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1. • CommentRowNumber1.
   • CommentAuthorDavidRoberts
   • CommentTimeMay 7th 2013
   • (edited May 7th 2013)
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   Author: DavidRoberts
   Format: MarkdownItexDoes the [[effective topos]] (and [[realisability topos]]es more generally) come with a geometric morphism to $Set$? I think it doesn't, but I'm struggling to find a proof of this fact.

   Does the effective topos (and realisability toposes more generally) come with a geometric morphism to $Set$? I think it doesn’t, but I’m struggling to find a proof of this fact.

2. • CommentRowNumber2.
   • CommentAuthorMike Shulman
   • CommentTimeMay 7th 2013
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   Author: Mike Shulman
   Format: MarkdownItexYou are correct, it doesn't. Instead it comes with a geometric morphism *from* Set: the global sections functor has a right adjoint rather than a left adjoint. I'm not an expert in realizability toposes, so I don't know offhand the easiest proof that no left adjoint exists. My instinct would be to try to derive a contradiction from the existence of a coproduct of uncountably many copies of 1.

   You are correct, it doesn’t. Instead it comes with a geometric morphism from Set: the global sections functor has a right adjoint rather than a left adjoint.

   I’m not an expert in realizability toposes, so I don’t know offhand the easiest proof that no left adjoint exists. My instinct would be to try to derive a contradiction from the existence of a coproduct of uncountably many copies of 1.

3. • CommentRowNumber3.
   • CommentAuthorZhen Lin
   • CommentTimeMay 7th 2013
   • (edited May 7th 2013)
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   Author: Zhen Lin
   Format: MarkdownItexHere is the argument presented by Johnstone: > **Remark F2.2.19.** (b) [...] $\mathbf{Eff}(\Lambda)$ cannot admit any geometric morphism to a Boolean topos, for any (nontrivial) $\Lambda$. For if such a morphism $f : \mathbf{Eff}(\Lambda) \to \mathcal{B}$ existed, we could take it to be surjective by A4.5.22; and then since all objects of the form $f^* B$ would be (decidable, and hence) modest, the cardinality of any hom-set $\mathcal{B}(B, B')$ would have to be bounded by that of $\Lambda$. But, for any set $A$, we have $\mathcal{B}(1, f_* (\nabla A)) \cong \mathbf{Eff}(\Lambda)(1, \nabla A) \cong A$. (An assembly $(A, \alpha)$ is **modest** if the relation $\alpha : A \looparrowright \Lambda$ is the opposite of a partial function $\Lambda \mathrel{⇁} A$.)

   Here is the argument presented by Johnstone:

   Remark F2.2.19. (b) […] $\mathbf{Eff}(\Lambda)$ cannot admit any geometric morphism to a Boolean topos, for any (nontrivial) $\Lambda$. For if such a morphism $f : \mathbf{Eff}(\Lambda) \to \mathcal{B}$ existed, we could take it to be surjective by A4.5.22; and then since all objects of the form $f^* B$ would be (decidable, and hence) modest, the cardinality of any hom-set $\mathcal{B}(B, B')$ would have to be bounded by that of $\Lambda$. But, for any set $A$, we have $\mathcal{B}(1, f_* (\nabla A)) \cong \mathbf{Eff}(\Lambda)(1, \nabla A) \cong A$.

   (An assembly $(A, \alpha)$ is modest if the relation $\alpha : A \looparrowright \Lambda$ is the opposite of a partial function $\Lambda \mathrel{⇁} A$.)

4. • CommentRowNumber4.
   • CommentAuthorMike Shulman
   • CommentTimeMay 7th 2013
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   Author: Mike Shulman
   Format: MarkdownItexBut what if Set is not Boolean? (-:

   But what if Set is not Boolean? (-:

5. • CommentRowNumber5.
   • CommentAuthorDavidRoberts
   • CommentTimeMay 8th 2013
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   Author: DavidRoberts
   Format: MarkdownItexWell, for my purposes I am assuming it to be so, answering my underlying question.

   Well, for my purposes I am assuming it to be so, answering my underlying question.