Goldblatt's "Topoi" shows that in $\mathbf{Set}$ all monic arrows are equalizers, using the following argument:
Let $f : A \rightarrow B$ be injective, let $C = \{ 0, 1 \}$ and define functions $g, h : B \rightarrow C$ as $g(b) = 1$ (so a constant function) and $h(b) = I[b \in \text{im} f]$, where $I$ is the indicator function. Then it's obvious that $g \circ f = h \circ f$.
But I'm not sure I intuitively accept it.
- How do we use injectivity of $f$?
- Shouldn't we require that $f$ is non-surjective? Otherwise $g = h$, and $f$ equalizes them "trivially". And we don't call an arrow an equalizer because it equalizes some other arrow with itself, do we?
Only monomorphisms are pullbacks of the $\top:1\to\Omega$ arrow. Suppose that $m:M\to B$ is a monomorphism classified by $\chi$; if $\chi\circ f:A\to B\to \Omega$ is equal to $\top\circ !_A$, then it's the fact that $m$ is a pullback of $\top$ along $\chi$ that gives us the unique arrow $e:A\to M$ with $m\circ e=f$. And this is precisely what makes $m$ an equalizer of $\chi$ and $\top\circ !_B$.
No, of course not. Why exclude the trivial case when doing so buys us nothing and complicates the definition?