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What does it mean for the adjunction isomorphism to be natural in A?

Naturality in AA means that the adjunction correspondence is compatible with precomposition by any morphism h:A′→Ah:A'\to A. Specifically, taking a morphism A→G(B)A\to G(B), precomposing with hh, and then translating across the adjunction gives the same result as first translating A→G(B)A\to G(B) to F(A)→BF(A)\to B and then precomposing with F(h):F(A′)→F(A)F(h):F(A')\to F(A).

Conditions

  • Fix an object B∈BB\in\mathcal{B}.
  • The statement quantifies over all morphisms h:A′→Ah:A'\to A in A\mathcal{A}.
  • The correspondence is between morphisms in A\mathcal{A} and morphisms in B\mathcal{B}.

Reasoning, step by step

  1. Start from the adjunction bijection B(F(A),B)≅A(A,G(B))\mathcal{B}(F(A),B)\cong\mathcal{A}(A,G(B)).
  2. Introduce an arbitrary morphism h:A′→Ah:A'\to A in A\mathcal{A}.
  3. Compose hh with a morphism A→G(B)A\to G(B) to obtain a morphism A′→G(B)A'\to G(B).
  4. Apply the functor FF to hh, obtaining F(h):F(A′)→F(A)F(h):F(A')\to F(A).
  5. Compose F(h)F(h) with the corresponding morphism F(A)→BF(A)\to B.
  6. Require that these two procedures agree under the adjunction bijection.

Example

The board writes: 'Naturally in A: for any h:A′→Ah:A'\to A, A′→hA→G(B)A'\xrightarrow{h}A\to G(B) in A\mathcal{A} is the same as F(A′)→F(h)F(A)→BF(A')\xrightarrow{F(h)}F(A)\to B in B\mathcal{B}.'

Common misconceptions

  • Thinking that adjunction only requires a set-theoretic bijection for each pair (A,B)(A,B) without respecting composition.
  • Confusing naturality in AA with naturality in BB.

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