math 216: foundations of algebraic geometry - Stanford University

28 Math 216: Foundations of Algebraic Geometry
that the “outside rectangle” (involving U, V, Y, and Z) is also a fiber diagram.
U
W
Y
2.3.Q. EXERCISE. Given morphisms X1 → Y, X2 → Y, and Y → Z, show that
there is a natural morphism X1 ×Y X2 → X1 ×Z X2, assuming that both fibered
products exist. (This is trivial once you figure out what it is saying. The point of
this exercise is to see why it is trivial.)
2.3.R. USEFUL EXERCISE: THE MAGIC DIAGRAM. Suppose we are given morphisms
X1, X2 → Y and Y → Z. Show that the following diagram is a fibered
square.
X1 ×Y X2
Y
V
X
Z
X1 ×Z X2
Y ×Z Y
Assume all relevant (fibered) products exist. This diagram is surprisingly useful
— so useful that we will call it the magic diagram.
2.3.7. Coproducts. Define coproduct in a category by reversing all the arrows in
the definition of product. Define fibered coproduct in a category by reversing all
the arrows in the definition of fibered product.
2.3.S. EXERCISE. Show that coproduct for Sets is disjoint union. This is why we
use the notation for disjoint union.
2.3.T. EXERCISE. Suppose A → B and A → C are two ring morphisms, so in
particular B and C are A-modules. Recall (Exercise 2.3.K) that B ⊗A C has a ring
structure. Show that there is a natural morphism B → B ⊗A C given by b ↦→ b ⊗ 1.
(This is not necessarily an inclusion; see Exercise 2.3.G.) Similarly, there is a natural
morphism C → B ⊗A C. Show that this gives a fibered coproduct on rings, i.e. that
B ⊗A C
C
satisfies the universal property of fibered coproduct.
2.3.8. Monomorphisms and epimorphisms.
B
2.3.9. Definition. A morphism f : X → Y is a monomorphism if any two morphisms
g1 : Z → X and g2 : Z → X such that f ◦ g1 = f ◦ g2 must satisfy g1 = g2.
In other words, there is at most one way of filling in the dotted arrow so that the
A