# Images of complemented subobjects in toposes

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 $$x$$?

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.

• 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?) – Mendieta Jun 22 at 23:43
• @Mendieta If $S$ is Boolean (which is weaker than having AC), every subobject of every object is complemented. – მამუკა ჯიბლაძე Jun 23 at 6:46
• @მამუკაჯიბლაძე What you say is true, but the question still makes sense because E need not be Boolean. – Mendieta Jun 23 at 17:29
• @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 $x$. – მამუკა ჯიბლაძე Jun 23 at 19:21
• @მამუკაჯიბლაძე, the subobject $u$ is complemented by hypothesis. – Mendieta Jun 23 at 19:30

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”.

• please confirm: you mean ${\coprod_n U_n \subseteq \Delta(\mathbb{N}) \times 1}$ ? – Mendieta Jun 23 at 22:13
• @Mendieta: Yes — sorry, am on mobile so editing/proofreading the maths is tricky. – Peter LeFanu Lumsdaine Jun 23 at 22:21
• 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. – Mendieta Jun 23 at 22:30