this entry is about the concept in topology, especially in point-free topology. For the concept in physics/chemistry see nucleus (physics), or see nucleus (disambiguation).

Recall that frames are dual to locales, and locales are kinds of spaces. So, if you adopt locales as models for spaces, then your models for subspaces are sublocales or, dually, quotient frames. However, much as a quotient set can be described by an equivalence relation on the original set, so a quotient frame may be described by an appropriate structure on the original frame. That structure is a *nucleus*.

Thus, nuclei correspond to sublocales.

Let $L$ be a frame, that is a suplattice satisfying the infinite distributivity law.

A **nucleus** on $L$ is a function $j\colon L \to L$ which is

- monotone, i.e. $j(a \wedge b) = j(a) \wedge j(b)$ for all $a,b\in L$,
- inflationary, $a \leq j(a)$ for all $a\in L$,
- satisfies $j(j(a)) \leq j(a)$ for all $a\in L$.

In other words, a nucleus on $L$ is a meet-preserving monad on $L$. If only the first two properties are satisfied, $j$ is sometimes called an **inflator**.

Note that the following properties of a nucleus might be included in the definition, but they follow from the above:

- $j(\top) = \top$,
- $j(a) \leq j(b)$ if $a \leq b$,
- (idempotency) $j(j(a)) = j(a)$.

Thus one sometimes defines a nucleus as a monotone, inflationary and idempotent (endo)function on $L$

Let $L$ be a frame.

As a nucleus $j$ on $L$ (being a monad on a poset) is a kind of Moore closure, we say that an element $a$ of $L$ is **$j$-closed** if $j(a) = a$. (But note that this has nothing to do with the closed subspaces of the locale $L$.)

We may equivalently define a nucleus on $L$ to be a subset $J$ of $L$ that satisfies certain conditions, namely these identities:

- $\bigwedge A \in J$ whenever $A \subseteq J$ (using that $L$ is a complete lattice),
- $a \Rightarrow b \in J$ whenever $b \in J$ (using that $L$ is a Heyting algebra).

Then we recover $j\colon L \to L$ by

$j(a) \coloneqq \bigwedge \{ b\colon L \;|\; b \in J,\; a \leq b \}$

and we have

$J = \{ a \colon L \;|\; j(a) = a \} .$

Check all this, and expand on it if necessary.

This approach to nuclei is not appropriate in a predicative approach to topology, where we want to use large (but accessible) frames, which may not be meet-complete.

Let $L$ be a frame, and let $j$ be a nucleus on $L$.

Let $L/j$ be the subset of $L$ consisting of the $j$-closed elements of $L$ (those elements $a$ such that $j(a) = a$). Note that, by property (3) above, we may interpret $j$ as a function $j^*\colon L \to L/j$, which is a surjective frame homomorphism. Since Frm is an algebraic category, this means that $L/j$ is a regular quotient of $L$.

Conversely, suppose that $M$ is any regular quotient of $L$; that is, we have a surjective frame homomorphism $k\colon L \to M$. Since $k$ is a frame homomorphism, it has (by the adjoint functor theorem) a right adjoint $k_*\colon M \to L$. Let $j\colon L \to L$ be the composite of $k$ followed by $k_*$. Then $j$ is a nucleus, and $k_*$ is an embedding (in Pos, not $Frm$) whose image is $L/j$.

In short, given a nucleus $j$, we have an adjunction $j^*\colon L \rightleftarrows L/j: j_*$, where $j^*$ is a surjective homomorphism and $j_*$ is the inclusion function; while, given a surjective homomorphism $k$, we have an adjunction $k\colon L \rightleftarrows M: k_*$, where $k_* \circ k$ is a nucleus and $k_*$ is an embedding.

If we think of $L$ as a locale, then we define a **sublocale** of $L$ to be a quotient frame of $L$, which corresponds to a nucleus on $L$ as above.

Quantales are a generalization of locales. The corresponding version of a nucleus is sometimes called quantic nucleus.

If you categorify from locales to toposes, then nuclei become Lawvere–Tierney topologies, and the operation of the nucleus becomes sheafification.

*Stone Spaces*, II.2- André Joyal,
*A crash course on topos theory*, part 1/4, IHES 2006 yt

Last revised on October 10, 2023 at 11:03:43. See the history of this page for a list of all contributions to it.