The naturality of the canonical projections? The first isomorphism theorem (for groups) states that $G/ \text{Ker} (\varphi) = \varphi(G)$.
This is called the canonical projection, but also sometimes the natural projection.
I wondered if this is because these canonical projections were actually natural transformation between appropriate functors.
 A: Here's what I tried (not a complete answer to my question)
$\require{AMScd}$

Given $G$ and $\varphi: G \to \varphi(G)$. We try to construct a functor $F: \mathbb{C} \to \mathbb{C}$, such that there is a natural transformation $\eta : \text{id} \to F$. Here $\text{id}$ is the identity functor and $\mathbb{C}$ is the category of all groups. To alleviate notation let $N=Ker(\varphi)$, also consider the following two properties a group $H$ may have:
$(1)$ There is at least some morphism $\alpha: H \to G$ such that $\alpha(H) - N \not = \emptyset$, i.e. there is some $h \in H$ such that $\alpha(h) \not \in N$
$(2)$ There is at least some morphism $\beta: G \to H$ such that $\beta(N) \not = 1$
Groups with these properties have some of the structure of their $\eta$-images determined by the naturality condition, and furthermore these are the only groups, such that there is some restriction in their $\eta$-images. To see these, take a group $K$ that does not have $(1)$ nor $(2)$. Then, naturality for a morphism $f: K \to G$ is:
\begin{CD}
H @>{f}>> G \\
@VVV @VVV \\
\eta H @>{\eta f}>> G/N;
\end{CD}
If $f$ is just the trivial homomorphism then this just forces $\eta f$ to be a trivial homomorphism so $\eta H$ is "free", meaning there's no restrictions on what $\eta H$ must be in this commutative square. To see that it is "free" in any commutative commutative square where $H$ is the domain and $G$ is the codomain, by the lack of property $1$, we have that $f(H) - N = \emptyset$, so the above commutative square looks like:
\begin{CD}
h @>{f}>> f(h) \\
@VVV @VVV \\
\eta h @>{\eta f}>> 1;
\end{CD}
Again, this forces $\eta f$ to be the trivial homomorphism so that $\eta H$ is free. The last case, where $H$ is the domain and there is a morphism to any group $K$ is similar. Similarly for when $H$ is the codomain.

This shows that for the natural transformation to exist, I only  need to check it exists for those groups satisfying $(1)$ or $(2)$. However these doubts remain:

*

*Is there actually a well defined $\eta$ on those groups such that the naturality condition holds?

*If yes to the above, is the naturality condition enough to uniquely determine the $\eta$ images of those groups?


Here are some last thoughts:

*

*Interestingly if this $\eta$ actually exists, then every group homomorphism is natural, since for $f: A \to B$ I can first get a natural transformation such that $A \to A/\text{Ker}(f)$ is a component of $\eta$, but by the first isomorphism theorem this is just the same as $A \to B$.

*The sort of construction that I tried to do reminds me of field extensions, I am not well versed in this topic but I think there's more than a vague connection.

