Products and sets of maps

Section 5.7 Products and sets of maps

Here's a basic property that relates the product of sets and the set of maps. It's a bit of a tongue-twister, but it's worth it to unpack. Let \(S\text{,}\) \(T\text{,}\) and \(U\) be three sets. Then define a map

\begin{equation*} \phi \colon \Map(S \times T, U) \to \Map(S, \Map(T, U)) \end{equation*}

that carries an element \(h \in \Map(S \times T, U)\) — that is, a map \(h \colon S \times T \to U\) — to the element \(\phi(h) \in \Map(S, \Map(T, U))\) — that is, the map \(\phi(h) \colon S \to \Map(T, U)\) — that carries an element \(s \in S\) to the element \(\phi(h)(s) \in Map(T,U)\) — that is, the map \(\phi(h)(s) \colon T \to U\) — that carries an element \(t \in T\) to the element

\begin{equation*} \phi(h)(s)(t) = h(\angs{s,t}) \in U\text{.} \end{equation*}

Did you catch that? Let's say it differently: we're starting with a map \(h \colon S \times T \to U\text{.}\) We want to get a map \(\phi(h) \colon S \to \Map(T,U)\text{.}\) To describe that, we start with an element \(s \in S\text{,}\) and we want to get a map \(\phi(h)(s) \colon T \to U\text{.}\) To define that, we start with an element \(t \in T\text{,}\) and we want to get an element of \(U\text{;}\) that element is \(h(\angs{s,t})\text{.}\) Some times the map \(\phi(h)(s) \colon T \to U\) is written \(h(\angs{s,-})\text{,}\) where the second position is treated as a blank where we can fill in \(t \in T\text{.}\)

Let's go the other way, and define a map

\begin{equation*} \psi \colon \Map(S, \Map(T, U)) \to \Map(S \times T, U)\text{.} \end{equation*}

For this, we're starting with a map \(k \colon S \to \Map(T, U)\text{,}\) and we want to define a map \(\psi(k) \colon S \times T \to U\text{.}\) This is defined by

\begin{equation*} \psi(k)(\angs{s,t}) \coloneq k(s)(t)\text{.} \end{equation*}

Now if you inspect thes formulas carefully, you'll see that in fact \(\phi \circ \psi = \id\) and \(\psi \circ \phi = \id\text{.}\) In other words, \(\phi\) is a bijection, and \(\psi\) is its inverse. In still other words, the sets \(\Map(S \times T, U)\) and \(\Map(S, \Map(T, U))\) are the “same”, up to relabelling.