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1. • CommentRowNumber1.
   • CommentAuthorUrs
   • CommentTimeJan 25th 2011
   • (edited Jan 25th 2011)
   • PermaLink
   Author: Urs
   Format: MarkdownItexLet $\mathcal{E}$ be a [[local topos]] or sheaf topos over a [[concrete site]] and $X \in \mathcal{E}$ a [[concrete sheaf]]. The concreteness-condition on a sheaf may be read as saying that $X$ has "enough points", in a sense. What about the slice topos $\mathcal{E}/X$? What can we say about its [[point of a topos|topos points]]? Under which conditions does it have enough?

   Let $\mathcal{E}$ be a local topos or sheaf topos over a concrete site and $X \in \mathcal{E}$ a concrete sheaf.

   The concreteness-condition on a sheaf may be read as saying that $X$ has “enough points”, in a sense.

   What about the slice topos $\mathcal{E}/X$? What can we say about its topos points? Under which conditions does it have enough?

2. • CommentRowNumber2.
   • CommentAuthorMike Shulman
   • CommentTimeJan 26th 2011
   • PermaLink
   Author: Mike Shulman
   Format: MarkdownItexAny slice topos of a topos with enough points has enough points. Proof: suppose $E$ has enough points. A point of $E/X $ is a point $e\colon Set \to E$ together with an element $x\in e^*(X)$. To prove that it has enough points, we must show that if $A,B\in E/X$ and $f:A\to B$ is a morphism in $E/X $ such that $(e,x)^*(f)$ is an isomorphism for every point $(e,x)$ of $E/X $, then $f$ is an isomorphism. But $(e,x)^*$ is given by applying $e^*$ and then taking the fiber over $x$, so if $(e,x)^*(f)$ is an isomorphism for all $e$ and $x$, then $e^*(f)$ is an isomorphism for all $e$, hence (since $E$ has enough points) $f$ is an isomorphism in $E$, hence also in $E/X $.

   Any slice topos of a topos with enough points has enough points. Proof: suppose $E$ has enough points. A point of $E/X$ is a point $e\colon Set \to E$ together with an element $x\in e^*(X)$. To prove that it has enough points, we must show that if $A,B\in E/X$ and $f:A\to B$ is a morphism in $E/X$ such that $(e,x)^*(f)$ is an isomorphism for every point $(e,x)$ of $E/X$, then $f$ is an isomorphism. But $(e,x)^*$ is given by applying $e^*$ and then taking the fiber over $x$, so if $(e,x)^*(f)$ is an isomorphism for all $e$ and $x$, then $e^*(f)$ is an isomorphism for all $e$, hence (since $E$ has enough points) $f$ is an isomorphism in $E$, hence also in $E/X$.

3. • CommentRowNumber3.
   • CommentAuthorUrs
   • CommentTimeJan 26th 2011
   • (edited Jan 26th 2011)
   • PermaLink
   Author: Urs
   Format: MarkdownItexThanks, Mike! I have added some discussion to [[over-topos]]. I see the points of the form $(e,x)$, they are the composites $$ (e,x) : Set \stackrel{x^*}{\to} Set/e^*(X) \stackrel{e/X}{\to} \mathcal{E}/X \,. $$ How do I see that every point of the slice topos arises this way? But more importantly: in which generality can we assume $\mathcal{E}$ to have enough points in the first place? To talk about concrete sheaves we need a local topos. What can we say about local toposes having enough points? (Certainly they have one important canonical point, by definition.) I need to think again about the standard example that I am interested in, the topos $Sh(CartSp) \simeq Sh(Mfd)$. It should have enough points, with one point $(n)$ per $n \in \mathbb{N}$ given by the stalk at any ordinary point in $\mathbb{R}^n$. Together with the above this would also show where my intuition about concrete sheaves inducing "enough points" was wrong: the topos points of $Sh(CartSp)/X$ are not related to global points $* \to X$ but to stalks of maps of disks into $X$, hence manifestly do not distinguish between concrete and non-concrete $X$. Is there maybe some other property of the slice $\mathcal{E}/X$ that witnesses the fact that $X$ is concrete? Let's see, if we start with a local topos $(p^* \dashv p_* \dashv p^!) : \mathcal{E} \to Set$ then we should maybe be looking at the canonical point $e_0 := (p_* \dashv p^!) : Set \to \mathcal{E}$ and the induced points of the special form $$ (e_0, x) : Set \to \mathcal{E}/X \,. $$ Now _these_ are really those given by the global points $x : * \to X$ of $X$. So concreteness of $X$ should say that $\mathcal{E}/X$ has in some sense enough of these $e_0$-points. Hm...

   Thanks, Mike!

   I have added some discussion to over-topos. I see the points of the form $(e,x)$, they are the composites

   $$(e,x) : Set \stackrel{x^*}{\to} Set/e^*(X) \stackrel{e/X}{\to} \mathcal{E}/X \,.$$

   How do I see that every point of the slice topos arises this way?

   But more importantly: in which generality can we assume $\mathcal{E}$ to have enough points in the first place? To talk about concrete sheaves we need a local topos. What can we say about local toposes having enough points? (Certainly they have one important canonical point, by definition.)

