Equalizer

Remarks

An equalizer is the limit of the diagram $X \rightrightarrows Y$ formed by $f$ and $g$, so it is unique up to unique isomorphism. It is a monomorphism: if $e \circ a = e \circ b$, then $a$ and $b$ both solve clause 2 for $k = e \circ a$, so $a = b$.

Example: in the category of sets, let $E = \set{x \in X : f(x) = g(x)}$ and let $e$ be the inclusion. Clause 1 holds by the choice of $E$. If $f \circ k = g \circ k$, every $k(z)$ lies in $E$, so $u(z) = k(z)$ works, and it is the only choice since $e$ is injective. In the category of abelian groups the equalizer of $f$ and $g$ is the kernel of $f - g$ with its inclusion.

Non-example: for $f, g \colon \R \to \R$ with $f(x) = x^2$ and $g(x) = x$, the inclusion $\{0\} \to \R$ satisfies clause 1 but not clause 2. The map $k \colon \{*\} \to \R$ with $k(*) = 1$ has $f \circ k = g \circ k$ but does not factor through $\{0\}$.