(The following definitions are meant to be standard and are reproduced for completeness of the question.) A **frame** is a partially ordered set in which every finite subset has a greatest lower bound (“meet”, denoted $x\_1\wedge\cdots\wedge x\_n$, or $\top$ for the empty meet) and every subset has a least upper bound (“join”, denoted $\bigvee\{x\_i : i\in I\}$), and such that the latter distributes over the former, which amounts to demanding $z \wedge \bigvee \{x\_i : i \in I\} = \bigvee\{(z\wedge x\_i) : i \in I\}$ (in fact, arbitrary meets exist in a frame, but they generally fail to distribute over joins). (An important source of examples of frames are the lattice of open sets on a topological space, in which case finite meets and arbitrary joins are simply finite intersections and arbitrary unions.) The **Heyting operation** on a frame $L$, denoted $\Rightarrow$, is defined as $(x\Rightarrow z) := \bigvee\{y : x\wedge y \leq z\}$ (this sup is, in fact, a max). A **nucleus** on a frame $L$ is a map $j\colon L\to L$ such that ① $j(x\wedge y) = j(x)\wedge j(y)$ (from which it follows that $j$ is order-preserving), ② $x \leq j(x)$, and ③ $j(j(x)) = j(x)$ (these three conditions are understood universally quantified over free variables in $L$). The set of nuclei on $L$, with the pointwise order (viꝫ., $j\_1\leq j\_2$ when $j\_1(x)\leq j\_2(x)$ for all $x\in L$) is itself a frame, denoted $N(L)$; its top element is $x \mapsto \top\_L$ and bottom $x \mapsto x$. Meet of nuclei can be defined pointwise: $(j\_1\wedge j\_2)(x) = j\_1(x) \wedge j\_2(x)$ (in fact, the same holds for arbitrary meets). Joins of nuclei, however, are more difficult to compute in general, although it is true that the fixset $L\_j := \operatorname{im} j = \{x\in L : j(x)=x\}$ of $j := \bigvee\{j\_i : i\in I\}$ is $\bigcap\_{i\in I} L\_{j\_i}$ (so $j(x)$ can be expressed as the smallest element $\geq x$ of $L$ which is fixed by every $j\_i$). See [Escardó, “Joins in the Frame of Nuclei” (2003)](https://doi.org/10.1023/A:1023555514029) for details and references about nuclei in general and how to compute their join.


My question concerns the computation of the Heyting operation in the frame $N(L)$ of nuclei over a frame $L$. So $j\_1\Rightarrow j\_2$ is defined as the largest nucleus $k$ such that $j\_1(x) \wedge k(x) \leq j\_2(x)$ for all $x$ (in particular, we have $(j\_1\Rightarrow j\_2)(x) \leq (j\_1(x) \Rightarrow j\_2(x))$). But this is not an equality in general.


To provide at least one interesting example of this, for an arbitrary element $a\in L$ we two naturally occurring nuclei, $j\_a\colon x\mapsto a\vee x$ (the “complementary closed sublocale” nucleus associated to $a$) and $j^a\colon x\mapsto (a\Rightarrow x)$. They turn out to be complementary in $N(L)$ (see, e.g., Fourman & Scott, “Sheaves in Logic”, p. 302–401 in Fourman, Mulvey & Scott, *Applications of Sheaves (Durham 1977)*, Springer LNM **753** (1979), §2.18), in the sense that $j\_a \vee j^a = \top\_{N(L)}$ and $j\_a \wedge j^a = \bot\_{N(L)}$; in particular, we have $j^a = (j\_a \Rightarrow \bot\_{N(L)})$, which is unsurprising, but also $j\_a = (j^a \Rightarrow \bot\_{N(L)})$, which is more surprising as the inequality $a\vee x \leq ((a\Rightarrow x) \Rightarrow x)$ is generally not an equality (the RHS here is not a nucleus in $x$, although it *is* a nucleus in $a$ which caused me some level of confusion).


**Question:** How can we compute the value $(j\_1\Rightarrow j\_2)(x)$ of the Heyting operation for $j\_1,j\_2 \in N(L)$ two nuclei (and $x\in L$)? Or how can we, at least, say something beyond the inequality $(j\_1\Rightarrow j\_2)(x) \leq (j\_1(x) \Rightarrow j\_2(x))$ remarked above? Pretty much anything that can be said about $(j\_1\Rightarrow j\_2)(x)$ interests me, including the special cases $j\_2 = \bot\_{N(L)}$, and/or when $L$ is the lattice of open sets on a topological space $X$ and $j\_1,j\_2$ are defined by reflecting to open subsets of subspaces $E\_1,E\_2 \subseteq X$.



Computing the Heyting operation on the frame of nuclei
First, every nucleus is the join of those of the form $j^x\land j\_y$.
Namely,
$$
j=\bigvee\_xj^x\land j\_{jx}
$$
Hm, this is too confusing. Let me change notation and write: $u\_a$ instead of $j^a$; $v\_a$ instead of $j\_a$. Thus,
$$
j=\bigvee\_xu\_x\land v\_{jx}.
$$


It then follows
$$
j\_1\Rightarrow j\_2=\bigwedge\_x[(u\_x\land v\_{j\_1x})\Rightarrow j\_2].
$$
Next, since $u\_x$ and $v\_x$ are complements of each other, $u\_x\Rightarrow j=v\_x\lor j$ and $v\_x\Rightarrow j=u\_x\lor j$. Thus $[u\_x\Rightarrow j]a=j(a\lor x)$ and $[v\_x\Rightarrow j]a=x\Rightarrow ja$ for every $a$, $x$ and every nucleus $j$.


To show that $j(x\lor-)$ is a nucleus,
$$
j(x\lor j(x\lor a))=jj(x\lor a)=j(x\lor a)
$$
since $x\leqslant jx\leqslant j(x\lor a)$.


And, to show that $x\Rightarrow j-$ is a nucleus,
$$
x\Rightarrow j(x\Rightarrow ja)=x\Rightarrow(x\Rightarrow ja)=x\Rightarrow ja
$$
since $L\_j$ is an exponential ideal, i. e. $b\in L\_j$ implies $x\Rightarrow b\in L\_j$ for any $x$.


This gives
$$
[j\_1\Rightarrow j\_2]a=\bigwedge\_x j\_1x\Rightarrow j\_2(a\lor x),
$$
which can be also replaced by
$$
\tag{\*}
[j\_1\Rightarrow j\_2]a=\bigwedge\_{x\geqslant a}j\_1x\Rightarrow j\_2x,
$$
as well as
$$
[j\_1\Rightarrow j\_2]a=\bigwedge\_x(a\Rightarrow x)\Rightarrow(j\_1x\Rightarrow j\_2x).
$$


One can also use that every nucleus is a meet of nuclei of the form $w\_x$, where $w\_x(a)=(a\Rightarrow x)\Rightarrow x$. Namely,
$$
j=\bigwedge\_{x\in L\_j}w\_x.
$$
Using this, we can further replace $(\*)$ with
$$
[j\_1\Rightarrow j\_2]a=\bigwedge\_{x\geqslant a,y\in L\_{j\_2}}j\_1x\Rightarrow((a\Rightarrow y)\Rightarrow y)$$
or as well with
$$
[j\_1\Rightarrow j\_2]a=\bigwedge\_{x\geqslant a,y\in L}j\_1x\Rightarrow((a\Rightarrow j\_2y)\Rightarrow j\_2y).$$


The version $(\*)$ is the most concise but sometimes other versions might be more useful.


For references - all this must be in the book by Picado and Pultr, for example.


