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1. • CommentRowNumber1.
   • CommentAuthorDavidRoberts
   • CommentTimeSep 4th 2014
   • (edited Sep 4th 2014)
   • PermaLink
   Author: DavidRoberts
   Format: MarkdownItexGiven a [[locally connected topos]] $\Gamma\colon E \leftrightarrows Set : \Delta$, we have the extra left adjoint $\pi_0 \dashv \Delta$. $\pi_0$ preserves epis, as any left adjoint does. If $E$ is boolean I'm sure that $\pi_0$ also reflects epis, at least assuming classical logic in the background. And I can't think of a non-boolean example where $\pi_0$ doesn't reflect epis (just my own ignorance, rather than any compelling mathematical reason). Alternatively, I'm suspicious of $\pi_0$ reflecting epis in the case of boolean $E$ without classical logic. To be concrete, say I've got an epi $T \to V$ in boolean locally connected $E$ where $V$ is connected (throw in enough colimits if you like). $T$ has a component (as $\pi_0(T)$ is inhabited assuming $Set$ is constructively well-pointed), say $T_0 \subset T$. Then $f\colon T_0 \to V$ induces an iso on applying $\pi_0$, but there's no immediately compelling reason I can see for it to be epi. Given that $E$ is boolean, I can see that $f$ is _not not_ epi, since if it were not epi, $V$ would not be connected since $V$ would split as the image of $T_0$ and its complement. Presumably the topos of directed graphs would be good to consider. Hmmm...

   Given a locally connected topos $\Gamma\colon E \leftrightarrows Set : \Delta$, we have the extra left adjoint $\pi_0 \dashv \Delta$. $\pi_0$ preserves epis, as any left adjoint does. If $E$ is boolean I’m sure that $\pi_0$ also reflects epis, at least assuming classical logic in the background. And I can’t think of a non-boolean example where $\pi_0$ doesn’t reflect epis (just my own ignorance, rather than any compelling mathematical reason).

   Alternatively, I’m suspicious of $\pi_0$ reflecting epis in the case of boolean $E$ without classical logic.

   To be concrete, say I’ve got an epi $T \to V$ in boolean locally connected $E$ where $V$ is connected (throw in enough colimits if you like). $T$ has a component (as $\pi_0(T)$ is inhabited assuming $Set$ is constructively well-pointed), say $T_0 \subset T$. Then $f\colon T_0 \to V$ induces an iso on applying $\pi_0$, but there’s no immediately compelling reason I can see for it to be epi. Given that $E$ is boolean, I can see that $f$ is not not epi, since if it were not epi, $V$ would not be connected since $V$ would split as the image of $T_0$ and its complement.

   Presumably the topos of directed graphs would be good to consider. Hmmm…

2. • CommentRowNumber2.
   • CommentAuthorDavidRoberts
   • CommentTimeSep 4th 2014
   • PermaLink
   Author: DavidRoberts
   Format: MarkdownItexIn the last comment that should be _reflexive_ directed graphs (see [[cohesive topos#Graphs]], and indeed, $\pi_0$ (there called $\Pi_0$) does _not_ reflect epis. I've probably implicitly used the atomicity in the specific topos I have at hand :-S

   In the last comment that should be reflexive directed graphs (see cohesive topos#Graphs, and indeed, $\pi_0$ (there called $\Pi_0$) does not reflect epis.

   I’ve probably implicitly used the atomicity in the specific topos I have at hand :-S

3. • CommentRowNumber3.
   • CommentAuthorZhen Lin
   • CommentTimeSep 4th 2014
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   Author: Zhen Lin
   Format: MarkdownItexThe topos of (reflexive directed) graphs is not boolean, though?

   The topos of (reflexive directed) graphs is not boolean, though?

4. • CommentRowNumber4.
   • CommentAuthorDavidRoberts
   • CommentTimeSep 4th 2014
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   Author: DavidRoberts
   Format: MarkdownItexYeah, but that addresses my first question as to a counterexample I was seeking.

   Yeah, but that addresses my first question as to a counterexample I was seeking.

5. • CommentRowNumber5.
   • CommentAuthorZhen Lin
   • CommentTimeSep 4th 2014
   • PermaLink
   Author: Zhen Lin
   Format: MarkdownItexReturning the penultimate paragraph of #1, it seems you have assumed that $\pi_0$ reflects initial objects. That is a weak form of Nullstellensatz, and I'm not sure it holds non-classically. Here's the classical proof: take a subcanonical site for the topos consisting of connected objects, which we may do because it is locally connected; then constant presheaves are sheaves, so $\pi_0$ can be computed as the colimit of the underlying presheaf and we are done. Of course, if you assume $\pi_0$ reflects epimorphisms, then it also reflects initial objects: indeed, in a topos, any epimorphism whose codomain is an initial object must be an isomorphism.

   Returning the penultimate paragraph of #1, it seems you have assumed that $\pi_0$ reflects initial objects. That is a weak form of Nullstellensatz, and I’m not sure it holds non-classically.

   Here’s the classical proof: take a subcanonical site for the topos consisting of connected objects, which we may do because it is locally connected; then constant presheaves are sheaves, so $\pi_0$ can be computed as the colimit of the underlying presheaf and we are done.

   Of course, if you assume $\pi_0$ reflects epimorphisms, then it also reflects initial objects: indeed, in a topos, any epimorphism whose codomain is an initial object must be an isomorphism.

6. • CommentRowNumber6.
   • CommentAuthorDavidRoberts
   • CommentTimeSep 4th 2014
   • PermaLink
   Author: DavidRoberts
   Format: MarkdownItexHmm, thanks. I've just taken $\pi_0$ reflecting epis as a hypothesis now. It will hold in the final construction I do, so even if it is restrictive it makes the hypotheses cleaner.

   Hmm, thanks. I’ve just taken $\pi_0$ reflecting epis as a hypothesis now. It will hold in the final construction I do, so even if it is restrictive it makes the hypotheses cleaner.