Contents

topos theory

# Contents

## Idea

The concept of dense subtopos generalizes the concept of a dense subspace from topology to toposes. A subtopos is dense if it contains the initial object $\emptyset$ of the ambient topos.

## Definition

Let $i:\mathcal{E}_j\hookrightarrow \mathcal{E}$ be a subtopos with corresponding Lawvere-Tierney topology $j$. $\mathcal{E}_j$ is called dense if the following equivalent conditions hold:

• $j\circ\bot=\bot$ (with $\bot: 1\to\Omega$ the classifying map of $\emptyset\rightarrowtail 1$).

• $\emptyset\rightarrowtail 1$ is $j$-closed.

• $\emptyset$ is a $j$-sheaf.

• the direct image satisfies $i_\ast (\emptyset)\simeq\emptyset$.

• the inverse image satisfies: from $i^\ast (Z)\simeq\emptyset$ follows $Z\simeq\emptyset$.

A topology $j$ satisfying these conditions is also called dense.

### Remark

$j\circ\bot$ classifies the $j$-closure $\bar{\emptyset}$ of $\emptyset$ whence $j\circ\bot = \bot$ iff $\bar{\emptyset}=\emptyset$ i.e. $\emptyset\rightarrowtail 1$ is $j$-closed. Since $1$ is a $j$-sheaf for any topology $j$ and subobjects of $j$-sheaves in general are $j$-closed precisely when they are $j$-sheaves, this is equivalent to $\emptyset$ being a $j$-sheaf. Another way to say this is that $\emptyset$ is preserved by $i_\ast$.

The equivalence between the last two formulations follows from the adjunction $i^\ast\dashv i_\ast$ and the strictness of $\emptyset$ in a topos: $id_{i_\ast(\emptyset)}$ corresponds under the adjunction to a map $i^\ast( i_\ast (\emptyset))\to \emptyset$ showing that $i^\ast (i_\ast(\emptyset))\simeq\emptyset$ in general. Conversely, $id_{i^\ast (Z)}$ corresponds to a map $Z\to i_\ast(i^\ast(Z))$ showing that $Z\simeq\emptyset$ provided $i^\ast(Z)\simeq\emptyset$ and $i_\ast(\emptyset)\simeq\emptyset$.

In SGA4 (p.430) another equivalent formulation is on offer, namely it suffices to check the last condition on subterminal objects $Z$ (because $i_\ast(\emptyset)$ is a subterminal in general since $i_\ast$ as a right adjoint preserves monos hence subterminals). An even more comprehensive list can be found in (Caramello 2012, p.9).

The last two conditions make sense not only for embeddings: general geometric morphisms fulfilling them are called dominant. So another way to express that $i:\mathcal{E}_j\hookrightarrow\mathcal{E}$ is a dense subtopos is to say that the inclusion $i$ is dominant.

Notice that there is also a certain Grothendieck topology on small categories $\mathcal{C}$ called the dense topology whose corresponding Lawvere-Tierney topology on $Set^{\mathcal{C}^{op}}$ is dense in the above sense, and coincides with the double-negation topology $\neg\neg$ on $Set^{\mathcal{C}^{op}}$.

## Properties

### Some basic observations

Of course, the composition $k\circ j$ of dense inclusions $j,k$ is again dense. Conversely, we have

###### Proposition

Given the following commutative diagram

$\array{ \mathcal{E}_i &\overset{j}{\to}& \mathcal{E} \\ &{i}{\searrow}& \uparrow k \\ & & \mathcal{E}_k }$

where $i,j,k$ are subtopos inclusions. When $j$ is dense, then $i$ and $k$ are dense as well.

Proof: Suppose $\emptyset =i^\ast (Z)$. Since $k$ is an inclusion we have that the counit $\epsilon:k^\ast k_\ast\to id_{\mathcal{E}_k}$ is a natural isomorphism whence $Z= k^\ast k_\ast (Z)$ and therefore

$\emptyset =i^\ast(Z)=i^\ast(k^\ast k_\ast (Z))=i^\ast k^\ast (k_\ast (Z))=j^\ast(k_\ast(Z))\quad .$

From which $k_\ast(Z)=\emptyset$ since $j$ is dense by assumption.

Then $Z=k^\ast k_\ast (Z)=k^\ast (\emptyset)=\emptyset$ since $k^\ast$ preserves colimits, in other words, we have shown that $i$ is dense.

But from $k_\ast (Z)=\emptyset$ and $Z=\emptyset$ follows that $k$ is dense as well. $\qed$

Given two dense topologies $j_1$, $j_2$ on a topos $\mathcal{E}$, their join $j_1\vee j_2$ is again dense.

