§ Fiber products of sets

If we have a set SS, we can form a category of bundles over SS. These are pairs (X,π:X→S)(X, \pi: X \rightarrow S). The morphisms between such objects (X,π:X→S)(X, \pi: X \rightarrow S) and (X′,π′:X′→S)(X', \pi': X' \rightarrow S) are arrows h:X→X′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×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×Y→S;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:P→S\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 (p∈P,πX−1(sp)⊆X)(p \in P, \pi_X^{-1}(s_p) \subseteq X). That is P×SX≃πX−1(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 E→πBE \xrightarrow{\pi} B. Now consider a new base space B′B' with a map f:B′→Bf: B' \rightarrow B. So we have the data: 

       E     pi|       vB'-f-> B

Given this, we would like to pullback the bundle EE along ff to get a new bundle over B′B'.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′:Ef′→Bpi': E'_f \rightarrow B, π′((b′,e))≡b′\pi'((b', e)) \equiv b'. The projection into the second factor gives a map h:Ef′→Eh: E'_f \rightarrow E, h((b′,e))≡eh((b', e)) \equiv e. This makes the obvious diagram commute: 

E' -h-> E|pi'    |piB' -f-> B

Any section σ:B→E\sigma: B \rightarrow E of EE induces a section of E′E' σ′:B′→E′\sigma': B' \rightarrow E', by producing the function (given as a relation):

σ′:B′→E′σ(b′)≡(b′,σ(f(b′))∈?E′≃B′×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 E′E'. To check, if (b′,σ(f(b′))(b', \sigma(f(b')) is in E′E', 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 B′B'. 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

Let R,A,BR, A, B be rings. consider A⊗RBA \otimes_R B. What is Spec(A⊗RB)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).