We have already essentially introduced locales in a previous post. Locales, as objects, are just complete Heyting algebras. However, the morphisms are different.
Before we begin in earnest, a good reference for more information about the contents of this post is the excellent book Stone Spaces by Johnstone.
First we start by defining another category whose objects are complete
Heyting algebras, $\mathbf{Frm}$, the category of frames. Since we already know that the objects are complete Heyting algebras, it suffices to define a frame homomorphism. This will be defined by abstracting the properties of the inverse image map $f^{-1}:\Omega(Y)\rightarrow \Omega(X)$ of a continuous function $f:X\rightarrow Y$, where $\Omega(X)$ means the topology on $X$ (which we know is a complete Heyting algebra (joins are unions, meets interior of intersection, etc.)). What properties does $f^{-1}$ satisfy? Well, it preserves arbitrary joins and finite meets. So we define a frame homomorphism $f:A\rightarrow B$ between frames to be a map preserving arbitrary joins and finite meets. Then we define the category of locales, $\mathbf{Loc}:=\mathbf{Frm}^{op}$. We must remember that the morphisms of locales are not functions, for instance if we ever need to check whether two locale morphisms are equal we often have to pass to the opposite category of frames for the computation.
Then almost by definition we have a functor: $\Omega:\mathbf{Top}\rightarrow \mathbf{Loc}$ given on objects by $X\mapsto \Omega(X)$, (where again $\Omega(X)$ is the topology of $X$ considered as a complete Heyting albegra) and on arrows by $f:X\rightarrow Y \mapsto \Omega(f):\Omega(X)\rightarrow \Omega(Y)$ where $\Omega(f)^{op}:=f^{-1}$.
How can we abstract the notion of point in a space to a locale? Well, first a point, as an element of a space $X$, is a morphism $x:1\rightarrow X$, where $1$ is the one-point space. Then the induced locale map is $\Omega(x):\Omega(1)=2\rightarrow \Omega(X)$, where $2$ is the lattice $0\rightarrow 1$. Hence, we define a point of a locale $A$ to be a locale morphism $2\rightarrow A$. Hence a point of a locale is the opposite of a frame morphism $A\rightarrow 2$, and if $p:2\rightarrow A$ is a point, we write $f^\star:A\rightarrow 2$ for the induced frame map (rather than $f^{op}$, since we would like to emphasize that $f^\star$ is actually a function). As a working definition, we define $pt(A)$ to be the set of all points of $A$.
If this notion of point is to be a suitable generalization of the notion of element of a space, then we should at least have that the points of $A$ form a space in a natural way. To this end, if $A$ is a locale, define a frame map $\phi_A: A\rightarrow P(pt(A)$ (where $P$ is the power-set functor) by $a\mapsto \phi(a):=\{p\;|\;p^\star(a)=1\}$. Note that since each $p^\star$ preserves finite meets and arbitrary joins, being a frame morphism, and since each $p^\star$ can only take values $0$ or $1$, we have $\phi_A(a\wedge b):=\{p\;|\;p^\star(a)\wedge p^\star(b)=1\}=\{p\;|\;p^\star(a)=1\}\cap\{p\;|\;p^\star(b)=1\}
= \phi_A(a)\wedge \phi_A(b)$ and
$\phi_A(\bigvee a_i) = \{p\;|\;\bigvee p^\star (a_i)=1\}=\bigcup\{p\;|\;p^\star(a_i)=1\} = \bigvee \phi_A(a_i)$. Hence, $\phi_A$ is indeed a frame homomorphism. In particular, $\phi_A(pt(A))$ is a topology. So we set $\Omega(pt(A)):= \phi_A(pt(A))$ and now consider $\phi_A:A\rightarrow \Omega(pt(A))$ to have co-domain $\Omega(pt(A))$. Hence $pt(A)$ is a space for each locale $A$. Hence we may revise our working definition of $pt(A)$ to reflect this fact.
Before we continue, a bit of notation: if $C$ is a category, we will replace the tedious notation hom$_C(\_,\_)$ for the hom functor, by $<\_,\_>_C$, reflecting the analogy with the inner product, often dropping the subscript $C$ when the context is obvious.
Note that $pt(A)$ is essentially $<2,A>_{\mathbf{Loc}}$ with the proviso that $pt(A)$ is a space and not just a set. Hence we should seek to define $pt:\mathbf{Loc}\rightarrow \mathbf{Top}$ using the representable functor $<2,\_>$ as a guide. So $pt$, on objects, $A\mapsto pt(A)$, has already been defined. And on arrows, we define $pt(f:A\rightarrow B):pt(A)\rightarrow pt(B)$ by $pt(f)(p):=fp$. That this is a functor is straightforward.
