A student argues: 'Since every group homomorphism is a set function, the forgetful functor U: Grp → Set must be full — every morphism in Grp maps to a morphism in Set, so nothing is lost.' What is wrong with this reasoning?
AThe reasoning is correct; U: Grp → Set is both faithful and full
BFullness would require every set function between underlying sets to be a morphism in Grp — that is, every function between groups to be a homomorphism — which is false; U is faithful but not full
CThe forgetful functor is not well-defined because groups have multiple valid underlying set representations
DFullness and faithfulness are equivalent for forgetful functors, so the distinction is irrelevant in this context
Fullness of U: Grp → Set would mean that for any two groups G, H, every set function from U(G) to U(H) is the image of some group homomorphism G → H. That is obviously false: most set functions between groups fail to preserve the group operation. Faithful means U is injective on each hom-set — distinct homomorphisms in Grp give distinct set functions, so morphisms are 'determined by their action on elements.' The student confused the direction: every morphism maps *to* a set function (that's just functoriality), but not every set function comes *from* a morphism (that would be fullness).
Question 2 Multiple Choice
The homotopy category hoTop — whose objects are topological spaces and whose morphisms are homotopy classes of continuous maps — is not a concrete category. What does this mean precisely?
AThere exists no faithful functor from hoTop to Set
BhoTop is not a legitimate category because homotopy classes are not well-defined morphisms
CThe objects of hoTop do not have underlying sets, so no forgetful functor can be defined
DhoTop is too large to be a category, since there are too many homotopy classes
By Freyd's theorem, no faithful functor hoTop → Set exists. This is a non-trivial result — it means homotopy classes of maps are not 'faithfully represented' by any assignment of elements to spaces. Contrast this with Top: the forgetful functor Top → Set is faithful because continuous maps are genuine set functions, and distinct continuous maps are distinct set functions. In hoTop, two homotopic maps become the same morphism even though they are different set functions — the identification collapses morphism sets in a way that is incompatible with faithfulness to Set.
Question 3 True / False
Faithfulness of a functor U: C → Set ensures that two distinct morphisms in C cannot have the same underlying set function — morphisms in C are completely determined by their action on elements.
TTrue
FFalse
Answer: True
Faithfulness means U is injective on each hom-set: if f ≠ g in C, then U(f) ≠ U(g) as set functions. This captures the intuition that morphisms in a concrete category 'are' structure-preserving functions — they cannot differ in some abstract categorical sense while agreeing on all elements. For the forgetful functor Grp → Set, two distinct homomorphisms must send some element to different images, since if they agree on all elements they are the same homomorphism.
Question 4 True / False
Concreteness is an intrinsic property of a category: if a category's objects 'have underlying sets' and its morphisms 'are functions,' it is automatically concrete without needing to specify any additional structure.
TTrue
FFalse
Answer: False
Concreteness is structure on a category, not a property of the abstract category. A concrete category is a pair (C, U) — the category together with a chosen faithful functor U: C → Set. The same abstract category can be made concrete in genuinely different ways: Grp can be concretized by sending each group to its underlying set (standard), or to its set of subgroups, or to its set of automorphisms. Different concretizations give different notions of 'elements.' Even when a canonical forgetful functor is obvious, it must be specified as part of the structure — concreteness is not automatic.
Question 5 Short Answer
Explain the difference between a faithful functor and a full functor, and give a concrete example showing why the forgetful functor Grp → Set is faithful but not full.
Think about your answer, then reveal below.
Model answer: A functor F: C → D is faithful if it is injective on each hom-set: distinct morphisms in C map to distinct morphisms in D. It is full if it is surjective on each hom-set: every morphism in D between images of C-objects comes from some morphism in C. The forgetful functor U: Grp → Set is faithful because two distinct group homomorphisms f, g: G → H must disagree on at least one element of G — so U(f) ≠ U(g) as set functions. It is not full because there exist set functions between groups that are not homomorphisms: for example, the function f: ℤ → ℤ defined by f(n) = n + 1 is a perfectly valid set function but does not preserve the group operation (f(a + b) = a + b + 1 ≠ f(a) + f(b) = a + b + 2). Such a function is not in the image of U on hom-sets.
The faithful-not-full situation is generic for forgetful functors: morphisms in the category must be structure-preserving functions, but not every function preserves the structure. The forgetful functor 'sees' all the morphisms correctly but also sees many extra set functions that are not morphisms. Fullness would be pathologically strong — it would mean every function is a homomorphism, that every function between topological spaces is continuous, etc.