Images of complemented subobjects in toposes

Question

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)

Answer 1

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.

Answer 2

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”.