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.
Proof
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 $m' \circ f = m = m' \circ g$.
Now $m'$ is a subobject, and a fortiori a monomorphism.
Hence $f = g$, and $\map {\mathbf{Sub}_{\mathbf C} } C$ is a preorder category.
$\blacksquare$
Also see
Sources
- 2010: Steve Awodey: Category Theory (2nd ed.) ... (previous) ... (next): $\S 5.1$