Describe the equalizer of $f,1:X\to X$ as explicitly as possible

Question

Exercise 5.2.22 from Leinster asks to describe the equalizer of $f,1:X\to X$ in $\mathbf{Set}$, where $f:X\to X$ is a map, as explicitly as possible.

What level of explicitness is expected? By definition, it is a pair $(S,h)$ where $S$ is a set and $S\to X$ is a map such that $fh=h$, and if $(S'g)$ is another such pair, then there is a unique $\bar g:S'\to S$ such that $h\bar g=g$. How can one make this more explicit?

Answer 1

To give a categorical construction explicitly in a given category is to give in terms of the concrete definition of the objects of the category. For sets, you give the elements, for algebraic objects, you give the elements and the operations; for spaces, the elements and the geometric or topological structure, and so on.

Answer 2

The equalizer of an endofunction $f:X \to X$ and the identity on $X$ is simply the set of fixed points of $f$. More generally, we could call the equalizer of an endomorphism $f$ and the identity in any category the "object of fixed points of $f$".

Answer 3

The equalizer of $s,t:X\to Y$ in $\mathbf{Set}$ is $(E,i)$ where $E=\{x\in X:s(x)=t(x)\}$ and $i:E\to X$ is the inclusion map. This is clearly a fork, and if $(A,g)$ is another fork, then define $\bar g:A\to E, a\mapsto g(a)$. Note that $g(a)$ indeed lives in $E$ because $sg=tg$, and $\bar g$ is the unique map $A\to E$ such that $i\bar g=g$.

Apply this to $Y=X,\ s=1,\ t=f$.