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The distributive and exponential laws of sets are natural isomorphisms

Example

The familiar distributive and exponential bijections of sets commute with functions in every variable, so they are natural isomorphisms.

Facts & Assumptions

Given: Sets $A, B, C$ and functions between such triples.

[L1]

Cartesian products, function sets, binary unions, singleton tags, ordered pairs, and the natural numbers $0, 1$ are available (The Cartesian product $A \times B : = {z \in P (P (A \cup B)) : \exists a \in A \exists b \in B z = (a, b)}$ , The set $B^{A}$ of all functions $A \to B$ , The union $⋃ x$ of a set, and the binary union $a \cup b : = ⋃ {a, b}$ , The unordered pair ${x, y}$ and the singleton ${x} = {x, x}$ , The Kuratowski ordered pair $(a, b) : = {{a}, {a, b}}$ , The natural numbers $N$ (von Neumann)).

[L2]

A function with a two-sided inverse is a bijection, and natural isomorphisms may be formed between functors on product categories ( $f : A \to B$ is a bijection if and only if there is a function $g : B \to A$ with $g \circ f = Δ_{A}$ and $f \circ g = Δ_{B}$ ; such a $g$ is unique, equals the inverse relation $f^{- 1}$ , and is itself a bijection, Natural isomorphism, Product category and its projection functors, Sets and functions form the large locally small category $S e t$ ).

Verification

technique · direct

1.1

Use the tagged union $B ⨿ C = (B \times {0}) \cup (C \times {1})$ . Define $A \times (B ⨿ C) \to (A \times B) ⨿ (A \times C)$ by $(a, (b, 0)) \mapsto ((a, b), 0)$ and $(a, (c, 1)) \mapsto ((a, c), 1)$ .

1.2

Define $A^{B ⨿ C} \to A^{B} \times A^{C}$ by restricting a function to the two tagged summands, and define $(A \times B)^{C} \to A^{C} \times B^{C}$ by composing with the two projections.

2.1

Untagging in step 1.1, joining two functions on disjoint tagged summands, and pairing two functions pointwise are respective two-sided inverses. Hence all three displayed maps are bijections.

step 1.1step 1.2L1L2

2.2

Applying functions to the named entries before or after any map in steps 1.1 and 1.2 produces the same tuple or function value. Precomposition behaves the same way in each exponent variable. Thus every naturality square commutes in all covariant and contravariant variables.

step 1.1step 1.2

3.1

The three componentwise bijections are natural by step 2.2, and their inverses are automatically natural. They therefore give the distributive and exponential natural isomorphisms.

step 2.1step 2.2L2∎

Depends on

Used by

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Sources

Saunders Mac Lane, Categories for the Working Mathematician, Chapter II (standard reference, not scraped)