Terminal objects and pullbacks 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? 
 A: 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.)
A: (Using the OP drawing/symbols)
You can see it easily this way:
1-A pullback is a terminal object in the category of cones to the diagram $D:X \stackrel{f}{\rightarrow} Z \stackrel{g}{\leftarrow} Y$
2- A product is a terminal object in the category of cones to the diagram $D':X,Y$ (just the 2 objects).
We are going to show that if $Z$ is a terminal object in $C$, then a cone to $D$ is the same as a cone to $D'$.
Indeed: a cone from vertex $Q$ to $D$ consists of:
any 3 arrows $q_1:Q\to X, q_2:Q\to Y, q_3:Q\to Z$, such that:
$f \circ q_1 = q_3 $
$g \circ q_2 = q_3$
This simply means that it consists of
any 2 arrows $q_1$ and $q_2$ such that:
(1)   $f \circ q_1 =g \circ q_2 $
Now if $Z$ is a terminal object, eq. 1 is automatically satisfied by any  $q_1:Q\to X, q_2:Q\to Y$, since the lhs and the rhs of eq. 1 will both be equal to the one and only morphism $Q\to Z$.
So our cone from $Q$ to $D$ reduces to: 
any 2 arrows $q_1:Q\to X, q_2:Q\to Y$.
(and no other condition)
This is the same as a cone from $Q$ to $D'$.
Conclusion: the cone categories to $D$ and to $D'$ are the same, so their terminals are the same and thus the pullback is the same as the product
