Category of Subobjects is Category

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 $C$.

Then $\mathbf{Sub}_{\mathbf C} \left({C}\right)$ is a metacategory.

Proof
Let us verify the axioms $(C1)$ up to $(C3)$ for a metacategory.

Let $f: m \to m'$ and $g: m' \to m''$ be morphisms of $\mathbf{Sub}_{\mathbf C} \left({C}\right)$.

That $g \circ f: m \to m''$ is again a morphism follows from the following commutative diagram in $\mathbf C$:


 * $\begin{xy}\xymatrix@+2em{

\operatorname{dom} m \ar[r]_*+{f} \ar[rd]_*+{m} \ar@/^1pc/[rr]^*+{g \circ f} & \operatorname{dom} m' \ar[r]_*+{g} \ar[d]^*+{m'} & \operatorname{dom} m'' \ar[ld]^*+{m''}

\\ & C }\end{xy}$

For every subobject $m$ of $C$, the diagram:


 * $\begin{xy}\xymatrix{

\operatorname{dom} m \ar[r]^*+{\operatorname{id}_m} \ar[rd]_*+{m} & \operatorname{dom} m \ar[d]^*+{m} \\ & C }\end{xy}$

in $\mathbf C$ commutes, for $\operatorname{id}_m = \operatorname{id}_{\operatorname{dom} m}$, and the latter is an identity morphism in $\mathbf C$.

This last property also immediately proves the required behaviour of $\operatorname{id}_m$ with respect to other morphisms of $\mathbf{Sub}_{\mathbf C} \left({C}\right)$.

Since the composition is inherited from $\mathbf C$, it is necessarily associative.

Hence $\mathbf{Sub}_{\mathbf C} \left({C}\right)$ is a metacategory.

Also see

 * Category of Subobjects is Preorder Category