I found a proposition and a proof in Categorical Closure Operators by Gabriele Castellini saying that:
Let $E \xrightarrow{e} X$, and $f, g : X \rightarrow Y$ be morphisms
such that $e$ is an equalizer of $f$ and $g$. Then $d$ is an
isomorphism if and only if $f=g$.
I understand the $\Rightarrow$ direction, but I'm not sure about the other way, showing if $f=g$, then $e$ is an isomorphism.
The proof in the book says the following:
If $f=g$, then $id_X$ satisfies $f \circ id_X = g \circ id_x$.
Consequently, by definition of equalizer, there is a morphism $X
\xrightarrow{t} E$ such that $ e \circ t = id_x$. Hence, e is a
monomorphism and a retraction and so an isomorphism.
Specifically, why is there a morphism $X \xrightarrow{t} E$ such that $ e \circ t = id_x$? I don't think that's a part of the definition of being an equalizer. Any guidance on how to understand this part?
Edit:
The definition of equalizer that I am using is the following:
An equalizer of a parallel pair of maps $f, g \rightrightarrows B$ in a cat $\mathbb{C}$ consists of an object E and an arrow $e: E \rightarrow A $ such that (1) $f \circ e = g \circ e$ (2) (the universal mapping property part) For all $h : X \rightarrow A$ with $f \circ h = g \circ h$, there exists a unique $m: X \rightarrow E$ with $e \circ m = h $
Best Answer
Note that the map $h: X \to X$ given by $1_X$ satisfies equation $(1)$:
$$ \begin{align} f \circ h &= f & \text{by identity axiom}\\ &= g &\text{by hypothesis }\\ &= g \circ h. & \text{by identity axiom} \tag{1} \end{align} $$ As $f \circ h = g \circ h$, it is then immediate by the universal mapping property that there is some (unique) $t:X \to E$ such that $e \circ t = h$. By definition of $h$ then $e \circ t = 1_X$. As required.