## § Examples of fiber products / pullbacks

#### § Fiber products of sets

If we have a set $S$, we can form a category of bundles over $S$. These are pairs $(X, \pi: X \rightarrow S)$. The morphisms between such objects $(X, \pi: X \rightarrow S)$ and $(X', \pi': X' \rightarrow S)$ are arrows $h: X \rightarrow X'$ that make the obvious diagram commute:
X ---h--> X'
\       /
pi     pi'
\   /
v v
S
The product of these objects is given by the fiber product or pullback:
$X \times_S Y \equiv \{ (x, y) \in X \times Y: \pi_X(x) = \pi_Y(y) \}$
along with the map $\pi_{X \times Y}: X \times_Y \rightarrow S; \pi((x, y)) \equiv \pi_X(x)$. See that for consistenty, we could also have defined this as $\pi((x, y)) \equiv \pi_Y(y)$. Since our condition is that $\pi_X(x) = \pi_Y(y)$, it all works out. Said differently, we consider the product of fibers over the same base-point.

#### § Fiber products of arbitrary bundle over a single-point base space

If $S \equiv \{ * \}$, then the projections are always $\pi(-) \equiv *$, and the fiber product is the usual product.

#### § Fiber products of singleton bundle over arbitrary base space

If $S$ is arbitrary while $P \equiv \{ p \}$ (for Point), then this bundle $P$ will lie over some point in $S$, given by $\pi: P \rightarrow S$, where the special point is chosen by $\pi(p) = s_p$. If we now consider some other bundle $X$ over $S$, Then $P \times_S X$ will pick the element $(p \in P, \pi_X^{-1}(s_p) \subseteq X)$. That is $P \times_S X \simeq \pi_X^{-1}(s_*)$, which is the fibre of $X$ over the special point $s_p = \pi(p)$. This explains the name.

#### § Fiber products of vector bundles

Consider a fiber bundle $E \xrightarrow{\pi} B$. Now consider a new base space $B'$ with a map $f: B' \rightarrow B$. So we have the data:
E
pi|
v
B'-f-> B
Given this, we would like to pullback the bundle $E$ along $f$ to get a new bundle over $B'$.This is defined by:
$E'_f \equiv \{ (b', e) : f(b') = \pi(e) \} \subseteq B' \times E$
This is equipped with the subspace topology. We have the projection map $pi': E'_f \rightarrow B$, $\pi'((b', e)) \equiv b'$. The projection into the second factor gives a map $h: E'_f \rightarrow E$, $h((b', e)) \equiv e$. This makes the obvious diagram commute:
E' -h-> E
|pi'    |pi
B' -f-> B
Any section $\sigma: B \rightarrow E$ of $E$ induces a section of $E'$ $\sigma': B' \rightarrow E'$, by producing the function (given as a relation):
\begin{aligned} \sigma': B' \rightarrow E' \\ \sigma(b') \equiv (b', \sigma(f(b')) \in_? E' \simeq B' \times E \end{aligned}
This has codomain $E'$. To check, if $(b', \sigma(f(b'))$ is in $E'$, we need $f(b') = \pi(\sigma(f(b'))$. But this is true since $\sigma$ is a section, and thus $\pi(\sigma(f(b')) = f(b')$. Moreover, we need to check that $\sigma'$ is indeed a section of $B'$. For this, we need to check that $\pi'(\sigma'(b')) = b'$. Chasing definitions, we find that this is:
\begin{aligned} &\pi'(\sigma'(b')) &= \pi'(b', \sigma(f(b'))) &= b' \end{aligned}
Hence we are done, we have indeed produced a legitimate section.

#### § Fiber products of Spec of affine scheme

Let $R, A, B$ be rings. consider $A \otimes_R B$. What is $Spec(A \otimes_R B)$, in terms of $Spec(A)$, $Spec(B)$, and whatever data you like about $R$? (Say I give you both $R$ and $Spec(R)$). The answer is that apparently, it's exactly $Spec(A) \times_{Spec(R)} Spec(B)$.