Proposition 1   If $\mathcal{C}$ has a terminal object $T$ and finite products, then $$A\times_T B = A\times B.$$

Proof. Since $T$ is terminal, we have the following commutative diagram

as the morphism $A\times B\to T$ is unique. Now, if $h_A:C\to A$ and $h_B:C\to B$ are morphisms such that $f\circ h_A = g \circ h_B$ , then there is a unique morphism $h:C\to A\times B$ such that $p_A\circ h=h_A$ and $p_B\circ h = h_B$ by the universal property of categorical direct products. This shows that $(A\times B,p_A,p_B)$ is the pullback of $f$ and $g$ .

Corollary 1   If $\mathcal{C}$ has a terminal object and pullbacks, then $\mathcal{C}$ has finite products.

Proof. A terminal object $T$ serves as empty product in $\mathcal{C}$ . Use induction, suppose now the product $C_1\times \cdots \times C_n$ exists, then the binary product $(C_1\times \cdots \times C_n)\times C_{n+1}$ exists by the proposition above, which is isomorphic to $C_1 \times \cdots \times C_{n+1}$ .

Proposition 2   If a category has finite products and equalizers, then it has pullbacks.

Proof. Let $f:A\to C$ and $g:B\to C$ be morphisms, and let $h:D\to A\times B$ be the equalizer of $f\circ \pi_A : A\times B\to C$ and $g\circ \pi_B: A\times B\to C$ :

We want to show that $p:=\pi_A\circ h: D\to A$ and $q:=\pi_B\circ h: D\to B$ is the pullback of $f$ and $g$ . First, we observe that $f\circ p= f\circ (\pi_A\circ h) = (f\circ \pi_A) \circ h = (g\circ \pi_B) = g\circ (\pi_B\circ h) = g\circ q$ . In other words, we have the commutative diagram

Suppose now we have another commutative diagram

By the universality of $A\times B$ , there is a unique morphism $z:E\to A\times B$ such that $\pi_A\circ z = x$ and $\pi_B \circ z = y$ :

Then $(f\circ \pi_A)\circ z = f\circ (\pi_A\circ z) = f\circ x= g\circ y = g\circ (\pi_B\circ z) = (g\circ \pi_B)\circ z$ , so that $z$ equalizes $f\circ \pi_A$ and $g\circ \pi_B$ :

By the universality of the equalizer $h:D\to A\times B$ , there is a unique morphism $s: E\to D$ such that $z = h\circ s$ . Finally $p\circ s = (\pi_A \circ h)\circ s = \pi_A \circ (h\circ s)=\pi_A \circ z = x$ . Similarly, $q\circ s = y$ . As a result, $(p,q)$ is the pullback of $(f,g)$ .