The forgetful functor U: Grp → Set is represented by ℤ. A student asks why a homomorphism φ: ℤ → G is completely determined by φ(1). What is the correct explanation?
Aℤ is the smallest group, so there are fewer homomorphisms to track than for any other choice of representing group
BEvery integer n equals 1+1+...+1 (or its negatives), so the homomorphism property forces φ(n) = φ(1)^n, making φ(1) the only free choice
CHomomorphisms from ℤ are always injective, so specifying any element of G determines the rest by injectivity
Dℤ is abelian, and homomorphisms from abelian groups are always determined by a single generator in any target group
ℤ is the free group on one generator (the integer 1). The homomorphism property requires φ(m + n) = φ(m)φ(n), so φ(n) = φ(1+1+...+1) = φ(1)^n for positive n, and the rule extends to negatives via φ(-n) = φ(n)⁻¹. This means once you choose where 1 maps, the values at all other integers are forced. Conversely, any element g ∈ G defines a valid homomorphism by φ(1) = g (no relations in ℤ can fail in G). So Hom_Grp(ℤ, G) ≅ G as sets, naturally in G — this is exactly representability. The key is that ℤ imposes no relations that might obstruct extension to a target group.
Question 2 Multiple Choice
You find a bijection α_B: F(B) ≅ Hom(A, B) for one specific object B in a category C. Is this sufficient to conclude that F is representable by A?
AYes — a bijection at any single object establishes representability, since functoriality will propagate the isomorphism
BNo — representability requires a natural isomorphism: the bijection must commute with all morphisms f: B → C in C, not just exist at one object
CYes — once the bijection holds at B, it holds at all objects isomorphic to B, which covers the important cases
DNo — you need bijections at two objects to confirm the pattern, then naturality follows automatically
Representability requires a *natural* isomorphism, not just a pointwise bijection. Naturality means: for every morphism f: B → C, the bijection α commutes with the induced maps — F(f) on the F-side must correspond to post-composition with f on the Hom-side. A bijection at a single object B might be completely ad hoc, failing to respect morphisms from B to other objects. Checking naturality is what distinguishes a genuine representation (the bijection 'works with the category structure') from an accidental set-level coincidence at one point.
Question 3 True / False
Nearly every functor F: C → Set is representable, since for any functor we can usually construct a representing object by taking a colimit.
TTrue
FFalse
Answer: False
Representability is a non-trivial condition that most functors do not satisfy. A functor F: C → Set is representable only if there exists an object A and a *natural* isomorphism F ≅ Hom(A, -). There is no general construction that produces a representing object for an arbitrary functor. The Yoneda lemma characterizes representability precisely: F is representable iff there is a universal element u ∈ F(A) such that every element of every F(B) is of the form F(f)(u) for a unique f: A → B. This is a strong condition that many functors fail. For example, a functor that sends all objects to a fixed set with more than one element may not be representable.
Question 4 True / False
If a functor F: C → Set is representable by both A and A', then A and A' must be isomorphic in C (though not necessarily equal as sets or constructions).
TTrue
FFalse
Answer: True
The representing object is unique up to unique isomorphism — a fundamental consequence of universal properties. If both A and A' represent F, then Hom(A, -) ≅ F ≅ Hom(A', -) naturally. By the Yoneda lemma, natural transformations between hom-functors correspond bijectively to morphisms in C: a natural isomorphism Hom(A, -) ≅ Hom(A', -) corresponds to a unique isomorphism A ≅ A' in C. This uniqueness-up-to-unique-isomorphism is a hallmark of categorical universal constructions: the representing object is determined by its universal property, not by any particular set-theoretic construction.
Question 5 Short Answer
In the representability setup, why is the element u = α_A(id_A) ∈ F(A) called a 'universal element,' and how does it generate all other elements of the functor F?
Think about your answer, then reveal below.
Model answer: The element u is universal because every element of F(B), for any object B, can be obtained from u by applying F to some morphism. Specifically, the natural bijection α_B: Hom(A, B) → F(B) sends each morphism f: A → B to F(f)(u) ∈ F(B). This means every element of every F(B) is the image of u under F(f) for a unique morphism f: A → B. The identity id_A maps to u itself under α_A. In this sense u is a single 'seed' element that generates the entire functor via the morphisms of C — which is why finding a universal element in F(A) is the standard way to prove a functor is representable and identify its representing object.
The universal element perspective reframes representability: instead of F being a family of sets (one per object) with a complicated functorial structure, everything is encoded in a single element u of a single set F(A). The morphisms of C 'transport' u to all other elements. This is analogous to how a group homomorphism from a free group is determined by one generator's image — the universal element plays the same role for functors that the generator plays for free groups. The Yoneda lemma formalizes this by showing that natural transformations Hom(A, -) → F correspond exactly to elements of F(A), with u corresponding to the identity natural transformation.