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    • CommentRowNumber1.
    • CommentAuthorDavidRoberts
    • CommentTimeMay 7th 2013
    • (edited May 7th 2013)

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

    • CommentRowNumber2.
    • CommentAuthorMike Shulman
    • CommentTimeMay 7th 2013

    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.

    • CommentRowNumber3.
    • CommentAuthorZhen Lin
    • CommentTimeMay 7th 2013
    • (edited May 7th 2013)

    Here is the argument presented by Johnstone:

    Remark F2.2.19. (b) […] Eff(Λ)\mathbf{Eff}(\Lambda) cannot admit any geometric morphism to a Boolean topos, for any (nontrivial) Λ\Lambda. For if such a morphism f:Eff(Λ)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 *Bf^* B would be (decidable, and hence) modest, the cardinality of any hom-set (B,B)\mathcal{B}(B, B') would have to be bounded by that of Λ\Lambda. But, for any set AA, we have (1,f *(A))Eff(Λ)(1,A)A\mathcal{B}(1, f_* (\nabla A)) \cong \mathbf{Eff}(\Lambda)(1, \nabla A) \cong A.

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

    • CommentRowNumber4.
    • CommentAuthorMike Shulman
    • CommentTimeMay 7th 2013

    But what if Set is not Boolean? (-:

    • CommentRowNumber5.
    • CommentAuthorDavidRoberts
    • CommentTimeMay 8th 2013

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