Every normal subgroup is the kernel of some homomorphism Clearly the kernel of a group homomorphism is normal (proof), but I often hear my professor mention that any normal subgroup is the kernel of some homomorphism.
This feels correct but isn't entirely obvious to me.

One thought I had is that for any normal subgroup $N$ of $G$, we could define the quotient homomorphism $\pi:G\to G/N$ since $G/N$ is a group.
I was imagining that we could consider $\pi^{-1}:G/N\to G$, whose kernel would then be $N$. However, $\pi^{-1}$ doesn't exist since $\pi$ is not a bijection in general.

So my question is this: is there an obvious way to define a homomorphism whose kernel is an arbitrary normal subgroup of $G$? Or does it depend on the particular group whether you can define such a homomorphism?
 A: You only have to consider that $\pi:G\to G/N$, defined by $\pi(x)=xN$, is a homomorphism and the $\ker \pi$ is precisely $N$.
A: Anyway, if the kernel is non-trivial, the homomorphism is not injective!
You're confusing the inverse of an isomorphism, and the inverse image of a subset by a map. So, yes, if $N$ is  a normal subgroup, it is the kernel of the canonical homomorphism :
\begin{align*}\pi\colon G&\longrightarrow G/N,\\
g&\longmapsto gN.
\end{align*}
Indeed, $\pi^{-1}(\bar 1)=\pi^{-1}(N)=N$.
A: Your intuition is correct, and unfortunately, yes, $\pi^{-1}$ isn't a function. Could we somehow "make" it one?
Well, one way this is often done with "ordinary" functions, is to treat $f^{-1}$, for a given function $f: A \to B$, as not something like this:
$f^{-1}: B \to A$
but instead considering $f^{-1}$ as a function between power sets:
$f^{-1}: \mathcal{P}(B) \to \mathcal{P}(A)$
where for a subset $Y \subseteq B$, we define $f^{-1}(Y) = X = \{x \in A: f(x) \in Y\}$.
Applying this definition to $\pi^{-1}$, we get:
$\pi^{-1}(e_{G/N}) = \pi^{-1}(N) = \{g \in G: \pi(g) \in N\}$
$= \{g \in G: gN = N\} = \{g \in G: ge^{-1} \in N\} = \{g \in G: g \in N\} = N$ 
