Product as Limit

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Theorem

Let $\mathbf C$ be a metacategory.

Let $C_1, C_2$ be objects of $\mathbf C$.

Let their binary product $C_1 \times C_2$ exist in $\mathbf C$.


Then $C_1 \times C_2$ is the limit of the diagram $D: 2 \to \mathbf C$ defined by:

$D_0 := C_1, D_1 := C_2$

where $2$ is the discrete category with two objects $0, 1$.


Proof

Since there are no non-identity morphisms, a cone to $D$ is simply a pair:

$\begin{xy}\[email protected][email protected]+3px{ C_1 & C \ar[l]_*+{f_1} \ar[r]^*+{f_2} & C_2 }\end{xy}$

of morphisms with common domain $C$.

By the UMP of the binary product $C_1 \times C_2$, for such a cone to $D$ there is a unique $u: C \to C_1 \times C_2$ making:

$\begin{xy}\[email protected][email protected]+1em{ & C \ar[dl]_*+{f_1} \[email protected]{-->}[d]^*{u} \ar[dr]^*+{f_2} \\ C_1 & C_1 \times C_2 \ar[l]_*+{p_1} \ar[r]^*+{p_2} & C_2 }\end{xy}$

a commutative diagram.

The conditions $p_1 \circ u = f_1$ and $p_2 \circ u = f_2$ precisely mean that $u: C \to C_1 \times C_2$ is a morphism of cones.


Thus for every cone $\left({C, f_1, f_2}\right)$ to $D$, there is a unique morphism of cones to the cone $\left({C_1 \times C_2, p_1, p_2}\right)$.

That is, $\left({C_1 \times C_2, p_1, p_2}\right)$ is a terminal object in $\mathbf{Cone} \left({D}\right)$, the category of cones to $D$.

Hence the result, by definition of limit.

$\blacksquare$


Sources