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Let $S$ be a Boolean topos.

Let ${f : E \rightarrow S}$ be a hyperconnected geometric morphism.

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 $\pi$ a complemented subobject of $x$?

(Note: this is a continuation of Images of complemented subobjects in toposes)

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No.

Take $S$ to be Sets, then for any set $s$, $f^* s \times x$ is the coproduct of $s$-copies of $X$, and a complemented subobject of $f^* s \times x$ is the same as an $s$ indexed collection of complemented subobject of $x$. The the image by the projection $f^*s \times x \to x$ is their union.

So, for $S= Set$, your question translate into: Are complemented subobject closed under union in hyperconnected topos ? This is not the case.

For example, the Jonsson-Tarski topos is an hyperconnected topos, which has a slice equivalent to Sh(K) for K the Cantor space. In particular it has an object $X$ such that the subobject of $X$ identifies with the open subset of the Cantor space and every open subset of the Cantor is a union of clopen (i.e. complemented open)

Edit: It was asked in the comment to give a description of the left adjoint $f^*$ from Sets to JT of the global section functor $f_*$ topos. It happens that there is a fairly explicit description of it. The trick is that the object $X$ such that $JT/X = Sh(K)$ is actually the free JT algebra on one generator. Meaning that given an object $Z$ of JT one recover the corresponding JT-algebra by looking at $Hom(X,Z)$.

So what you are asking is what is $Hom(X,f^*s)$, Let me call $h$ the geometric morphisms $JT/X \to JT$ and $g:JT/X = Sh(K) \to Sets$ the unique geometric morphisms. Now, as $h$ is étale, it has an exceptional further left adjoint $h_!$ and $X= h_! 1$, so, $Hom(X,Z)$ can be more generally written as $Hom(h_! 1 ,Z ) = Hom(1,h^*Z)$ and hence $Hom(X,f^*s) = Hom(1,h^* f^* s) = Hom(1,g^* s) $

And $g^*s$ is the sheaf of locally constant functions with values in $s$ on $K$, so $Hom(1,g^*s)$ is the set of locally constant functions from $K$ to $s$.

So at the end of the day one has:

$$ f^* (s) = \{ \text{locally constant functions } K \to s\} $$

where $K=2^\mathbb{N}$ is the Cantor space. With some additional calculation (related to how $JT/X$ identifies with $Sh(K)$), one can show that the JT algebra structure on $f^*(s)$ is obtained from the homeomorphisms $K = K \coprod K$ obtained by separating along the first component of $K = 2^\mathbb{N}$

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  • $\begingroup$ Simon Henry, how do you prove that the JT topos is hyperconnected? $\endgroup$
    – Mendieta
    Commented Jun 23, 2021 at 23:16
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    $\begingroup$ Its terminal object is the unique Jonsson-Tarski algebra structure on the singleton, and its only subobjects are $0$ or itself. Having exactly two subterminal objects is the same as being hyperconnected. If you prefer groupoids, You can also use its description as an étendu given by Freyd (see the nLab link) and observe that the corresponding groupoid has no non-trivial invariant open subsets, but that much more invovled. $\endgroup$
    – Simon Henry
    Commented Jun 24, 2021 at 2:23
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    $\begingroup$ The localic reflection of a topos is given by $f_*(\Omega) = Sub(1)$. So a topos has a trivial localic reflection if and only if the terminal object as exactly two subobject $\emptyset$ and $1$ (and $\emptyset \neq 1$). (see also Prop A.4.6.6.(v) in Sketches of an elephant that directly claim that a morphism is hyperconnected if and only if $f_*$ preserve the subobject classifier) $\endgroup$
    – Simon Henry
    Commented Jun 24, 2021 at 14:53
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    $\begingroup$ Subobjects of $X$ are the same as subterminal object of the topos $JT/X$ and subterminal objects of a sheaf topos Sh(K) (for any space $K$) corresponds to open subsets of $K$. So from the equicalence $JT/X \simeq Sh(K)$ you can identify open subsets of $K$ with subobjects of $X$. $\endgroup$
    – Simon Henry
    Commented Jun 24, 2021 at 17:15
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    $\begingroup$ This was too long for a comment, so I've answered as an edit. $\endgroup$
    – Simon Henry
    Commented Jun 24, 2021 at 20:07

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