A little more detail on my comment.
There is a general notion of ‘pullback’ used informally throughout mathematics to refer to, well, pulling back some structure or other through a given map. Category-theoretically speaking, ‘pulling back through/along $f : A \to B$’ usually corresponds to a map $F f : FB \to FA$ where $F$ is some contravariant functor. Explicitly:
The inverse image map, a.k.a. ‘pulling a subset of $B$ back along $f$’, is the map $f^* : \mathscr{P} B \to \mathscr{P} A$ where $\mathscr{P} : \textbf{Set} \to \textbf{Set}^\textrm{op}$ is the (contravariant) power set functor.
Generalising slightly, pulling back along an arrow $f : A \to B$ in a category $\mathbf{C}$ with all pullbacks induces a functor $f^* : \mathbf{C} / B \to \mathbf{C} / A$, where $\mathbf{C} / A$ denotes the slice category over $A$.
Specialising the above example, if we take $\mathbf{C} = \textbf{Top}$, then we get the notion of pullback bundles. For example, if $TN \to N$ is the tangent bundle over a manifold $N$, and $f : M \to N$ is smooth, then we have a bundle $f^* TN \to M$; the differential of $f$ then becomes a map $df : TM \to f^* TN$ of bundles over $M$.
Finally, consider the functor $\textrm{Hom}(- , Z)$. If we have a map $f : A \to B$, then we get the precomposition map $f^* : \textrm{Hom}(B, Z) \to \textrm{Hom}(A, Z)$.
Returning to the example of bundles, consider the space of global sections $\Gamma(N, T^* N)$ of the cotangent bundle $T^* N \to N$. This is a subset of $\textrm{Hom}(N, T^* N)$, so precomposing with $f$ yields a subset of $\textrm{Hom}(M, T^* N)$. But these factor through $f^* T^*N$, so we get a subspace of $\Gamma (M, f^* T^* N)$. And $f^* T^* N$ is simply the dual bundle of $f^* TN$ so we have a dual map $df^* : f^* T^*N \to T^* M$, so postcomposing with this yields a subspace of $\Gamma(M, T^* M)$. This map $\Gamma(N, T^* N) \to \Gamma(M, T^* M)$ is what geometers mean by ‘pullback of differential forms’, and as you can see, there are lots of contravariant functors involved!
A pullback as just a certain subset of the product: those pairs $(a,b)$ on which the two given maps on coordinates agree, i.e. $f(a)=g(b)$.
This needs an explicit example, which I'll build on the lovely product diagram we already have available. So map $X=\{1,2,3,4,5,6\}\to \{\text{even},\text{odd}\}$ in the obvious way, and $Y=\{\clubsuit,\diamondsuit,\heartsuit,\spadesuit\}$ by saying, for instance, the black suits $\clubsuit,\spadesuit$ are "even" and the others are "odd." Then the pullback becomes a subset of $X\times Y$ given by pairs $(n,s)$ where $n$ and $s$ have the same image in $\{\text{even},\text{odd}\}$. That is, the pullback has the twelve elements $\{(1,\heartsuit),(1,\diamondsuit),(3,\heartsuit),...,(5,\diamondsuit),(2,\spadesuit),(2,\clubsuit),...,(6,\clubsuit)\}$. If this begins to make some sense, see if you can compute
- the pullback of $X$ and $Y$ over the unique maps sending $X$ and $Y$ to a one-element set
- the pullback over the identity map $i:Y\to Y$ and any map $f:X\to Y$.
Best Answer
Just for you, and it turns out my answer has to contain at least 30 characters, so let's make it 100.