Local Membership of Equalizer

Theorem
Let $\mathbf C$ be a metacategory.

Let $e: E \to C$ be the equalizer of $f, g : C \to D$.

Then a variable element $z: Z \to C$ is a local member of $e$ $f \circ z = g \circ z$:


 * $z \in_C e \iff f \circ z = g \circ z$

Proof
Firstly, note that by Equalizer is Monomorphism, local membership of $e$ is defined.

Necessary Condition
Suppose that $z \in_C e$.

By definition of local membership, there is an $h: Z \to E$ such that $z = e \circ h$.

Then since $e$ is the equalizer of $f$ and $g$:


 * $f \circ e = g \circ e$

from which we deduce:


 * $f \circ z = f \circ e \circ h = g \circ e \circ h = g \circ z$

by using metacategory axiom $(\text C 3)$, that is associativity of $\circ$.

Sufficient Condition
Suppose that $f \circ z = g \circ z$.

Since $e$ is the equalizer of $f$ and $g$, we find $h: Z \to E$ such that:


 * $z = e \circ h$

Hence $z \in_C e$, by definition of local membership.