This follows from the general fact that $Sh_{j_1\vee j_2}(\mathcal{E})$ corresponds to the meet, i.e. the intersection of the corresponding subtoposes $Sh_{j_1}(\mathcal{E})\cap Sh_{j_2}(\mathcal{E})$ in the lattice of subtoposes, and this obviously contains $\emptyset_\mathcal{E}$ for $j_1$, $j_2$ dense.

In other words, the intersection of two dense subtoposes is still dense!

Somewhat surprisingly, this still holds if one takes the intersection of all dense subtoposes, as the next section details.

### Relation to double-negation topology

For any topos $\mathcal{E}$, its double negation topology gives the smallest dense subtopos. This agrees with the situation for locales but contrasts with the situation for topological spaces where, in general, smallest dense subspaces do not exist.

###### Proposition

$Sh_{\not\not}(\mathcal{E}) \hookrightarrow \mathcal{E}$ is the smallest dense subtopos.

In fact, dense topologies are characterized by their relation to $\neg\neg$:

###### Proposition

Let $\mathcal{E}$ be a topos. A topology $j$ satisfies $j\le\neg\neg$ , i.e. $j$ is dense, iff $(\mathcal{E}_j)_{\not\not}=\mathcal{E}_{\not\not}$.

From this and the fact that $\mathcal{E}$ is trivially dense, follows:

###### Proposition

A topos $\mathcal{E}$ is Boolean iff $\mathcal{E}$ has exactly one dense subtopos, namely $\mathcal{E}_{\neg\neg}=\mathcal{E}$.

Notice that, though these results prevent a topos from having more than one dense Boolean subtopos, nothing prevents a topos from having more than one Boolean subtopos e.g. the Sierpinski topos $Set^{\to}$ has two non trivial ones that complement each other in the lattice of subtoposes. This example, incidentally, also shows that in the above proposition just $(\mathcal{E}_j)_{\neg\neg}\cong\mathcal{E}_{\neg\neg}$ wouldn’t do.

### The (dense,closed)-factorization

$Sh_j(\mathcal{E}) \hookrightarrow \mathcal{E}$

factors as

$Sh_j(\mathcal{E}) \hookrightarrow Sh_{c(ext(j))}(\mathcal{E}) \hookrightarrow \mathcal{E}$

where $ext(j)$ (the “exterior” of $j$) denotes the $j$-closure of $\emptyset \rightarrowtail 1$ and

$\bar j \coloneqq c(ext(j))$

the closed topology corresponding to the subterminal object $ext(j)$.

Here the first inclusion exhibits a dense subtopos and the second a closed subtopos.

This is the so called (dense,closed)-factorization and implies e.g. that proper dense subtoposes aren’t closed.

Dense inclusions participate also in the description of skeletal inclusions as the closure of open inclusions under composition with dense inclusions.

### Some parallels to topology

The above terminology suggests to view a dense subtopos as one with an empty exterior.

This analogy to topology is pursued further in (SGA4, p.462) where a dense subtopos is characterized as a subtopos $Sh_j(\mathcal{E})$ whose ‘exterior’ $Ext(Sh_j(\mathcal{E}))$ (i.e. the open subtopos that corresponds to the subterminal object $ext(j)$) is trivial and whose ‘closure’ $Cl(Sh_j(\mathcal{E})):=Sh_{c(ext(j))}(\mathcal{E})$ (i.e. the closed subtopos corresponding to $ext(j)$) coincides with the ‘whole space’ $\mathcal{E}$.

Let’s have a look at some of the details:

Due to the construction of open subtoposes we know that the objects of $Ext(Sh_j(\mathcal{E}))$ have the form $X^{ext(j)}$ for some $X\in\mathcal{E}$. Hence the exterior is trivial, i.e. $X^{ext(j)}=1$ for all $X\in\mathcal{E}$, precisely when $\bar\emptyset =ext(j)\simeq \emptyset$ which means that $Sh_j(\mathcal{E})$ is dense. By construction $Cl(Sh_j(\mathcal{E}))$ is the complement of $Ext(Sh_j(\mathcal{E}))$ in the lattice of subtoposes hence $Cl(Sh_j(\mathcal{E}))=\mathcal{E}$ in case the latter is trivial. This follows also directly from the description of objects in $Cl(Sh_j(\mathcal{E}))$ as those objects $X\in\mathcal{E}$ with $X\times ext(j)\cong ext(j)$.

E.g. let $Sh_k(\mathcal{E})$ be a subtopos that has a trivial intersection with a non-trivial open subtopos $Sh_o(\mathcal{E})$. Then $Sh_k(\mathcal{E})$ is contained in the (closed) complement of $Sh_o(\mathcal{E})$ hence $Cl(Sh_k(\mathcal{E}))\neq \mathcal{E}$ and we see that $Sh_k(\mathcal{E})$ cannot be dense: we have recuperated the familiar fact from point-set topology that a dense subset intersects all non-trivial open sets non-trivially.

