Given sets $A, B, C$ and set functions $f: A \to C$ and $g: B \to C$, everywhere I look seems to be telling me that the pullback is $A \times_C B = \left\{(a, b) \mid f(a) = f(b)\right\}$ (together with the usual projective maps). I do not understand why we are not required in the construction to quotient this set by the relation $(a, b) \sim (a', b')$ if and only if $f(a) = f(a')$ (equivalently, if and only if $g(b) = g'(b)$). Why do I say this? Because if we want $A \times_C B$ to be a pullback, then we want it to admit a unique set function from any other set $Z$ to $A \times_C B$ (of course, we assume $Z$ comes together with $f' : Z \to A$ and $g' : Z \to B$ such that the diagram commutes). One can easily imagine that for a given $x \in Z$ and setting $c_x \equiv f(f'(x)) = g(g'(x)$ that $A_{c_x} \equiv f^{-1}(c_x)$ or $B_{c_x} \equiv g^{-1}(c_x)$ may have more than one element. Then in general, one can define many possible functions from $Z \to A \times_C B$ by sending $x$ to any element of $A_{c_x} \times B_{c_x}$ and the diagram will commute, thus violating the desired property that such a map is unique. Indeed, as far as I can see, the only way to avoid this is to quotient out by the relation presented above.
Thanks for your help.
You say that in general, one can send $x \in Z$ to many different elements in $A \times_C B$. How is that true? Remember that the diagrams involving $Z$, $A \times_C B$ and $A$, as well as $Z$, $A \times_C B$ and $B$ are required to commute. Thus the only option is to send $x$ to $(f'(x),g'(x))$.