Why Kernels are special case of equalizers in $\textbf{Grp}$?

Question

Here is the part of Tom Leinster's book I am reading:

Let $\theta: G \to H$ be a homomorphism of groups. As in Example 0.8, the homomorphism $\theta$ gives rise to a fork

But I do not understand why showing that the given shape an equalizer amounts to showing that $\bar{f}$ is a homomorphism and how to show this actually. Could someone explain this to me please?

Here is the definition of an equalizer:

Here is Example 0.8 (until I typeset it):

Answer 1

The key obeservation is that the category $\bf Grp$ has a zero object (object that is both initial and terminal), $\{e\}$, the trivial group. The consequence is that for any two groups $G$ and $H$ there is a unique homomorphism $\varepsilon\colon G\to H$ given as a composition $G\to \{e\} \to H$. Of course, this is just the homomorphism $\varepsilon(g) = e_H$, for all $g\in G$.

So, let $f\colon G\to H$ be a homomorphism. Then, $$g\in\ker f\iff f(g) = e_H\iff f(g) = \varepsilon(g)\iff g\in \operatorname{eq}(f,\varepsilon).$$

The same is true for any category with zero object.

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Here we are using set theoretical definitions of kernel and equalizer that works for $\bf Grp$. To make this formal in sense of category theory, you have to show that the sets $$\ker f = \{ g\in G\mid f(g) = e_h\}$$ $$\operatorname{eq}(f_1,f_2) = \{ g\in G\mid f_1(g) = f_2(g) \}$$

equipped with inclusion maps to the group $G$ indeed satisfy the universal property of kernel and equalizer. I can write it down, but I'm sure it's written somewhere in Leinster.

Now, what I've shown is that sets $\ker f$ and $\operatorname{eq}(f,\varepsilon)$ are equal, so the identity map $$\operatorname{id}\colon \ker f\to \operatorname{eq}(f,\varepsilon)$$ is isomorphism in the category of groups that commutes with inclusions, so they both satisfy the same universal properties by abstract nonsense.

Answer 2

This answer is just to get you un-stuck on the language.

But I do not understand why showing that the given shape an equalizer amounts to showing that $f¯$ is a homomorphism

This doesn't quite make sense, but anyway. If $f^-$ is a homomorphism then the analogous diagram to $(5.5)$ will be a valid diagram in the category of groups. The analogue of $(5.4)$ is also a diagram of groups since the inclusion and the trivial map are homomorphisms. That's important: the equaliser in Set isn't necessarily an equaliser in your target category. In this case it is, but that is just "lucky" (this is false in the category of spaces, iirc, so it's worth being careful).

It remains to check that $f^-$ always exists and that it is always unique, which is not so bad. Leinster seems to think you've already done that part.