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switched notation for frames
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seldon
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$\newcommand{\Psh}{\operatorname{Psh}} \newcommand{\Sh}{\operatorname{Sh}} \newcommand{\op}{{\operatorname{op}}}$$\newcommand{\Psh}{\operatorname{Psh}} \newcommand{\Sh}{\operatorname{Sh}} \newcommand{\O}{{\mathcal{O}}}$ Let $X$ be a locale, $X^\op$$\O(X)$ the corresponding frame.

  1. What's the localic reflection of $\Psh X$?

We know that $$ X^\op \cong \mathrm{Sub}_{\Sh X}(1) $$$$ \O(X) \cong \mathrm{Sub}_{\Sh X}(1) $$ Call $Y = \mathrm{Sub}_{\Psh X}(1)$ the localic reflection of $\Psh X$. In particular, oneOne has $\Psh X \cong \Sh Y$$\Psh X \simeq \Sh Y$.

Since $\Sh X$ is a subtopos of $\Psh X$, $X$ should be a sublocale of $Y$, i.e. there should be a nucleus $j$ on $Y^\op$$\O(Y)$ such that $X^\op = Y^\op / j$$\O(X) = \O(Y) / j$.

  1. Who's $j$?

$\newcommand{\Psh}{\operatorname{Psh}} \newcommand{\Sh}{\operatorname{Sh}} \newcommand{\op}{{\operatorname{op}}}$ Let $X$ be a locale, $X^\op$ the corresponding frame.

  1. What's the localic reflection of $\Psh X$?

We know that $$ X^\op \cong \mathrm{Sub}_{\Sh X}(1) $$ Call $Y = \mathrm{Sub}_{\Psh X}(1)$ the localic reflection of $\Psh X$. In particular, one has $\Psh X \cong \Sh Y$.

Since $\Sh X$ is a subtopos of $\Psh X$, $X$ should be a sublocale of $Y$, i.e. there should be a nucleus $j$ on $Y^\op$ such that $X^\op = Y^\op / j$.

  1. Who's $j$?

$\newcommand{\Psh}{\operatorname{Psh}} \newcommand{\Sh}{\operatorname{Sh}} \newcommand{\O}{{\mathcal{O}}}$ Let $X$ be a locale, $\O(X)$ the corresponding frame.

  1. What's the localic reflection of $\Psh X$?

We know that $$ \O(X) \cong \mathrm{Sub}_{\Sh X}(1) $$ Call $Y = \mathrm{Sub}_{\Psh X}(1)$ the localic reflection of $\Psh X$. One has $\Psh X \simeq \Sh Y$.

Since $\Sh X$ is a subtopos of $\Psh X$, $X$ should be a sublocale of $Y$, i.e. there should be a nucleus $j$ on $\O(Y)$ such that $\O(X) = \O(Y) / j$.

  1. Who's $j$?
added 47 characters in body
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seldon
  • 1.1k
  • 8
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$\newcommand{\Psh}{\operatorname{Psh}} \newcommand{\Sh}{\operatorname{Sh}} \newcommand{\op}{{\operatorname{op}}}$ Let $X$ be a locale, $X^\op$ the corresponding frame.

  1. What's the localic reflection of $\Psh X$?

We know that $$ X^\op \cong \mathrm{Sub}_{\Sh X}(1) $$ Call $Y = \mathrm{Sub}_{\Psh X}(1)$ the localic reflection of $\Psh X$. In particular, one has $\Psh X \cong \Sh Y$.

Since $\Sh X$ is a subtopos of $\Psh X$, $X$ should be a sublocale of $Y$, i.e. there should be a nucleus $j$ on $Y^\op$ such that $X^\op = Y^\op / j$.

  1. Who's $j$?

$\newcommand{\Psh}{\operatorname{Psh}} \newcommand{\Sh}{\operatorname{Sh}} \newcommand{\op}{{\operatorname{op}}}$ Let $X$ be a locale, $X^\op$ the corresponding frame.

  1. What's the localic reflection of $\Psh X$?

We know that $$ X^\op \cong \mathrm{Sub}_{\Sh X}(1) $$ Call $Y = \mathrm{Sub}_{\Psh X}(1)$ the localic reflection of $\Psh X$. Since $\Sh X$ is a subtopos of $\Psh X$, $X$ should be a sublocale of $Y$, i.e. there should be a nucleus $j$ on $Y^\op$ such that $X^\op = Y^\op / j$.

  1. Who's $j$?

$\newcommand{\Psh}{\operatorname{Psh}} \newcommand{\Sh}{\operatorname{Sh}} \newcommand{\op}{{\operatorname{op}}}$ Let $X$ be a locale, $X^\op$ the corresponding frame.

  1. What's the localic reflection of $\Psh X$?

We know that $$ X^\op \cong \mathrm{Sub}_{\Sh X}(1) $$ Call $Y = \mathrm{Sub}_{\Psh X}(1)$ the localic reflection of $\Psh X$. In particular, one has $\Psh X \cong \Sh Y$.

Since $\Sh X$ is a subtopos of $\Psh X$, $X$ should be a sublocale of $Y$, i.e. there should be a nucleus $j$ on $Y^\op$ such that $X^\op = Y^\op / j$.

  1. Who's $j$?
corrected minor imprecision
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seldon
  • 1.1k
  • 8
  • 16

$\newcommand{\Psh}{\operatorname{Psh}} \newcommand{\Sh}{\operatorname{Sh}} \newcommand{\op}{{\operatorname{op}}}$ Let $X$ be a locale, $X^\op$ the corresponding frame.

  1. What's the localic reflection of $\Psh X$?

We know that $$ X^\op \cong \mathrm{Sub}_{\Sh X}(1) $$ Call $Y = \mathrm{Sub}_{\Psh X}(1)$ the localic reflection of $\Psh X$. Since $\Sh X$ is a subtopos of $\Psh X$, $X$ should be a sublocale of $Y$, i.e. there should be a nucleus $j$ on $Y^\op$ such that $X = Y / j$$X^\op = Y^\op / j$.

  1. Who's $j$?

$\newcommand{\Psh}{\operatorname{Psh}} \newcommand{\Sh}{\operatorname{Sh}} \newcommand{\op}{{\operatorname{op}}}$ Let $X$ be a locale, $X^\op$ the corresponding frame.

  1. What's the localic reflection of $\Psh X$?

We know that $$ X^\op \cong \mathrm{Sub}_{\Sh X}(1) $$ Call $Y = \mathrm{Sub}_{\Psh X}(1)$ the localic reflection of $\Psh X$. Since $\Sh X$ is a subtopos of $\Psh X$, $X$ should be a sublocale of $Y$, i.e. there should be a nucleus $j$ on $Y^\op$ such that $X = Y / j$.

  1. Who's $j$?

$\newcommand{\Psh}{\operatorname{Psh}} \newcommand{\Sh}{\operatorname{Sh}} \newcommand{\op}{{\operatorname{op}}}$ Let $X$ be a locale, $X^\op$ the corresponding frame.

  1. What's the localic reflection of $\Psh X$?

We know that $$ X^\op \cong \mathrm{Sub}_{\Sh X}(1) $$ Call $Y = \mathrm{Sub}_{\Psh X}(1)$ the localic reflection of $\Psh X$. Since $\Sh X$ is a subtopos of $\Psh X$, $X$ should be a sublocale of $Y$, i.e. there should be a nucleus $j$ on $Y^\op$ such that $X^\op = Y^\op / j$.

  1. Who's $j$?
Became Hot Network Question
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seldon
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