§ Examples of fiber products / pullbacks

§ Fiber products of sets

If we have a set SS, we can form a category of bundles over SS. These are pairs (X,π:XS)(X, \pi: X \rightarrow S). The morphisms between such objects (X,π:XS)(X, \pi: X \rightarrow S) and (X,π:XS)(X', \pi': X' \rightarrow S) are arrows h:XXh: 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×SY{(x,y)X×Y:πX(x)=πY(y)} X \times_S Y \equiv \{ (x, y) \in X \times Y: \pi_X(x) = \pi_Y(y) \}
along with the map πX×Y:X×YS;π((x,y))πX(x)\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 π((x,y))πY(y)\pi((x, y)) \equiv \pi_Y(y). Since our condition is that πX(x)=πY(y)\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{}S \equiv \{ * \}, then the projections are always π()\pi(-) \equiv *, and the fiber product is the usual product.

§ Fiber products of singleton bundle over arbitrary base space

If SS is arbitrary while P{p}P \equiv \{ p \} (for Point), then this bundle PP will lie over some point in SS, given by π:PS\pi: P \rightarrow S, where the special point is chosen by π(p)=sp\pi(p) = s_p. If we now consider some other bundle XX over SS, Then P×SXP \times_S X will pick the element (pP,πX1(sp)X)(p \in P, \pi_X^{-1}(s_p) \subseteq X). That is P×SXπX1(s)P \times_S X \simeq \pi_X^{-1}(s_*), which is the fibre of XX over the special point sp=π(p)s_p = \pi(p). This explains the name.

§ Fiber products of vector bundles

Consider a fiber bundle EπBE \xrightarrow{\pi} B. Now consider a new base space BB' with a map f:BBf: B' \rightarrow B. So we have the data:
       E
     pi|
       v
B'-f-> B 
Given this, we would like to pullback the bundle EE along ff to get a new bundle over BB'.This is defined by:
Ef{(b,e):f(b)=π(e)}B×E 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:EfBpi': E'_f \rightarrow B, π((b,e))b\pi'((b', e)) \equiv b'. The projection into the second factor gives a map h:EfEh: E'_f \rightarrow E, h((b,e))eh((b', e)) \equiv e. This makes the obvious diagram commute:
E' -h-> E
|pi'    |pi
B' -f-> B
Any section σ:BE\sigma: B \rightarrow E of EE induces a section of EE' σ:BE\sigma': B' \rightarrow E', by producing the function (given as a relation):
σ:BEσ(b)(b,σ(f(b))?EB×E \begin{aligned} \sigma': B' \rightarrow E' \\ \sigma(b') \equiv (b', \sigma(f(b')) \in_? E' \simeq B' \times E \end{aligned}
This has codomain EE'. To check, if (b,σ(f(b))(b', \sigma(f(b')) is in EE', we need f(b)=π(σ(f(b))f(b') = \pi(\sigma(f(b')). But this is true since σ\sigma is a section, and thus π(σ(f(b))=f(b)\pi(\sigma(f(b')) = f(b'). Moreover, we need to check that σ\sigma' is indeed a section of BB'. For this, we need to check that π(σ(b))=b\pi'(\sigma'(b')) = b'. Chasing definitions, we find that this is:
π(σ(b))=π(b,σ(f(b)))=b \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 (WIP)

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