# Is the collage of two spatial toposes a spatial topos?

Consider the collage operation along a profunctor, defined between two categories ${\bf C}, \bf D$.

Suppose now that the two categories are toposes, say $Sh(X), Sh(Y)$ for two topological spaces $X,Y$ and that the profunctor between them is the terminal one (sending every object $(C,D)$ to the singleton set): this collage is precisely the join of the two categories in study. It's rather intuitive to represent this collage as the disjoint union of two "layers": the upper category $\mathbf C$ and the lower category $\mathbf D$; between any object in the upper layer and any object in the lower layer there is exactly one arrow.

Now there is a chain of functors $$\mathbf{Topos}\times \mathbf{Topos}\xrightarrow{\circledast} \mathbf{Cat}\xrightarrow{N_s}\mathbf{sSet}\xrightarrow{R_t} \mathbf{Topos}$$ where

1. $N_s$ is the classical simplicial set taking the nerve of a category, whose $n$-simplices are $n$-tuples of composable arrows in that category;
2. $R_t$ is the functor which topos-theoretically realizes (clic) a simplicial set; this functor is obtained by Yoneda-extending the functor $j\colon\Delta\to \mathbf{Topos}$ sending $[n]$ to $\text{Sh}(\Delta^n)$, or in other words $$(P\in \mathbf{sSet})\mapsto \int^{n:\Delta} \mathbf{sSet}([n], P)\cdot \text{Sh}(\Delta^n)\cong \int^{n:\Delta} P_n \cdot \text{Sh}(\Delta^n)$$

Rather surprisingly, it's simple to determine the $n$-simplices of the category $\mathcal E\circledast \mathcal F$: whenever a sequence of $n$ arrows "descends" to the lower category, obviously there's now ay it can go up again, so from now on the sequence of composable arrows is completely determined by its terminal point.

This leads to represent the $n$-simplices as $$N(\mathcal E\circledast\mathcal F)_n \cong N\mathcal F_n\cup N\mathcal E_n\cup\bigcup_{k=1}^{n-1}\Big(N\mathcal E_k\times N\mathcal E_0\Big)$$ In such a way that the coend in study simplifies to \begin{align*} R_t(N(\mathcal E\circledast \mathcal F)) &\cong \int^{n:\Delta} N(\mathcal E\circledast \mathcal F))_n\cdot \text{Sh}(\Delta^n)\\ &\cong \int^{n:\Delta}\Big(\Big( \underbrace{N\mathcal E_0 \times \bigcup_{k=1}^{n-1} N\mathcal E_k }_{A_n}\Big)\cup N\mathcal F_n\cup N\mathcal E_n\Big)\cdot \text{Sh}(\Delta^n)\\ &\cong \int^{n:\Delta} A_n\cdot \text{Sh}(\Delta^n) \cup \int^{n:\Delta} N\mathcal E_n\cdot \text{Sh}(\Delta^n)\cup \int^{n:\Delta} N\mathcal F_n\cdot \text{Sh}(\Delta^n)\\ & = R_t(N\mathcal E)\amalg R_t(N\mathcal F)\amalg \mathcal K_{\mathcal E,\mathcal F} \end{align*} I think this is a good point to stop: I'm simply doodled with symbols, and I got this fascinanting thing. But now I have to come up with some questions:

1. The coend manipulation seems to be right, but I am too amazed by this "inspiring" result to rely on my ability. Do you see any evident flaw?
2. Has this "curvature" (!) term $\mathcal K_{\mathcal E,\mathcal F}$ any meaning? If $A_n$ happens to be a simplicial set, then it is the toposophic-realization of a third simplicial set $K_{\mathcal E,\cal F}$, which should be interpreted as...?
3. Does this construction (provided it is meaningful) can be taken as "the best topos associated to the category $\mathcal E\circledast \mathcal F$"?
4. Does this topos happen to be spatial? And how can I prove that it is? Has the space $X\boxplus Y$ such that $Sh(X)\circledast Sh(Y)\cong Sh(X\boxplus Y)$ the meaning of a "gluing" of the spaces?
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Why not just try this: take disjoint union of X and Y and declare opens to be those of X together with the sets $X\cup U$ for all opens $U$ of $Y$. –  მამუკა ჯიბლაძე Feb 16 '14 at 11:46
I am pleased to have an answer from you, prof. Jibladze! And I like the idea very much. Is there a name for this topology on $X\amalg Y$? Does it realize some universal property? –  tetrapharmakon Feb 16 '14 at 12:19
Again, you are neglecting unavoidable size issues... –  Zhen Lin Feb 16 '14 at 12:34
@tetrapharmakon Putting several posets (lattices, Heyting algebras, ...) on top of one another is what Esakia liked to call a cascade; it is though a particular case of what you call join. As for this case I believe it is nothing but gluing along the functor which sends 1 to 1 and everything else to 0. If correct, this then gives the usual universal property enjoyed by glueings. –  მამუკა ჯიბლაძე Feb 16 '14 at 13:26
@ZhenLin: I remember the problem in the other thread was the lack of (1-)cocompleteness for $\mathbf{Topos}$; but maybe I misunderstood the solution, as I remember that you proposed to bound the functor $j$ to spatial toposes (and your objection involved the "toposic nerve" more than the realization)... I'm glad to accept objections and advices! –  tetrapharmakon Feb 16 '14 at 14:46