Another easy result in this vein is

###### Proposition

Let $i:Sh_j(\mathcal{E})\hookrightarrow\mathcal{E}$ be a dense subtopos that is connected in the sense that $1$ is indecomposable: if $W\coprod Z=1$ then $W=\emptyset$ or $Z=\emptyset$. Then $\mathcal{E}$ is connected as well.

Proof: Let $X\coprod Y = 1$ be a decomposition of $1$ in $\mathcal{E}$. Since $i^\ast$ is a left exact left adjoint, it preserves coproducts and the terminal object and $i^\ast(X)\coprod i^\ast (Y)$ is therefore a decomposition of $1$ in $Sh_j(\mathcal{E})$ hence trivial by assumption. Let’s say $i^\ast(X)\simeq\emptyset$ but $Sh_j(\mathcal{E})$ is dense and therefore we can conclude $X\simeq\emptyset$ hence $X\coprod Y = 1$ is trivial as well. $\qed$

### Example I: two-valued toposes

Presheaf toposes $Set^{M^{op}}$ of actions of a monoid $M$ are classical examples of toposes whose truth value objects $\Omega$ have exactly two global points $1\to\Omega$ without the toposes being necessarily Boolean. In fact they are Boolean precisely when $M$ is a group.

As the next proposition shows, they are also instances of toposes in which only the degenerate subtopos fails to be dense:

###### Proposition

The non-degenerate subtoposes $Sh_j(\mathcal{E})$ of a two-valued topos $\mathcal{E}$ are precisely the dense subtoposes of $\mathcal{E}$.

Proof: Truth values $1\to\Omega$ correspond precisely to subobjects of $1$. Hence the $j$-closure $ext(j)$ of $\emptyset\rightarrowtail 1$ is either $\emptyset$ or $1$. In the first case, $Sh_j(\mathcal{E})$ is dense, in the second, from $X\times 1=1$ for $X\in Cl(Sh_j(\mathcal{E}))$ follows triviality. $\qed$

Combining this with the above shows that two-valued and Boolean toposes are opposite extremes when it comes to dense subtoposes and the following observation (cf. Caramello (2009); prop. 10.1) follows immediately:

###### Proposition

A topos $\mathcal{E}$ that is two-valued and Boolean has no non-trivial subtoposes. $\qed$

In other words, two-valued Boolean toposes are atoms in the lattice of subtoposes. Notice that this applies e.g. to well-pointed toposes.

### Example II: persistent localizations

Recall that a persistent localization is given by a Lawvere-Tierney topology $j$ with the property that every $j$-separated object is a $j$-sheaf. But separated objects are closed under taking subobjects and therefore in the case of persistent $j$, subobjects of $j$-sheaves are themselves $j$-sheaves.

In particular, this applies to $\emptyset\rightarrowtail 1$, since $1$ is always a sheaf. Whence $\emptyset$ is a $j$-sheaf and we see that persistent localizations are dense. This includes e.g. ‘quintessential localizations’ aka quality types.

This observation is due to Johnstone (1996).

By the above proposition it follows immediately that every persistent localization of a Boolean topos is trivial.

### Relation to Aufhebung

Notice that, since the localization $L$ corresponding to a subtopos is a left exact functor, all subtoposes necessarily contain the terminal object $\ast$ of the ambient topos. Moreover, the idempotent comonad and idempotent monad constant on the initial object and terminal object, respectively, are adjoint to each other (forming an adjoint modality). Denoting by “$\vee$” the inclusion of modal objects, then the general situation for any subtopos localized on by $L$ is depicted by

$\array{ && L \\ && \vee \\ \emptyset &\dashv& \ast } \,.$

In view of this, the subtopos being dense says that not only $\ast$, but this whole adjoint modality that it participates in sits inside the subtopos. Lawvere had proposed to call this situation resolution or (a special minimal version of it) Aufhebung of the unity of opposites expressed by $\emptyset \dashv \ast$ (“becoming”).

In other words, for an essential subtopos being dense is equivalent to resolve $\emptyset \dashv \ast$ in the Hegelian calculus of levels!

#### Example: essential subtoposes of the topos of globular sets

Kennett-Riehl-Roy-Zaks (2011) show that in the gros topos $Set^{\mathcal{G}^{op}}$ of reflexive globular sets essential subtoposes correspond to dimensional truncations (plus the level ${Set^{\mathcal{G}^{op}}}$ ‘at infinity’). Then level $n+1$ is the Aufhebung of $n$ starting from $\emptyset\dashv\ast$ at level $0$. In general, the Aufhebung $\bar{l}$ of a level $l$ resolves all the levels that $l$ resolves. Therefore in $Set^{\mathcal{G}^{op}}$ all essential subtoposes (above 0) resolve $\emptyset\dashv\ast$ and hence are dense!

## References

Last revised on May 20, 2018 at 08:40:47. See the history of this page for a list of all contributions to it.