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Suppose we say a monomorphism is regular if it is the equalizer of some pair of parallel arrows.

The reference(s) I found just give the category SET as an example of a category where everyone mono is regular, regrettably, without much explanation. Which leads me to believe this must be a trivial result. However I'm having some trouble proving it, thus I am wondering if there are some really obvious reason to why this is true, that I may have missed.

Cheers

PIandpie
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In fact, in any pretopos every monomorphism is regular (nLab link). However, I think that an explicit construction for the category of sets will be more insightful in this case.

Let $f: X \to Y$ be a monomorphism in $\mathbf{Set}$, so that means that $f$ is injective. Now we define two functions $Y \to \{0,1\}$. The first function $g: Y \to \{0,1\}$ will just be the constant function taking value 1, so $g(y) = 1$ for all $y \in Y$. Then for the second function $h: Y \to \{0,1\}$, we set $h(y) = 1$ if $y$ is in the image of $f$, and $h(y) = 0$ otherwise. So if $f$ is just an inclusion of sets, then $h$ is the characteristic function of $X$ as a subset of $Y$.

Now the equalizer of $g$ and $h$ is calculated as $E = \{y \in Y : h(y) = g(y)\} = \{y \in Y : h(y) = 1\}$ (the equalizer is then the inclusion $E \hookrightarrow Y$). So $E$ is precisely the image of $f$. Thus $f$ induces a bijection $X \to E$ and so we see that $f$ itself is an equalizer of $g$ and $h$.

Mark Kamsma
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  • The equalizer of $g,h$ is the inclusion of $E$ into $Y$. – Paul Frost Oct 16 '19 at 12:25
  • @PaulFrost Yes, of course, I clarified that in an edit now. Thanks! – Mark Kamsma Oct 16 '19 at 12:28
  • @MarkKamsma Thank you, this is indeed helpful. But while we are on the topic, I have to ask, what are the thought process that goes on when constructing $g$ and $h, and choosing {0,1} as the codomain. – PIandpie Oct 16 '19 at 13:20
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    @some1fromhell The first idea is that monos in $\mathbf{Set}$ are just injective functions, and so we may as well think of them as subsets. So then we need to 'encode' a subset using a parallel pair of arrows, and the easiest way to 'encode' a subset using functions is by using the characteristic function. This explains how we get $h$. Then we just need something to pick the element 1 (the element that $h$ uses to indicate which elements should be in the subset), so this is how we arrive at $g$. – Mark Kamsma Oct 16 '19 at 13:35