Posted on 25th June 2023

A long-time-coming comment on a post by John Baez about "bargin-basement mathematics". (But let's not do ourselves down-- the rest of the system is more than happy enough to do that for us-- so I shall say "cheap" not quite "bargain-basement").

The setup here is a (monotone Galois connection), which consists of two posets \( X \) and \( Y \), and order-preserving maps \[ L:X\rightarrow Y, \qquad R:Y\rightarrow X, \] which satisfy \( L(x)\leq y \) if and only if \( x\leq R(y) \). The exercise, whose solution I'll give below, is that while \( L \) and \( R \) may of course fail to be mutual inverses, if we define \( X'=R(Y) \) and \( Y'=L(X) \) then \( L,R \) restrict to \( X',Y' \) respectively to become mutual inverses.

Lemma 1:For any \( x\in X \) we have \( x\leq R(L(x)) \), and for any \( y\in Y \) with have \( L(R(y)) \leq y \).

Proof:Given \( x \), set \( y=L(x) \) so certainly \( L(x) \leq y \), and hence by the defining assumption, \( x \leq R(y) = R(L(x)) \). Similarly, \( R(y) \leq R(y) \) implies that \( L(R(y)) \leq y \).

Lemma 2:For any \( x\in X' \) we have that \( x=R(L(x)) \). For any \( y\in Y' \) we have that \( y=L(R(y)) \).

Proof:Given \( x\in X' \) there is \( y\in Y \) with \( x=R(y) \). For any \( z \), by Lemma 1, we have \( L(R(z)) \leq z \) and so \( L(R(L(x))) \leq L(x) \) and \( x\leq R(y) \) so \( L(x)\leq y \), hence conclude that \( L(R(L(x))) \leq y \). This implies that \( R(L(x)) \leq R(y) = x \). By Lemma 1, \( x \leq R(L(x)) \), and so \( R(L(x)) = x \).Now given \( y\in Y' \) there is \( x\in X \) with \( L(x)=y \), which implies that \( x\leq R(y) \). By Lemma 1, for any \( z \), we have that \( z\leq R(L(z)) \) and so \( R(y) \leq R(L(R(y))) \) so \( x \leq R(L(R(y))) \). This implies that \( y = L(x) \leq L(R(y)) \). Lemma 1 tells us that \( y \geq L(R(y)) \), and so we conclude that \( y = L(R(y)) \).

We can now prove the main claim.

Claim:\( L \) and \( R \) restrict to mutually inverse bijections between \( X' \) and \( Y' \).

Proof:By definition, \( L \) maps into \( Y' \) and \( R \) maps into \( X' \), so we obtain restricted maps \( L:X'\rightarrow Y' \) and \( R:Y'\rightarrow X' \). We have shown that \( x=R(L(x)) \) for \( x\in X' \) and \( y=L(R(y)) \) for \( y\in Y' \). We claim that \( L \) is bijective. This is clear, as \( L(R(y))=y \) for any \( y\in Y' \) shows \( L \) is surjective, while \( R(L(x))=x \) for any \( x\in X' \) shows that \( L \) injects. Similarly \( R \) is a bijection.

Notice that we never used that \( L \) and \( R \) preserve order, only that \( L(x)\leq y \Leftrightarrow x\leq R(y) \).

The wikipedia article says a lot more, of course.

This is all quite neat, but I have struggled to think of a naturally occurring example in the sort of algebraic functional analysis which I study.