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.
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