Questions: Hom-Functors and Representability

5 questions to test your understanding

Score: 0 / 5
Question 1 Multiple Choice

In the category Grp, the underlying-set functor U: Grp → Set (which sends a group to its underlying set) is representable, with representing object ℤ. What does this mean concretely?

AEvery group can be generated by a single element, just as ℤ is generated by 1
BThere is a natural bijection between group homomorphisms ℤ → G and elements of the underlying set of G, for every group G
CThe integers ℤ are the initial object in Grp, so every group has a unique map from ℤ
DEvery group is isomorphic to a quotient of ℤ, showing ℤ generates all groups
Question 2 Multiple Choice

A functor F: C → Set has a universal element u ∈ F(A). What does universality of u mean, and why does it imply F is representable?

Au is the largest element in F(A) under some ordering, ensuring F is bounded
BFor every object X and every element x ∈ F(X), there is a unique morphism f: A → X such that F(f)(u) = x — every element of F everywhere is uniquely 'generated' from u
Cu is preserved by all natural transformations from F, making it a fixed point
Du corresponds to the identity morphism on A, which generates all endomorphisms of A
Question 3 True / False

If a functor F: C → Set is representable by object A, then the representing object A is unique up to unique isomorphism.

TTrue
FFalse
Question 4 True / False

Nearly every functor F: C → Set is representable, since for any set-valued functor we can usually construct a suitable hom-functor that matches its values.

TTrue
FFalse
Question 5 Short Answer

Why does the Yoneda lemma imply that recognizing a functor as representable is equivalent to finding a universal property, and why is this organizationally powerful in mathematics?

Think about your answer, then reveal below.