Why pullback in Set is defined through equality $f(a) = g(b)$

Question

There's a recurring example of a pullback in Set category, mentioned at least on Wikipedia, here, and in some book I've read.

It is: take sets $A$, $B$, $Z$, and functions $f: A → Z$ and $g: B → Z$. The pullback $Q = A×_Z B$ then defined to be $\{(a,b) ∈ A×B | f(a) = g(b)\}$.

The problem is that the categorical definition says nothing about equality, so I can take any single pair $(a,b) ∈ A×B$, with injections to $A$ and $B$.

$$ \begin{array}{} \begin{align} & \quad \; \; (a,b) \\ & \; \, \swarrow \quad \quad \searrow \\ & A \quad \quad \quad \; \; B \\ & _f\searrow \quad \quad \swarrow _g \\ & \quad \quad Z \end{align} \end{array} $$

There is a function $Q → (a,b)$, and an injection $(a,b) → Q$, thus they're both satisfying to the universal property, and $(a,b)$ is a pullback too.

Is my reasoning correct, and the recurring example is just a special case of a pullback in Set (and authors not mention it as such just to confuse everyone), or do I miss something important?

Answer 1

The functions between $Q$ and $(a,b)$ aren't isomorphisms in most cases, so actually $(a,b)$ doesn't satisfy the universal property. In other words: you need there to be a unique function from $(a,b)$ to $Q$ making everything commute, which isn't generally true, and a unique function in the other direction making everything commute, but generally your function $Q\to (a,b)$ won't actually do that.

Answer 2

Equality is mentioned in the definition of the pullback, in the key assumption that your diagrams must commute. The definition of a pullback is:

A diagram $A\stackrel{p}\leftarrow C\stackrel{q}\to B$ is a pullback of $A\stackrel{f}\to Z\stackrel{g}\leftarrow B$ if:

1. $fp=gq$ and
2. Given any diagram $A\stackrel{r}\leftarrow D\stackrel{s}\to B$ such that $fr=gs$ there is a unique map $u:D\to C$ such that $pu=r$ and $qu=s$.

In particular, the condition that $fp=gq$ immediately rules out an arbitrary singleton $\{(a,b)\}$ being the pullback in general. Indeed, if you were to take $\{(a,b)\}$ to be the pullback with $p$ and $q$ the projections, then $f(p(a,b))=f(a)$ and $g(q(a,b))=g(b)$, so you need $f(a)=g(b)$ to be true.

So this is where the equality $f(a)=g(b)$ comes from. But you still can't just pick a single pair $(a,b)$ such that $f(a)=g(b)$ is true and conclude that $\{(a,b)\}$ is the pullback, since you still need condition (2) to be true. In particular, if you take $D$ to be an arbitrary singleton $\{(a,b)\}$ such that $f(a)=g(b)$, the existence and uniqueness of the map $u$ tells you that $C$ must contain exactly one point corresponding to $(a,b)$. And in order for $fp=gq$ to be true, every point of $C$ must correspond to a pair $(a,b)$ such that $f(a)=g(b)$. So $C$ must be in bijection with the set of all pairs $(a,b)$ such that $f(a)=g(b)$. (To make this a formal proof, you need to precisely say what "correspond" means: an element $c\in C$ "corresponds" to $(a,b)\in A\times B$ if $p(c)=a$ and $q(c)=b$.)