Questions: Fibered Categories

Question 1

In the codomain fibration cod: Arr(C) → C, morphisms are commutative squares. Under what condition is a commutative square a cartesian morphism?

• A. When the square is commutative — any commutative square is automatically cartesian in the codomain fibration
• B. When the horizontal arrows of the square are isomorphisms
• C. When the square is a pullback square — satisfying the universal property that any competing morphism factors uniquely through it
• D. When the domain objects of the top and bottom arrows coincide

Answer: (C)

In the codomain fibration, a morphism in Arr(C) is cartesian if and only if it is a pullback square. This is the key example making the abstract definition concrete: the universal property of a cartesian morphism — any morphism in E with the same base morphism in B factors uniquely through it — becomes exactly the universal property of the pullback. Not every commutative square is a pullback (option A is the most common error), and isomorphisms of horizontal arrows (option B) would be a much stronger, and incorrect, condition.

Question 2

What is the precise relationship between a Grothendieck fibration p: E → B and a pseudofunctor F: B^{op} → Cat?

• A. They are the same structure: every fibration is a pseudofunctor and every pseudofunctor is a fibration, with no translation needed
• B. They are equivalent via the Grothendieck construction, which provides a 2-categorical equivalence between fibrations over B and pseudofunctors B^{op} → Cat
• C. A pseudofunctor is a special case of a fibration: every pseudofunctor gives a fibration, but not every fibration comes from a pseudofunctor
• D. A fibration is strictly stronger: fibrations always yield split functorial compositions, which pseudofunctors only satisfy up to isomorphism

Answer: (B)

The Grothendieck construction gives a 2-categorical equivalence between fibrations over B and pseudofunctors B^{op} → Cat. Given a fibration, one extracts a pseudofunctor by b ↦ E_b and f ↦ f* (the pullback functor); given a pseudofunctor, one constructs the total category. They are equivalent — two different but compatible ways to describe 'a family of categories varying over B' — but not identical. A fibration is a single functor p: E → B; a pseudofunctor is a contravariant assignment B → Cat. Conflating them (option A) is the most common misconception in the topic.

Question 3

In the codomain fibration cod: Arr(C) → C, the fiber over an object b ∈ C is the slice category C/b.

• T. True
• F. False

Answer: True

The fiber over b consists of all objects e in Arr(C) with cod(e) = b — that is, all morphisms in C with codomain b — together with all morphisms between them that project to the identity on b. This is exactly the slice category C/b: objects are arrows f: a → b in C, and morphisms from f: a → b to g: a' → b are arrows h: a → a' making the triangle commute. The codomain fibration is the canonical first example for building intuition about fibered categories.

Question 4

A fibered category p: E → B is the same mathematical object as a pseudofunctor B^{op} → Cat, so there is no need to distinguish between them.

• T. True
• F. False

Answer: False

A fibration p: E → B is a single functor satisfying the cartesian lifting property. A pseudofunctor F: B^{op} → Cat is a contravariant assignment of categories and functors to objects and morphisms of B. They are *equivalent* structures via the Grothendieck construction, but they are not the same object. In particular, a fibration does not come with a canonical choice of pullback functors (that requires choosing a cleavage), while a pseudofunctor specifies F(f) = f* explicitly. Most natural fibrations are not split, meaning the induced pseudofunctor satisfies (f ∘ g)* ≅ g* ∘ f* only up to coherent isomorphism, not on the nose.

Question 5

Why is the universal property essential in the definition of a cartesian morphism, rather than simply requiring φ: e' → e to be any morphism in E lying over f: b' → b in B?

Model answer: Without the universal property, there would be no canonical way to relate fibers across morphisms in B — any morphism over f would qualify as a 'lifting,' and different choices would give incompatible fiber relationships. The universal property says φ is the unique best lifting of f above a fixed target e: any other morphism ψ over a factorization of f factors uniquely through φ. This uniqueness makes the pullback functor f*: E_b → E_{b'} well-defined up to unique isomorphism, which is what makes descent theory coherent and the Grothendieck construction work.

If we only required existence of some morphism over f, different choices of liftings would yield non-canonically related fibers, and the pullback functors f* could not be assembled into a well-defined pseudofunctor. The universal property is precisely what characterizes each cartesian lifting uniquely (up to isomorphism), ensuring that two different cleavages give equivalent, not merely isomorphic, total categories.