Category of Subobjects is Preorder 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 preorder category.

Proof
By Category of Subobjects is Category, we know $\mathbf{Sub}_{\mathbf C} \left({C}\right)$ 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 $\mathbf C$:


 * $\begin{xy}\xymatrix@+1em{

\operatorname{dom} m \ar[r] ^*+{f} \ar[r]<-2pt>_*+{g} \ar[rd]_*+{m} & \operatorname{dom} m' \ar[d]^*+{m'} \\ & C }\end{xy}$

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

Now $m'$ is a subobject, and a fortiori a monomorphism.

Hence $f = g$, and $\mathbf{Sub}_{\mathbf C} \left({C}\right)$ is a preorder category.

Also see

 * Inclusion Relation on Subobjects