Pullbacks

A pullback is a finite limit in Set. It uses the machinery of the fiber product to create a commutative structure.

In the world of Spivak, where Ologs reign, one sees the concept of a pullback used to define new concepts. For example: “a cell phone that has a bad battery is defined as a cell phone that has a battery which remains charged for less than one hour.” The notions of a cell phone having a battery and the notions of a battery lasting a duration of time in some small interval are joined using a fiber product to give rise to a set isomorphic to entities which have a battery which also performs poorly. This build-up of a new concept actually can rely on pasting pullbacks.

Pullback

First, we need the machinery of the fiber product.

Fiber product

$$let f := (x) => z let g := (y) => z X fiber_Z Y is { (x, z, y) | f(x) = z = g(y) }$$

Intuitively, the fiber product is all fibers which connect X to Y when strung through Z given the two functions f and g.

The pullback of X and Y over Z is any set W for which we have an isomorphism between W and the fiber product X fiber_Z Y.

The commutative structure arises from projecting the pullback onto X and Y and then taking the respective projections to Z using the functions f (from X) and g (from Y). The projections ensure the pullback gives us elements of X and Y which can go to Z. Since the pullback is isomorphic to X fiber_Z Y, we know the resulting diagram would actually commute. The fiber product, as the name implies, strings X and Y together and ensures we only travel along fibers.

Universal property for pullbacks

The dotted arrow indicates the unique function which can be created to make everything commute if the rest of the diagram (before adding projections) already commutes.

Fun Facts

• If one of X or Y happens to be the empty set, then:

$$X fiber_Z Y is the empty set.$$

Intuitively, any fiber will fail to have an end, so it can't be a fiber...

• If X and Y are arbitrary sets and Z is a set with exactly one element, then:

$$X fiber_Z Y is isomorphic to X x Y$$

Intuitively, any fiber must route through Z, which is like routing directly from X to Y.

Pasting

If one extends a pullback diagram with further pullbacks, ambiguity does not exist with respect to small and large rectangles: if a small commutative rectangle is a pullback, so is the larger, and vice versa.

An isomorphism exists between left-hand small rectangles and their enclosing rectangles. By "induction" this ambiguity can be resolved by successively recognizing isomorphisms between enclosing and successfully larger commutative rectangles.

A proof that the left square and "larger" rectangle have isomorphisms between their pullbacks is shown, roughly, by recognizing the b in (a, b (b, c, c')) pairs (fiber products) can be ignored.

An example using an Olog (ontological log):