In what sense is the kernel of a group homomorphism a universal arrow in a comma category?

Question

Let $K$ denote the kernel of the group homomorphism $f : G \rightarrow H$. Let $i : K \rightarrow G$ denote the inclusion homomorphism.

Then evidently $(K, i)$ is a universal arrow in the comma category $(S \downarrow G)$ for some functor $S$.

Question: How so? What is this $S$?

Attempt: I know that if $T$ is another group and $j$ another mapping that goes from $T$ to $G$ with the property that $j(T) \in K$, then there exists a unique mapping $j' : T \rightarrow K$ s.t. $i \circ j' = j$. Namely, $j'$ is just $j$ with codomain narrowed to $K$.

Answer 1

The inclusion of a kernel is simply a universal element of the contravariant functor $F:=\mathrm{Hom}_{f,0}(-,G)$ sending a group $L$ to the set of homomorphisms $L\to G$ vanishing upon composition with $f$. It looks to me like this is what MacLane had in mind. This can be viewed as an initial object of the comma category $\mathbf{Gp}^{\mathrm{op}}/F$ (EDIT:) that is, the category of natural transformations from representables into $F$. This is isomorphic, using Yoneda, to the category $*\downarrow F$ of elements of $F$, and since $F$ is a subfunctor of the representable on $G$, we see that $*\downarrow F$ is the corresponding subcategory of the slice category $*\downarrow yG=\mathbf{Gp}\downarrow G$, which are two other descriptions of the appropriate category given on this page.

Answer 2

The arrow $i$ is universal among all arrows $j$ into $G$ which equalize $f$ and $0 : G \to H$. So it is universal in the subcategory of the slice over $G$ on arrows that equalize $f$ and $0$. The answer seems to be that $i$ is universal in a full subcategory of the slice over $G$ (which is a comma category of course), but not in a straight comma category, unless there's some trick to present a full subcategory of a slice as a comma.