   I need to think again about the standard example that I am interested in, the topos $Sh(CartSp) \simeq Sh(Mfd)$. It should have enough points, with one point $(n)$ per $n \in \mathbb{N}$ given by the stalk at any ordinary point in $\mathbb{R}^n$.

   Together with the above this would also show where my intuition about concrete sheaves inducing “enough points” was wrong: the topos points of $Sh(CartSp)/X$ are not related to global points $* \to X$ but to stalks of maps of disks into $X$, hence manifestly do not distinguish between concrete and non-concrete $X$.

   Is there maybe some other property of the slice $\mathcal{E}/X$ that witnesses the fact that $X$ is concrete?

   Let’s see, if we start with a local topos $(p^* \dashv p_* \dashv p^!) : \mathcal{E} \to Set$ then we should maybe be looking at the canonical point $e_0 := (p_* \dashv p^!) : Set \to \mathcal{E}$ and the induced points of the special form

   $$(e_0, x) : Set \to \mathcal{E}/X \,.$$

   Now these are really those given by the global points $x : * \to X$ of $X$. So concreteness of $X$ should say that $\mathcal{E}/X$ has in some sense enough of these $e_0$-points.

   Hm…

4. • CommentRowNumber4.
   • CommentAuthorUrs
   • CommentTimeJan 26th 2011
   • (edited Jan 26th 2011)
   • PermaLink
   Author: Urs
   Format: MarkdownItexNot sure if it helps, but just an observation: if $\mathcal{E}$ is a local topos and $X \in \mathcal{E}$ a concrete object, then the "global points"-monad corresponding to the adjunction $Set/p_*(X) \stackrel{\overset{p_*/X}{\leftarrow}}{\underset{p^!/X}{\to}} \mathcal{E}/X$ acts on an object $(A \to X) \in \mathcal{E}/X$ by concretifying $A$ relative to $X$: by construction of the slice geometric morphism $(p_*/X \dashv p^!/X)$ we have that $ p^!/X \circ p_*/X (A \to X)$ is the pullback $\tilde A$ in $$ \array{ \tilde A &\to& p^! p_* A \\ \downarrow && \downarrow \\ X &\to& p^! p_* X } \,, $$ where the bottom morphism is the unit. By definition of concrete objects the bottom morphism is a mono precisely if $X$ is concrete. Since monos are stable under pullback, also the top morphism is a mono. Applying $p_*$ to the top morphism and using that $p^!$ is by definition full and faithful, we find that $\Gamma \tilde A \hookrightarrow \Gamma A$ is a mono. By the universal property of the unit this implies that $\tilde A \to p_! p^* \tilde A $ is a mono, hence that $\tilde A$ itself is concrete. Now, on the objects $\tilde A \to X$ in the image of this "relative concretization"-monad it is certainly true that the $(e_0,x)$-points detect isomorphisms. Hm...

   Not sure if it helps, but just an observation: if $\mathcal{E}$ is a local topos and $X \in \mathcal{E}$ a concrete object, then the “global points”-monad corresponding to the adjunction $Set/p_*(X) \stackrel{\overset{p_*/X}{\leftarrow}}{\underset{p^!/X}{\to}} \mathcal{E}/X$ acts on an object $(A \to X) \in \mathcal{E}/X$ by concretifying $A$ relative to $X$:

   by construction of the slice geometric morphism $(p_*/X \dashv p^!/X)$ we have that $p^!/X \circ p_*/X (A \to X)$ is the pullback $\tilde A$ in

   $$\array{ \tilde A &\to& p^! p_* A \\ \downarrow && \downarrow \\ X &\to& p^! p_* X } \,,$$

   where the bottom morphism is the unit. By definition of concrete objects the bottom morphism is a mono precisely if $X$ is concrete. Since monos are stable under pullback, also the top morphism is a mono. Applying $p_*$ to the top morphism and using that $p^!$ is by definition full and faithful, we find that $\Gamma \tilde A \hookrightarrow \Gamma A$ is a mono. By the universal property of the unit this implies that $\tilde A \to p_! p^* \tilde A$ is a mono, hence that $\tilde A$ itself is concrete.

   Now, on the objects $\tilde A \to X$ in the image of this “relative concretization”-monad it is certainly true that the $(e_0,x)$-points detect isomorphisms.

   Hm…

5. • CommentRowNumber5.
   • CommentAuthorMike Shulman
   • CommentTimeJan 26th 2011
   • PermaLink
   Author: Mike Shulman
   Format: MarkdownItex> How do I see that every point of the slice topos arises this way? By pullback, it suffices to show that every section of $E/X\to E$ arises from a map $1\to X$ in E. You can regard that as a special case of the full embedding of internal locales in E into localic geometric morphisms over E, where X is regarded as a discrete internal locale in E.

   How do I see that every point of the slice topos arises this way?

   By pullback, it suffices to show that every section of $E/X\to E$ arises from a map $1\to X$ in E. You can regard that as a special case of the full embedding of internal locales in E into localic geometric morphisms over E, where X is regarded as a discrete internal locale in E.