The naturality square for η: F ⇒ G at a morphism f: A → B requires which condition to hold?
Aη_A ∘ F(f) = G(f) ∘ η_B
Bη_B ∘ F(f) = G(f) ∘ η_A
CF(f) ∘ η_A = η_A ∘ G(f)
DG(f) = F(f) for all f
The naturality condition is η_B ∘ F(f) = G(f) ∘ η_A. Reading left to right: going from F(A) to F(B) via F(f) and then applying η_B must equal applying η_A first (from F(A) to G(A)) and then going from G(A) to G(B) via G(f). The square commutes when both paths — 'apply η after F' and 'apply G before η' — give the same composite morphism.
Question 2 True / False
Any family of morphisms {φ_A: F(A) → G(A)} indexed by objects of C is automatically a natural transformation as long as each φ_A has the correct type.
TTrue
FFalse
Answer: False
Having the correct type signature is necessary but not sufficient. The naturality condition — that η_B ∘ F(f) = G(f) ∘ η_A for every morphism f: A → B — must also be verified. Many 'obvious' component-wise maps fail this coherence check. For example, the isomorphism V ≅ V* (dual vector space) that requires choosing a basis is not natural: applying it to a linear map T and then dualizing does not commute with dualizing first and then applying T.
Question 3 Short Answer
The double dual embedding V → V** is considered 'natural' while the isomorphism V ≅ V* is not, even though both are isomorphisms of vector spaces of the same dimension. What makes the difference?
Think about your answer, then reveal below.
Model answer: The map V → V** (sending v to the evaluation functional ev_v: φ ↦ φ(v)) is defined without choosing a basis and commutes with every linear map T: the naturality square holds for all T. In contrast, V ≅ V* requires picking an inner product or a basis to define the isomorphism, and the resulting map changes when the basis changes — it does not commute uniformly with all linear maps.
Naturality formalizes 'canonical' or 'basis-free': a natural transformation is the same construction applied uniformly across all objects, independent of arbitrary choices. The V → V** map needs nothing extra because evaluation is intrinsically defined. The V ≅ V* isomorphism is not intrinsic — it depends on external data (a bilinear form or basis), so it cannot satisfy the naturality square for all morphisms in a uniform way.