Questions: Presheaves

Question 1

In the category Open(X) of open sets of a topological space, with morphisms being inclusions U ↪ V whenever U ⊆ V, a presheaf F assigns data to each open set. When U ⊆ V, what map does F provide between F(U) and F(V)?

• A. A map F(U) → F(V), sending local data on U to global data on V
• B. A map F(V) → F(U), restricting data defined on the larger set V to the smaller set U
• C. A bijection between F(U) and F(V), since they contain the same data up to restriction
• D. No map between them, since U and V are not directly comparable in C^op

Answer: (B)

A presheaf is a contravariant functor F: C^op → Set. In C = Open(X), a morphism U ↪ V (inclusion) becomes a morphism V → U in C^op (morphisms reverse). The functor F sends this to a map F(V) → F(U) — a restriction map from larger sets to smaller sets. This is the direction data actually flows: a function defined on all of V can be restricted to the subset U, but not conversely. The contravariance of the functor exactly captures this restriction direction. Option A gets the direction backwards and would describe a covariant functor (a 'copresheaf').

Question 2

Which statement best describes the relationship between representable presheaves and all presheaves on a category C?

• A. Every presheaf on C is representable — all presheaves arise as Hom(−, A) for some object A ∈ C
• B. No presheaf is representable unless C is a small complete category
• C. The representable presheaves Hom(−, A) correspond to objects of C, but many presheaves exist that encode 'generalized objects' with no representative in C itself
• D. Representable presheaves only exist when the category C has a terminal object

Answer: (C)

The Yoneda embedding shows that every object A ∈ C gives a representable presheaf Hom(−, A), and these are faithfully encoded in [C^op, Set]. But the presheaf category contains far more: non-representable presheaves exist that do not arise from any single object of C. These encode 'generalized objects' — for instance, in algebraic geometry, moduli problems that have no representing scheme still define perfectly good presheaves. Non-representability is not a deficiency but a feature: it means the presheaf category freely extends C with all the 'formal colimits' that C might be missing.

Question 3

The Yoneda embedding y: C → [C^op, Set], which sends each object A to the representable presheaf Hom(−, A), is full and faithful — meaning C embeds as a full subcategory of its presheaf category.

• T. True
• F. False

Answer: True

Fullness means every natural transformation between representable presheaves Hom(−, A) → Hom(−, B) arises from a unique morphism A → B in C. Faithfulness means distinct morphisms in C give distinct natural transformations. Together, they say that the embedding y: C → [C^op, Set] is injective on both objects and morphisms — C sits inside its presheaf category without distortion. This is the Yoneda lemma's payoff: an object is completely determined by the maps into it, so we can replace the object with its 'generalized points' functor without losing information.

Question 4

The presheaf category [C^op, Set] and the functor category [C, Set] are the same category — the notational difference is purely conventional.

• T. True
• F. False

Answer: False

These are genuinely distinct categories. [C^op, Set] consists of contravariant functors from C to Set (equivalently, covariant functors from C^op to Set), while [C, Set] consists of covariant functors from C to Set. The two categories are related by the opposite category construction: [C^op, Set] ≅ [C, Set]^op as categories (roughly). In the geometric setting, presheaves on Open(X) assign data that restricts to smaller open sets; covariant functors would assign data that extends to larger sets (a very different structure). The distinction is not notational — it reflects a fundamental difference in the direction of the data flow.

Question 5

What does it mean to say that the presheaf category [C^op, Set] is the 'free cocompletion' of C, and why is this a useful way to think about presheaves?

Model answer: Free cocompletion means that [C^op, Set] is the universal way to add all small colimits to C. More precisely: given any functor F: C → D where D is cocomplete (has all small colimits), there exists a unique (up to unique natural isomorphism) colimit-preserving extension F̃: [C^op, Set] → D through the Yoneda embedding y: C → [C^op, Set], so that F̃ ∘ y ≅ F. In other words, presheaves are the 'formal colimits' you get by freely adjoining all colimits to C. This is useful because it means any cocomplete target category can receive a unique colimit-preserving functor from the presheaf category, making [C^op, Set] the universal cocomplete extension of C.

The practical implication: if you want to build colimits out of objects of C (formal quotients, pushouts, filtered colimits), you can work in [C^op, Set] where all colimits exist, do your construction there, and then ask whether the result is representable (i.e., lies in the image of C under Yoneda). This is exactly the strategy used in algebraic geometry when constructing moduli spaces and algebraic spaces.