PlanetMath: category with arbitrary products and pullbacks is complete

In the parent entry, it is stated that a complete category can be characterized as being a category with arbitrary products and equalizers. In this entry, we show, as a corollary, that every category with arbitrary products and pullbacks is complete. We begin with the following observation:

Proof. Suppose we have a pair of morphisms $f,g: A\to B$ . Given the product $A\times B$ , there are unique morphisms $f', g': A\to A\times B$ with the following commutative diagrams

$\displaystyle \xymatrix@+=1.5cm{ & A \ar[dr]^f \ar[d]_{f'} \ar[dl]_{1_A} & \ ... ...d]_{g'} \ar[dl]_{1_A} & \ A & A\times B \ar[l]^-{\pi_A} \ar[r]_-{\pi_B} & B }$

For the pair $f',g': A\to A\times B$ , let

$\displaystyle \xymatrix@+=2cm{ P\ar[d]_p \ar[r]^q & A\ar[d]^{g'} \ A\ar[r]_-{f'} & A\times B }$

be the pullback diagram, which, after combining with the two small commutative triangles containing the edge $\pi_A$ above, produces the following commutative diagram

$% latex2html id marker 258 $\displaystyle \xymatrix@+=1.5cm{ P\ar[d]_p \ar[r]^q... ...r[r]^-{f'} \ar@/_1ex/[drr]_{1_A} & A\times B \ar[dr]\vert-{\pi_A} & \ & & A }$$

This implies that $p=q$ . This result, together with the pullback diagram combined with the remaining commutative triangles (containing the edge $\pi_B$ )

$% latex2html id marker 260 $\displaystyle \xymatrix@+=1.5cm{ P\ar[d]_p \ar[r]^p... ...\ar[r]^-{f'} \ar@/_1ex/[drr]_{f} & A\times B \ar[dr]\vert-{\pi_B} & \ & & B }$$

we see that $p$ equalizes $f$ and $g$ . Suppose now that $r:R\to A$ also equalizes $f$ and $g$ : $f\circ r = g\circ r$ . Then we get two commutative diagrams

$% latex2html id marker 262 $\displaystyle \xymatrix@+=1.5cm{ R\ar[d]_r \ar[r]^r... ...r[r]^-{f'} \ar@/_1ex/[drr]_{1_A} & A\times B \ar[dr]\vert-{\pi_A} & \ & & A }$$

first of which comes from the equation $f\circ r = g\circ r$ and the second one is obvious. By the universality of the product $A\times B$ , we have the commutative diagram

$\displaystyle \xymatrix@+=2cm{ R\ar[d]_r \ar[r]^r & A\ar[d]^{g'} \ A\ar[r]_-{f'} & A\times B }$

By the universality of the pullback diagram, there is a unique morphism $s: R\to P$ so that $r = p\circ s$ , which implies that $p$ is the equalizer of $f$ and $g$ . $\qedsymbol$