Equivalent Subobjects have Isomorphic Domains

Theorem
Let $\mathbf C$ be a metacategory.

Let $C$ be an object of $\mathbf C$.

Let $\mathbf{Sub}_{\mathbf C} \left({C}\right)$ be the category of subobjects of $C$.

Let $m, m'$ be equivalent subobjects of $C$.

Then there exists an isomorphism $f: m \to m'$.

Proof
For brevity, let us write $\operatorname{dom} m = B$ and $\operatorname{dom} m' = B'$.

Since $m$ and $m'$ are equivalent subobjects of $C$, we have that:


 * $m \subseteq m'$ and $m' \subseteq m$

where $\subseteq$ denotes the inclusion relation on subobjects.

That is, there exist morphisms $f: m \to m'$ and $f': m' \to m$ of $\mathbf{Sub}_{\mathbf C} \left({C}\right)$.

Hence, we have morphisms:


 * $f \circ f': m' \to m'$
 * $f' \circ f: m \to m$

By Category of Subobjects is Preorder Category, these are the unique such morphisms.

It follows that both are identity morphisms:


 * $f \circ f' = \operatorname{id}_{m'}$
 * $f' \circ f = \operatorname{id}_m$

That is, $f: m \to m'$ is an isomorphism.