Morphism Class Equivalence is Equivalence

Theorem
Let $\mathbf C$ be a metacategory.

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

Let $\map {\mathbf {Sub}_{\mathbf C} } C$ be the category of subobjects of $C$.

Let $\sim$ denote morphism class equivalence on the morphisms of $\map {\mathbf {Sub}_{\mathbf C} } C$.

Then $\sim$ is an equivalence.

Proof
Let us check the three conditions for an equivalence in turn.

Reflexivity
To show that $\sim$ is reflexive, note that for $f: m \to m'$:


 * $f \sim f$

holds $m \sim m$ and $m' \sim m'$.

By Equivalence of Subobjects is Equivalence, these conditions are satisfied.

Symmetry
It is to be shown that for $f: m \to n$ and $f': m' \to n'$:


 * $f \sim f'$ implies $f' \sim f$

By definition of $\sim$, this is equivalent to:


 * $m \sim m' \land n \sim n'$ implies $m' \sim m \land n' \sim n$

Again by Equivalence of Subobjects is Equivalence, this is satisfied.

Transitivity
Given $f: m \to n$, $f': m' \to n'$ and $f: m \to n''$, it is to be shown that:


 * $f \sim f' \land f' \sim f$ implies $f \sim f$

By definition of $\sim$, this is equivalent to:


 * $m \sim m' \land n \sim n' \land m' \sim m \land n' \sim n$ implies $m \sim m \land n \sim n$

which also follows from Equivalence of Subobjects is Equivalence.

Thence $\sim$ is proven an equivalence.