Questions: Yoneda Embedding and Full Faithfulness

5 questions to test your understanding

Score: 0 / 5
Question 1 Multiple Choice

A student argues: 'Since the Yoneda embedding is fully faithful, every presheaf F: C^op → Set must be naturally isomorphic to Hom(−, X) for some object X in C.' What is wrong with this reasoning?

AFull faithfulness does imply that every presheaf is representable — the student is correct
BFull faithfulness means the embedding preserves and reflects morphisms between representables, but says nothing about presheaves outside the image of Y — many presheaves are non-representable
CThe error is that Hom(−, X) is a covariant functor, but presheaves are required to be contravariant
DFull faithfulness only holds when C is a small category; the claim fails in general
Question 2 Multiple Choice

The Yoneda embedding proves that if Hom(−, X) ≅ Hom(−, Y) as functors, then X ≅ Y in C. Which property of the embedding does this follow from?

AFull faithfulness — the embedding is bijective on hom-sets, so isomorphic hom-functors force the representing objects to be isomorphic
BEssential surjectivity — every presheaf is representable by a unique object, so isomorphic functors must have the same representing object
CCocompleteness of the presheaf category — colimits force the objects to coincide
DThe fact that Hom(−, X) is always a sheaf, making representable functors uniquely determined
Question 3 True / False

The Yoneda embedding Y: C → [C^op, Set] is an equivalence of categories — nearly every presheaf is naturally isomorphic to Hom(−, X) for some X.

TTrue
FFalse
Question 4 True / False

Two objects X and Y in a category C are isomorphic if and only if their representable functors Hom(−, X) and Hom(−, Y) are naturally isomorphic.

TTrue
FFalse
Question 5 Short Answer

What does it mean for the Yoneda embedding to be 'fully faithful,' and what philosophical conclusion does this imply about how objects are determined by their relationships?

Think about your answer, then reveal below.