Slice Category is Category

Theorem

Let $\mathbf C$ be a metacategory.

Let $C \in \mathbf C_0$ be a object of $\mathbf C$.

Let $\mathbf C / C$ be the slice category of $\mathbf C$ over $C$.

Then $\mathbf C / C$ is a metacategory.

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

Suppose that $a: f \to g$ and $b: g \to h$ are morphisms in $\mathbf C / C$.

To show that $b \circ a: f \to h$ is a morphism as well, compute:

$\ds \left({b \circ a}\right) \circ f$

$=$

$\ds b \circ \left({a \circ f}\right)$

$\ds $

$=$

$\ds b \circ g$

$a$ is a morphism

$\ds $

$=$

$\ds h$

$b$ is a morphism

Hence $b \circ a: f \to h$ is a morphism.

For $(C2)$, observe that for $a: f \to g$, with objects $f: X \to C$ and $g: Y \to C$, we have:

$\ds a \circ \operatorname{id}_f$

$=$

$\ds a \circ \operatorname{id}_X$

Definition of $\operatorname{id}_f$

$\ds $

$=$

$\ds a$

Since $\mathbf C$ is a metacategory

$\ds $

$=$

$\ds \operatorname{id}_Y \circ a$

$\ds $

$=$

$\ds \operatorname{id}_g \circ a$

Hence $(C2)$ is shown to hold.

Since the composition in $\mathbf C / C$ is inherited from $\mathbf C$, satisfaction of the associative property $(C3)$ is also inherited.

Hence $\mathbf C / C$ is a metacategory.

$\blacksquare$