H$ so
$$\Phi(x) = \Phi(x)1_H = \Phi(x)\Phi(n) = \Phi(xn) = \Phi(x')$$