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definition of a fiber product

DEFINITION 6.1   Let $ \mathcal{C}$ be a category. Let $ X,Y,Z$ be objects of $ \mathcal{C}$ . Assume that morphisms $ f: X\to Z$ and $ g:Y\to Z$ are given. Then the fiber product $ X\times_Z Y$ (more precisely,

$\displaystyle X\times_{f,Z,g}Y
$

) is defined as an object $ W$ together with morphisms

$\displaystyle p:W \to X, \quad q: W\to Y
$

such that

$\displaystyle p \circ f= q\circ g
$

which is universal in the following sense.

For any $ W_1 \in \operatorname{Ob}(\mathcal{C})$ together with morphisms

$\displaystyle p_1:W_1 \to X, \quad q_1: W_1\to Y
$

such that

$\displaystyle p_1 \circ f= q_1\circ g,
$

there exists a unique morphism $ h: W_1\to W$ such that

$\displaystyle p\circ h=p_1 , q\circ h=q_1
$

holds.

Using the usual universality argument we may easily see that the fiber product is, if exists, unique up to a unique isomorphism.

EXAMPLE 6.2   Fiber products always exists in the category $ (\operatorname{TOP})$ of topological spaces. Namely, let $ X,Y,Z \in \operatorname{Ob}(\operatorname{TOP})$ . Assume that morphisms(=continuous maps) $ f: X\to Z$ and $ g:Y\to Z$ are given. Then we consider the following subset $ S$ of $ X\times Y$ .

    $\displaystyle S=$ $\displaystyle (f,g)^{-1} (\Delta_Z) (\subset (X\times Y))$
    $\displaystyle =$ $\displaystyle \{(x,y)\in X\times Y; f(x)=g(y)\}.$

We equip the set $ S$ with the relative topology. Then $ S$ plays the role of the fiber product $ X\times_Z Y$ . (The morphisms $ p,q$ being the (restriction of) projections.)


next up previous
Next: tensor products of algebras Up: fiber product Previous: fiber product
2007-12-11