Yes the issue is whether or not $\phi$ is well-defined.

If we assume $\phi$ IS well-defined, this means that $\phi$ is constant on any coset of $N$ (it has to be, or else $\Phi$ is not a function).

In particular, $\phi$ must be constant on the coset of the identity, $N$. That is, for any $n \in N, \phi(N) = \phi(nN) = \Phi(n)$. Since $e \in N$, we must have $\phi(N) = \phi(eN) = \Phi(e)$, which shows that $\Phi(n) = \Phi(e) = e_H$

that is: $N \subseteq \text{ker }\Phi$.

One way to do this without "element-chasing" (using the $g_1,g_2$) is to say $\Phi$ factors through $\pi$, that is, assert there is a homomorphism $\phi$ such that: $\Phi = \phi\pi$. What we have just shown is that this happens iff $\Phi$ annihilates $N$.

The "canonical nature" of $\pi$ is then expressed as: $\pi$ is UNIVERSAL among homomorphisms that annihilate $N$.

(The term "annihilate" is used because the trivial group is a "zero object" for the class of all groups, in the sense that there is always UNIQUE homomorphisms:

$\iota: \{e\} \to G$ <--this says the trivial group is an "initial group"

$\tau: G \to \{e\}$ <--this says the trivial group is a "terminal group"

for any group $G$.

In other words, a trivial group acts much the same as a trivial vector space (consisting just of a 0-vector), which has similar properties with "linear mapping" substituted for "homomorphism").