Product Category is Product in Category of Categories

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Theorem

Let $\mathbf{Cat}$ be the category of categories.

Let $\mathbf C$ and $\mathbf D$ be small categories, and let $\mathbf C \times \mathbf D$ be their product category.


Then $\mathbf C \times \mathbf D$ is a binary product of $\mathbf C$ and $\mathbf D$ in $\mathbf{Cat}$.


Proof

Let $F: \mathbf A \to \mathbf C$ and $G: \mathbf A \to \mathbf D$ be morphisms in $\mathbf{Cat}$, i.e. functors.

Suppose $X: \mathbf A \to \mathbf C \times \mathbf D$ is a functor such that $\textrm{pr}_1 \circ X = F$ and $\textrm{pr}_2 \circ X = G$:

$\begin{xy}\[email protected][email protected][email protected]+3mu{ & \mathbf A \ar[dl]_*{F} \ar[d]^*{X} \ar[dr]^*{G} \\ \mathbf C & \mathbf C \times \mathbf D \ar[l]^*{\operatorname{pr}_1} \ar[r]_*{\operatorname{pr}_2} & \mathbf D }\end{xy}$

Here, $\operatorname{pr}_1$ and $\operatorname{pr}_2$ are projection functors.


Thus, for all objects $A$ of $\mathbf A$, it is required that:

$\operatorname{pr}_1 \left({X A}\right) = F A$
$\operatorname{pr}_2 \left({X A}\right) = G A$

meaning that $X A = \left({F A, G A}\right)$ by definition of the objects of the product category $\mathbf C \times \mathbf D$.

Similarly, for all morphisms $f$ of $\mathbf A$, it is required that:

$\operatorname{pr}_1 \left({X f}\right) = F f$
$\operatorname{pr}_2 \left({X f}\right) = G f$

meaning that $X f = \left({F f, G f}\right)$ by definition of the morphisms of the product category $\mathbf C \times \mathbf D$.


This uniquely determines the functor $X$, if it exists.

That this definition actually yields a functor is shown on Functor to Product Category.


Hence $\mathbf C \times \mathbf D$ is a binary product of $\mathbf C$ and $\mathbf D$ in $\mathbf{Cat}$.

$\blacksquare$


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