Functions and product bundles

We prove the simple, standard bijection between functions $f : A \to B$ and sections of the projection $π_{A} : A \times B \to A$ . We write $π_{A}$ (or simply $π$ ) for the first projection and $π_{B}$ for the second projection.

1. From a function to a section

Given any function $f : A \to B$ , define $s_{f} : A ⟶ A \times B, s_{f} (a) := (a, f (a)) .$ For every $a \in A$ we have $π_{A} (s_{f} (a)) = π_{A} (a, f (a)) = a$ . Hence $π_{A} ˆ s_{f} = {id}_{A}$ , so $s_{f}$ is a section of $π_{A}$ .

2. From a section to a function

Conversely, suppose $s : A \to A \times B$ is a section of $π_{A}$ ; i.e. $π_{A} ˆ s = {id}_{A}$ . Compose $s$ with the second projection to get a function $f_{s} := π_{B} ˆ s : A ⟶ B .$ Thus $s$ determines a function $f_{s}$ .

3. These constructions are inverse

We check the two assignments are inverse to each other.

Start with $f : A \to B$ . Form $s_{f} (a) = (a, f (a))$ and then $f_{s_{f}} = π_{B} ˆ s_{f}$ . For each $a$ , $f_{s_{f}} (a) = π_{B} (s_{f} (a)) = π_{B} (a, f (a)) = f (a),$ so $f_{s_{f}} = f$ .
Start with a section $s$ . Let $f = π_{B} ˆ s$ . By the universal property of the product (or by pairing), the unique map $⟨ {id}_{A}, f ⟩ : A \to A \times B$ satisfies $π_{A} ˆ ⟨ {id}_{A}, f ⟩ = {id}_{A}$ and $π_{B} ˆ ⟨ {id}_{A}, f ⟩ = f$ . But $s$ also satisfies $π_{A} ˆ s = {id}_{A}$ and $π_{B} ˆ s = f$ , so by uniqueness $s = ⟨ {id}_{A}, f ⟩$ . In particular $s_{f} = s$ .

Hence the map $f \mapsto s_{f}$ is a bijection between $Hom (A, B)$ and the set of sections of $π_{A}$ .

4. Conclusion

We have shown canonically and explicitly that every function $A \to B$ corresponds exactly to a section of the trivial product bundle $A \times B \to A$ ; the correspondence is $f ⟷ s_{f} = ⟨ {id}_{A}, f ⟩$ .

(If $(A, B)$ lie in a category with products—e.g. $Set$ , $Top$ , a topos—exactly the same argument holds, with the extra requirement that functions/sections be morphisms in that category.)