A terminal object $T$ in a category $\mathcal C$ is an object such that for every object $X$ there exists a unique morphism $X \to T$.
The pullback of two morphisms $f: X \to Z$ and $g: Y \to Z$ is the unique object $P$ with morphisms $p_1 : P \to X$ and $p_2: P \to Y$ such that for every object $Q$ and morphisms $q_1$, $q_2$ there exists a unique morphism $u: Q \to P$ such that the following diagram commutes:
If $Z$ is terminal then $P = X \times Y$. I think the way to see this is to apply a forgetful functor $F: \mathcal C \to \mathbf{Set}$. Then $Z$ maps to a one element set so that $f,g$ become the constant maps and then $P = \{(x,y) \mid f(x) = g(y) \} = X \times Y$.
Is there a different way to see that if $Z$ is terminal then $P = X \times Y$, not involving knowledge of what terminal objects in $\mathbf{Set}$ look like? Maybe not involving $\mathbf{Set}$ at all?
Best Answer
Here is an easy way to "see it." The pullback is the limit of the diagram $X \stackrel{f}{\rightarrow} Z \stackrel{g}{\leftarrow} Y$. It is unique up to isomorphism. If $Z$ is the terminal object, there is exactly one $f$ and one $g$. So, such diagrams are one-to-one with pairs $(X,Y)$ of objects, or discrete diagrams with just $X$ and $Y$. Since the product is the limit of the discrete diagram, the pullback is a product (which is again unique up to isomorphism).
(By the way, categories in general do not have forgetful functors to Set, let alone the fact that "forgetful functor" is not a formally defined concept.)