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Let ${f : E \rightarrow S}$ be a geometric morphism (between toposes).

For $s$ in $S$ and $x$ in $E$ let ${\pi : f^* s \times x \rightarrow x}$ be the obvious projection in $E$.

Let ${u \rightarrow f^* s \times x}$ be a complemented subobject of ${f^* s \times x}$.

Is the image of $u$ along $\pi$ complemented as a subobject of $x$?

(See also Images of complemented subobjects in hyperconnected toposes over Boolean bases)

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No, not even if $E=S$, $f$ is the identity morphism, and $x=1$. In that special case, your question asks whether $\forall z\in s\,\big((z\in u)\lor \neg(z\in u)\big)$ (in the internal language of $S$) implies $(\exists z\in s\,z\in u)\lor\neg(\exists z\in s\,z\in u)$. When $s$ is $\mathbb N$, this is the limited principle of omniscience, which is not intuitionistically valid.

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  • $\begingroup$ Thanks very much for your reply, @Andreas Blass. I'd be happy to mark my question as answered; but I think it would be more interesting to edit it. What if we assume that $S$ is Boolean or satisfies Choice? (Shouldn't $\pi$ behave as a codiagonal?) $\endgroup$ – Mendieta Jun 22 at 23:43
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    $\begingroup$ @Mendieta If $S$ is Boolean (which is weaker than having AC), every subobject of every object is complemented. $\endgroup$ – მამუკა ჯიბლაძე Jun 23 at 6:46
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    $\begingroup$ @მამუკაჯიბლაძე What you say is true, but the question still makes sense because E need not be Boolean. $\endgroup$ – Mendieta Jun 23 at 17:29
  • $\begingroup$ @Mendieta Sorry you are right. What I said does not matter. What matters is that one can modify the answer to include any non-Boolean topos $E$ that admits a geometric morphism to a Boolean topos $S$. Take $u$ any non-complemented subobject of some object $x$ in $E$, and the rest as in the answer: take $s$ the terminal object of $S$ and $\pi$ the identity morphism of $x$. $\endgroup$ – მამუკა ჯიბლაძე Jun 23 at 19:21
  • $\begingroup$ @მამუკაჯიბლაძე, the subobject $u$ is complemented by hypothesis. $\endgroup$ – Mendieta Jun 23 at 19:30
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A different flavour of counterexample from Andreas Blass’s answer, showing this can fail when $S$ is Boolean and satisfies choice: Take $S$ to be sets, and $E = Sh(2^{\mathbb{N}})$, where $2^{\mathbb{N}}$ is the Cantor space (and $f$ is the unique geometric morphism $(\Gamma,\Delta) : Sh(2^{\mathbb{N}}) \to S$).

Now for $n \in \mathbb{N}$, we have $U_n \subseteq 1$ in $E$ corresponding to the clopen $\{ x \in 2^\mathbb{N}\ |\ x_n = 1 \}$. Now each $U_n \subseteq 1$ is complemented, so $\coprod_n U_n \subseteq \Delta(n) \times 1$ is complemented; but $\pi_!(\coprod_n U_n) = \bigcup_n U_n \subseteq 1$ is not complemented, since it’s dense but not equal to $1$.

Generally, sheaf toposes give many counterexamples to the principle “set-indexed unions of complemented subobjects are complemented”.

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  • $\begingroup$ please confirm: you mean ${\coprod_n U_n \subseteq \Delta(\mathbb{N}) \times 1}$ ? $\endgroup$ – Mendieta Jun 23 at 22:13
  • $\begingroup$ @Mendieta: Yes — sorry, am on mobile so editing/proofreading the maths is tricky. $\endgroup$ – Peter LeFanu Lumsdaine Jun 23 at 22:21
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    $\begingroup$ ok. Good example, many thanks. Yet, it is the sort of thing I wanted to avoid with hyperconnectedness. I am not sure how MO works so I hope the following is appropriate: I'll arrow up your answer, consider the question answered by Blass and ask a new question with the further hypotheses. $\endgroup$ – Mendieta Jun 23 at 22:30

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