In $\textbf{Set}$, does every monomorphism fit into *a* or *any* equaliser diagram?

Question

The question is posed as 'a' diagram. In this case, given a mono $f : X \to Y$ and parallel arrows $g, h : Y \to Z$, we would have the freedom to choose $Z = Y$ and $h = \text{id}_Y$. This would make things a lot easier, because then we would have $g \circ f = f$ and we only have to prove that for any set $W$ and any function $w : W \to Y$ satisfying $g \circ w = \text{id}_Y \circ w = w$, there exists a unique $s : W \to X$ such that $f \circ s = w$? Uniqueness of $s$ then readily follows from $f$ being mono.

I haven't proved the existence of such an $s$ yet, and in another post on seemingly the same topic (but in jargon I could not read), they were talking about pullbacks.

So the second question is: can this even be solved without pullbacks, or will they in some way ultimately be involved?

Cheers!

Answer 1

Let $f \colon X \to Y$ be a monomorphism in $\bf Set$.

Does $f$ is an equaliser of any pair of parallel arrows with domain $Y$?

Of course not. For example, if we take $x \in X$, two parallel arrows with domain $Y$ which doesn’t have the same value at $f(x)$ cannot have $f$ as an equaliser.

Does $f$ is an equaliser of some pair of parallel arrows with domain $Y$?

The answer is yes. As is proven here, $f$ is an equaliser of the constant function ${\bf 1} \colon Y \to \{0,1\}$ with value $1$ and the characteristic function $\chi_{\operatorname{im} f} \colon Y \to \{0,1\}$ of the image of $f$. Let's prove it with the definition, as opposed to how it was done in the link.

First, since $\chi_{\operatorname{im} f}(f(x)) = 1$ for all $x \in X$, we have $$ \chi_{\operatorname{im} f} \circ f = {\bf 1} = {\bf 1} \circ f. $$ Next, given any set $W$ and any function $g \colon W \to Y$ satisfying $$ \chi_{\operatorname{im} f} \circ g = {\bf 1} \circ g (= {\bf 1}), $$ we want to show that there is a function $s \colon W \to X$ with $f \circ s = g$. Indeed, the above equality means $\chi_{\operatorname{im} f}(g(w)) = 1$ for all $w \in W$; so, for each $w \in W$, $g(w)$ is in the image of $f$, and then there is a unique $x_w \in X$ such that $f(x_w)=g(w)$ (uniqueness is due to the injectivity of $f$). So, define $s \colon W \to X$ by $w \mapsto x_w$.