Hence we have functors $\Omega:\mathbf{Top}\rightarrow \mathbf{Loc}$ and $pt:\mathbf{Loc}\rightarrow \mathbf{Top}$. We conclude this post by showing that $\Omega \dashv pt$.
There's a sort of "shortcut theorem" for showing that a functor has an adjoint which says that solutions to a certain universal mapping problem for one functor is the same as an adjoint functor pair with another functor. I may cover this theorem in a later post. Johnstone uses this theorem in his proof that $\Omega \dashv pt$, and as a result obtains a much shorter proof than ours. We will use the "hom functor definition" of adjoint, i.e. we will show $<\Omega(\_),\_>\cong<\_,pt(\_)>$.
If $X$ is a space and $A$ a locale and $h:\Omega(X)\rightarrow A$, how can we define the proposed transpose $\bar{h}:X\rightarrow pt(A)$? Well, if $x:1\rightarrow X$ is an element of $X$ then $\Omega(x)=2\rightarrow \Omega(X)$ is a locale map. Hence $h\Omega(x)$ is a locale map from $2$ to $A$, a.k.a. a point of $A$, precisely the form that $\bar{h}(x)$ should take. So set $\bar{h}(x):h\Omega(x)$. We must check $\bar{h}$ is continuous. But $\bar{h}^{-1}(\phi_A)(a)=\{x\;|\;(h\Omega(x))^\star(a)=1\}$. But $(h\Omega(x))^\star(a) = \Omega(x)^\star h^\star(a)=x^{-1}(h^\star(a))=1 \iff x\in h^\star(a)$. So $\bar{h}^{-1}(\phi_A)(a)=h^\star(a)\in\Omega(X)$. Hence $\bar{h}$ is continuous. We show that this transpose operation defines a natural transformation.
Let $f:Y\rightarrow X$ be a space morphism (continuous map) and $g:A\rightarrow B$ be a locale homomorphism (opposite of a frame homomorphism). Then we must show that the "naturality square" commutes, i.e. if $h:\Omega(X)\rightarrow A$ then $<f,pt(g)>(\bar{h})= pt(g)\bar{h}f= \bar{(<\Omega(f),g>(h))}=\bar{(gh\Omega(f))}$. But for $y\in Y$, $\bar{(gh\Omega(f))}(y)=gh\Omega(f)\Omega(y)=gh\Omega(fy)$, and $pt(g)\bar{h}f(y)=pt(g)h\Omega(f(y))=gh\Omega(fy)$.
Hence the naturality square commutes, as desired.
Now we define the inverse of $\bar{()}$, completing the proof (since then $\bar{()}$ will be a natural equivalence). So for each $h:X\rightarrow pt(A)$, we seek to define a proposed transpose under the adjunction $\hat{h}:\Omega(X)\rightarrow A$. Note that $\phi_A:A\rightarrow \Omega(pt(A))$ is a frame map. Hence $\phi_A^{op}:\Omega(pt(A))\rightarrow A$ is a locale morphism. Hence, set $\hat{h}:=\phi_A^{op} \Omega(h)$. Note that for $a\in A$, $\hat{h}^\star(a) = (\phi_A^{op}\Omega(h))^\star(a) = h^{-1}(\phi_A(a))=\{x\;|\:h(x)^\star(a)=1\}.$
Then if $h:X\rightarrow pt(A)$ is continuous,$x\in X, \bar{\hat{h}}(x)=\hat{h}\Omega(x)=\phi_A^{op}\Omega(h)\Omega(x)=\phi_A^{op}\Omega(hx)=\hat{(hx)}$. But for $a\in A$, $\hat{(hx)}^\star(a) = \{t\in 1\;|\;hx(t)^\star(a)=1\} = 1$ when $h(x)(a)=1$ and $=\emptyset = 0$ when $h(x)(a)=0$ (since $x(1)=x$). Hence $\hat{(hx)}^\star = h(x)^\star.$ So $\hat{(hx)}=h(x)$. Then $\bar{\hat{h}}=h$.
If $h:\Omega(X)\rightarrow A$ is a locale morphism, $a\in A$, then $\hat{\bar{h}}^\star(a) = \{x\;|\;(\bar{h}(x))^\star(a)=1\}=\{x\;|\;(h\Omega(x))^\star(a)=1\} = \{x\;|\;x\in h^\star(a)\} = h^\star(a).$ Hence $\hat{\bar{h}}=h$. Therefore the proof is complete. We have shown $\Omega \dashv pt$.
In the next post we examine the unit and co-unit of the adjuntion and deduce a Stone-type duality between so called spatial locales and sober spaces.
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