What is a natural bijection?

Question

Here is the definition of adjoint functors taken from here:

and here is a property about universal arrows:

To me the concept of natural bijection is imprecise. Does it coincide with the notion of natural isomorphism (so a natural transformation that has an inverse)? Bijection because in the associated commutative diagram we get mappings between sets which should be perhaps bijections?

Answer 1

If $F,G:\mathbf{C}\to\mathbf{D}$ are functors, a natural transformation $\alpha:F\to G$ has an inverse if and only if the component $\alpha_A$ is an isomorphism for each $A\in\mathbf{C}$. In $\mathbf{Set}$ the isomorphisms are precisely the bijections. So "natural bijection" just means a natural isomorphism in the case where the target category is $\mathbf{Set}$.

In particular homsets are sets, so natural isomorphisms involving them may be called natural bijections.

Answer 2

It is, as you said, a bijective natural transformation and the naturality condition is one and the same as the defining condition for it to be a natural transformation.

In the following, I will use the conventions that for a category $$:

• $||$ = the objects of $$,
• for $X, X' ∈ ||$, $(X,X')$ = the set of arrows/morphisms $X → X'$,

along with the definitions

• $^+$ is the dual category:
  • $|^+| = ||$,
  • for $X, X' ∈ ||$, $^+(X,X') = (X',X)$,
  • identities in $^+$ are the same as those in $$,
  • compositions in $^+$ are the reverse of those in $$.
• $×$ is the direct product category:
  • $|×| = ||×||$,
  • for $(X,Y),(X',Y') ∈ |×|$, $(×)((X,Y),(X',Y')) = (X,X')×(Y,Y')$.

For there to be an adjunction relation between categories $$ and $$ with functors $$: → ,\quad : → $$ means that you have polymorphic operators $$⋀: (X, Y) → (X, Y),$$ $$⋁: (X, Y) → (X, Y),$$ such that for ($X ∈ ||$, $Y ∈ ||$) $$f: X → Y ⇒ ⋁⋀f = f,$$ $$g: X → Y ⇒ ⋀⋁g = g,$$ thereby making them inverses, such that the following naturality conditions hold $$⋀(l∘f∘k) = l∘⋀f∘k,\quad ⋁(l∘g∘k) = l∘⋁g∘k,$$ where $$k: X' → X,\quad l:Y → Y'.$$

Make the parametrization of the polymorphic operators explicit $⋀_{X,Y}$ and $⋁_{X,Y}$ and there's your natural transformation.

The natural transformations are thus $⋀: → $, $⋁: → $, where the functors $$,: ^+× → $$ are given by $$(X,Y) ∈ |^+×| ↦ (X,Y) ≡ (X,Y),$$ $$(k,l) ∈ (^+×)((X,Y),(X',Y')) ↦ ((k,l): f ∈ (X,Y) ↦ (l∘f∘k) ∈ (X',Y')),$$ and $$(X,Y) ∈ |^+×| ↦ (X,Y) ≡ (X,Y),$$ $$(k,l) ∈ (^+×)((X,Y),(X',Y')) ↦ ((k,l): g ∈ (X,Y) ↦ (l∘g∘k) ∈ (X',Y')).$$ The defining conditions for $⋀$ and $⋁$ to be natural transforms are: $$⋀_{X',Y'}∘(k,l) = (k,l)∘⋀_{X,Y},$$ $$⋁_{X',Y'}∘(k,l) = (k,l)∘⋁_{X,Y},$$ where $$(k,l) ∈ (^+×)((X,Y),(X',Y')) = (X',X) × (Y,Y'),$$ i.e. $k: X' → X$ and $l: Y → Y'$.

When applied to $$f ∈ (X,Y),\quad g ∈ (X, Y),$$ i.e. to $f: X → Y$ in $$ and $g: X → Y$ in $$, this yields the result: $$⋀_{X',Y'} (k,l)(f) = (k,l)\left(⋀_{X,Y} f\right),$$ $$⋁_{X',Y'} (k,l)(g) = (k,l)\left(⋁_{X,Y} g\right),$$ or, upon application of the definitions: $$⋀(l∘f∘k) = l∘⋀f∘k,\quad ⋁(l∘g∘k) = l∘⋁g∘k.$$