Local Membership of Equalizer
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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 $e$ if and only if $f \circ z = g \circ z$:
- $z \in_C e \iff f \circ z = g \circ z$
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 $\circ$.
$\Box$
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.
$\blacksquare$
Sources
- 2010: Steve Awodey: Category Theory (2nd ed.) ... (previous) ... (next): $\S 5.1$: Example $5.3$