In the Grothendieck construction for a pseudofunctor F: B → Cat, what is a morphism from (b, x) to (b', x') in the total category ∫F?
AA morphism f: b → b' in B such that F(f)(x) = x'
BA pair (f, φ) where f: b → b' in B and φ: F(f)(x) → x' in the fiber F(b')
CA pair (f, φ) where f: b → b' in B and φ: x → F(f)(x') in the fiber F(b)
DA morphism φ: x → x' in some common fiber, together with a proof that b = b'
The morphism structure is the key technical point of the construction. First, f: b → b' in B specifies how the base indices relate. Then F(f): F(b) → F(b') reindexes x from the source fiber to the target fiber, giving F(f)(x) ∈ F(b'). Only THEN does φ: F(f)(x) → x' operate — it is an internal morphism in the TARGET fiber F(b'). The reindexing must happen first: you can't compose a morphism in F(b) with a morphism in F(b') without first transporting across the fiber. Option C reverses this order, which does not give a well-typed morphism.
Question 2 Multiple Choice
For the category of elements of a set-valued functor F: C → Set, which of the following correctly describes the morphisms?
AA morphism (c, x) → (c', x') is any function from F(c) to F(c') that sends x to x'
BA morphism (c, x) → (c', x') is an arrow f: c → c' in C such that F(f)(x) = x', with no additional data
CA morphism (c, x) → (c', x') is a pair consisting of an arrow f: c → c' in C and a separate morphism x → x' in the fiber
DMorphisms only exist between (c, x) and (c, x') when the base objects are equal
In the Set-valued case, each fiber F(c) is a discrete category — its objects are set elements and the only morphisms are identities. So the φ component in the general construction is forced to be an identity: φ = id_{x'}. A morphism (c, x) → (c', x') is just an arrow f: c → c' in C satisfying F(f)(x) = x'. There is no additional data because there are no non-identity morphisms within any fiber. This is the simplest case and a good starting point for building intuition about the general construction.
Question 3 True / False
The total category ∫F of the Grothendieck construction is simply the disjoint union of most of the fiber categories F(b) — its morphisms primarily go between objects in the same fiber.
TTrue
FFalse
Answer: False
This is the most important misconception to avoid. The total category ∫F has morphisms between DIFFERENT fibers: a morphism (b, x) → (b', x') for b ≠ b' consists of a base morphism f: b → b' and a fiber morphism φ: F(f)(x) → x' in F(b'). These cross-fiber morphisms are precisely what encodes the functorial action of F — they connect the fibers in a coherent way. If ∫F were just a disjoint union, it would forget all the information about how F acts on morphisms in B, which is the essential structure. The whole point of the construction is to ASSEMBLE the fibers into a single category with connecting morphisms.
Question 4 True / False
The Grothendieck construction establishes a genuine equivalence between pseudofunctors B → Cat and Grothendieck fibrations over B — the two descriptions are interchangeable representations of the same mathematical structure.
TTrue
FFalse
Answer: True
This equivalence is the central theorem. Given a pseudofunctor F: B → Cat, the construction produces a Grothendieck fibration p: ∫F → B. Conversely, given a fibration p: E → B, one recovers a pseudofunctor by taking fibers F(b) = p⁻¹(b) and using Cartesian lifts to define the reindexing functors F(f). The two directions are inverse up to equivalence, and the equivalence is of 2-categories: pseudonatural transformations between pseudofunctors correspond to morphisms of fibrations (functors preserving Cartesian morphisms). This is why the choice of 'indexed' or 'fibered' perspective is purely one of convenience.
Question 5 Short Answer
What is the key asymmetry in the definition of morphisms in ∫F — why must reindexing happen before the internal fiber morphism, rather than simultaneously or in the reverse order?
Think about your answer, then reveal below.
Model answer: A morphism (b, x) → (b', x') requires combining information from two different fibers: x lives in F(b) and x' lives in F(b'). These are different categories, so you cannot directly state a morphism between x and x' without first transporting one into the other's world. Reindexing via F(f): F(b) → F(b') brings x into F(b') as F(f)(x), after which φ: F(f)(x) → x' is a well-typed morphism within F(b'). Reversing the order would require φ: x → F(f)(x'), which lives in F(b) — then there's no natural way to compose with base morphisms consistently and get associative composition in ∫F.
This asymmetry corresponds precisely to the direction of Cartesian morphisms in the fibration picture: Cartesian lifts go 'upward' from the base and 'across' fibers in a specific direction. The choice of direction (reindex then move, not move then reindex) is what makes ∫F a fibration over B rather than an opfibration. If you use the reverse convention (φ: x → F(f)(x')), you get the Grothendieck construction for an opfibration. Both are valid; the convention determines the variance of the associated pseudofunctor.