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.
Proof
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:
\(\ds \left({b \circ a}\right) \circ f\) | \(=\) | \(\ds b \circ \left({a \circ f}\right)\) | composition in $\mathbf C$ is associative | |||||||||||
\(\ds \) | \(=\) | \(\ds b \circ g\) | $a$ 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.
$\blacksquare$