Let $F:\mathbf{C}\rightarrow \mathbf{D}$ be left adjoint to $G:\mathbf{D}\rightarrow \mathbf{C}$. Let $\eta:1_\mathbf{C}\rightarrow GF, \varepsilon:FG \rightarrow 1_\mathbf{D}$ be the unit and co-unit, respectively, of the adjunction (see earlier post). Then we know that the triangle identities hold: $\varepsilon F \circ F\eta = 1_F, G\varepsilon \circ \eta G = 1_G$. Let $\mathbf{C_0}$ be the full subcategory of $\mathbf{C}$ generated by the objects $X$ of $\mathbf{C}$ for which $\eta_X:X\rightarrow GFX$ is an isomorphism, $\mathbf{D_0}$ be the full subcategory of $\mathbf{D}$ generated by the
objects $Y$ of $\mathbf{D}$ for which $\varepsilon_Y:FG Y\rightarrow Y$ is an
isomorphism (full means that for any two objects of the subcategory, all arrows in the ambient category are arrows of the subcategory). Then $F$ restricts to an equivalence between $\mathbf{C_0}$ and $\mathbf{D_0} $.
Indeed, let $F_0: = F|_{\mathbf{C_0}}, G_0:=G|_{\mathbf{D_0}}$. We show $G_0F_0 \cong 1_\mathbf{C_0}, F_0G_0\cong 1_\mathbf{D_0}$. But that will be obvious from the definition of $\mathbf{C_0}$ and $\mathbf{D_0}$ if we know that for $X\in \mathbf{C_0}, FX\in \mathbf{D_0}$ and for $Y\in \mathbf{D_0}, GY\in \mathbf{C_0}$. We prove the claim for $X\in \mathbf{C_0}$. The other may be proven analogously (using the other triangle identity). Let $X\in \mathbf{C_0}$. Then $\eta_X$ is iso. So $F\eta_X$ is iso. But by the first triangle identity, $\varepsilon_{FX}F\eta_X=1_{FX}$. Multiplying on the right of both sides by $(F\eta_X)^{-1}$ (since it is iso), yields $\varepsilon_{FX} = (F\eta_X)^{-1}$. So $\varepsilon_{FX}$ is iso as well. Hence $FX$ is in $\mathbf{D_0}$.
We now apply this result to the particular adjunction between $\mathbf{Top}$ and $\mathbf{Loc}$ from the previous post. Recall that in the adjunction $\Omega \dashv pt$, we had that the transpose under the adjunction of a locale morphism $h:\Omega X \rightarrow A$ is $\bar{h}:X\rightarrow ptA$, where $\bar{h}(x)=h\Omega(x)$, and the transpose under the adjunction of a continuous map $h:X\rightarrow ptA$ is $\hat{h}=\phi_A^{op}\Omega(h)$, where, recall, $\phi_A$ is a frame morphism defined by $\phi_A(a)=\{p\;|\;p^\star(a)=1\}$ and $\Omega ptA$ is defined by the requirement that $\phi_A$ is surjective.
Hence for $X \in \mathbf{Top}, \eta_X=\bar{1_FX}$ and so $\eta_X(x) = \Omega(x)$. What does it mean for $\eta_X$ to be surjective? It means every point of the locale $\Omega X$ (recall a point $p$ of a locale $A$ is a locale morphism $2\rightarrow A$ and $p$ is the opposite of a frame morphism $p^\star$) is of the form $\Omega(x)$ for some $x\in X$. What does it mean for $\eta_X$ to be injective? It means that if a point of $\Omega(X)$ is of the form $\Omega(x)$ for some $x\in X$ then that $x$ is unique. So $\eta_X$ is a bijection when every point of $\Omega(X)$ is of the form $\Omega(x)$ for a unique $x\in X$. Suppose $\eta_X$ is a bijection. Then is it an isomorphism? Clearly, the inverse, if it exists will be the map $p=\Omega(x)\mapsto x$, where we have used the assumption that we can write any point uniquely as $\Omega(x)$ . For definiteness let's call the proposed inverse $\lambda_X$. Clearly it's a function and the inverse of $\eta_X$ in $\mathbf{Set}$. We just need to show it's continuous. So let $U \in \Omega X$. Then $\lambda_X^{-1}(U)=\{\Omega(x)\;|\;x\in U\} = \{\Omega(x)\;|\;\Omega(x)^\star(U)=x^{-1}(U)=1\} = \{p\;|\;p^\star(U) = 1\} = \phi_{\Omega X}(U)\in \Omega PtX$. Hence $\eta_X$ is iso iff it is bijective iff every point of $\Omega X$ is of the form $\Omega(x)$ for a unique $x\in X$. We call such spaces $X$ sober.
Similarly, for $Y\in \mathbf{Loc}, \varepsilon_Y = \hat{1_{GY}}=\phi_A^{op}$. Hence $\varepsilon_Y$ is iso iff $\varepsilon_Y^\star = \phi_A$ is a frame isomorphism. But clearly if $\phi_A$ has an inverse in $\mathbf{Set}$ then the inverse also preserves meets and finite joins since $\phi_A$ does. So it will be a frame homomorphism as well. Hence $\phi_A$ will be invertible iff it is bijective. But $\phi_A$ is onto $\Omega pt A$ by definition of $\Omega pt A$. So $\phi_A$ will be invertible iff it is injective. When is $\phi_A$ injective? When elements of $A$ are uniquely determined by the points that they "contain", where we might say that $a$ "contains" a point $p:2\rightarrow A$ if $p^\star(a)=1$ (for intuition think of the case that $A$ is of the form $\Omega(X)$. Then $U\in \Omega (X)$ "contains" a point of the form $\Omega(x)$ if and only if $x\in U$ and $U$ is certainly determined by those "representable points".). Locales whose elements are uniquely determined by the points that they "contain" (a.k.a. $\phi_A$ injective) are called spatial.
Then by the theorem we proved at the beginning of the post, $\Omega$ restricts to an equivalence between the full subcategories of sober topological spaces and spatial locales.
Tuesday, October 15, 2013
Saturday, October 12, 2013
Locales and spaces: generalized points
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.
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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