Category of Subobjects is Preorder Category

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 $C$.

Then $\map {\mathbf{Sub}_{\mathbf C} } C$ is a preorder category.

By Category of Subobjects is Category, we know $\map {\mathbf{Sub}_{\mathbf C} } C$ is a metacategory.

By definition of preorder category, it suffices to show that if $f, g: m \to m'$ are morphisms with the same domain and codomain, then $f = g$.

The situation is sketched by the following commutative diagram in $\mathbf C$:

$\quad\quad \begin{xy}\xymatrix@+1em{ \Dom m \ar[r]<2pt>^*+{f} \ar[r]<-2pt>_*+{g} \ar[rd]_*+{m} & \Dom {m'} \ar[d]^*+{m'} \\ & C }\end{xy}$

Thus, we see that $m' \circ f = m = m' \circ g$.

Hence $f = g$, and $\map {\mathbf{Sub}_{\mathbf C} } C$ is a preorder category.

$\blacksquare$