The canonical topology on a complete heyting algebra

category-theorygrothendieck-topologiesheyting-algebra

I have troubles with catching the dividing line on Grothendieck topologies over posets between subcanonical and not subcanonical. In particular I have the following very specific basic question which I have not been able to trace in the literature:

Is the sup-topology on a complete Heyting algebra the canonical Grothendieck topology attached to it?

My understanding is that this is the case. Below I sketch an argument of why this topology is subcanonical, I still miss the argument of why it should be the canonical topology. I appreciate even proof-checks that my argument below is correct if not a complete answer to my question.

My understanding is that if $(H,\wedge,\bigvee,\leq,0,1)$ is a complete Heyting algebra, the non-trivial sieves on $0$ are $\emptyset$ and $\{0\}$.

Now the request that a Grothendieck topology $J:H\to \mathcal{P}(\mathcal{P}(H))$ is sub-canonical amounts to ask that $\mathrm{Y}(p)$ is a $J$-sheaf for any $p\in H$, where
$$
\mathrm{Y}(p)=\{\mathrm{Hom}_H(q,p):q\in H\}
\cong
\{q\mapsto (q\leq p): q\leq p\}
\cup
\{q\mapsto\emptyset: q\not\leq p\}
$$

is the Yoneda embedding $\mathrm{Y}:H\to\mathrm{Set}^{H^{op}}$ evaluated at $p$.
In case of the sup-topology $J$ on $H$ defined by
$$
J(p)=\{D\subseteq p: D \text{ is downward closed and} \bigvee D=p\},
$$

it appears to me that the unique non-trivial check occurs for $p=0$.

In this case, we get that $\emptyset$ is in $J(0)$, as $\bigvee\emptyset=0$ and "$\emptyset$ is downward closed" both hold
by definition: for example $\bigvee \emptyset$ is the infimum of the set of $p\in H$ such that $\forall x\in H (x\in\emptyset\rightarrow x\leq p)$; this set is $H$, since the universal formula trivially holds for all $p\in H$ as the premise of the quantified implication is always false. Similarly one deals with "$\emptyset$ is downward closed".

Now
$\mathrm{Y}(0):H^{op}\to\mathrm{Set}$ is the functor defined by $p\mapsto\emptyset$ for $p\neq 0$ and by $0\mapsto (0\leq 0)\cong \{*\}$.
$\mathrm{Y}(0)$ is a $J$-sheaf if:

  • any collation $\{x_f:f\in\emptyset\}$ of a matching family indexed by the empty sieve $\emptyset\in J(0)$ is in $\mathrm{Y}(0)\cong\{*\}$.
  • any collation $\{x_f:f\in\{0\}\}$ of a matching family for $\mathrm{Y}(0)$ indexed by the sieve $\{0\}=\downarrow 0\in J(0)$ is in $\mathrm{Y}(0)\cong\{*\}$.

Now:

  • The unique matching family indexed by $\emptyset$ is the empty matching family $\emptyset$ whose unique collation is the empty function $\emptyset$.
    Since $\{*\}\cong\{\emptyset\}$ in the category $\mathrm{Set}$, we get that $\mathrm{Y}(0)$ has a collation for the empty matching family.
  • The unique matching family indexed by the sieve $\{0\}$ has its unique element as its collation, so in this second case there is nothing to prove.

We conclude that $\mathrm{Y}(0)$ is a $J$-sheaf.

What could be a non subcanonical topology on a complete Heyting algebra, if any exists?
I am surprised that such basic questions are not treated in a standard textbook such as Moerdijk-Mac Lane or in online libraries on category theory such as n-lab. May be I'm not able to parse correctly through these sources?

Best Answer

Let $A$ be a preordered set and, for each $a \in A$, let $\Omega (a)$ be the set of downward-closed subsets of ${\downarrow} ( a ) = \{ a' \in A : a' \le a \}$. As you know, a Grothendieck topology on $A$ is the same thing as an assignment to each $a \in A$ of a subset $J (a) \subseteq \Omega (a)$ that satisfies these axioms:

  • For all $a \in A$, ${\downarrow} (a) \in J (a)$.
  • For all $a \in A$ and $S \in J (a)$, if $a' \le a$, then $S \cap {\downarrow} (a') \in J (a')$.
  • For all $a \in A$ and $S \in J (a)$, if $S' \in \Omega (a)$ and, for each $a' \in S$ we have $S' \cap {\downarrow} (a') \in J (a')$, then $S' \in J (a)$.

Say a downward-closed subset $S \subseteq A$ is $J$-closed if, for all $a \in A$, $S \cap {\downarrow} (a) \in J (a)$ implies $a \in S$.

Lemma. The following are equivalent:

  • $J$ is subcanonical.
  • For every $a \in A$, ${\downarrow} (a)$ is $J$-closed.
  • For all $a$ and $a'$ in $A$, if ${\downarrow} (a) \cap {\downarrow} (a') \in J (a')$, then $a' \le a$.  ◼

Now, let $K (a)$ be the set of all $S \in \Omega (a)$ such that, for all $a' \le a$, $a'$ is a supremum of $S \cap {\downarrow} (a')$.

Lemma. Let $S \in \Omega (a)$. Assuming $A$ is a complete lattice, the following are equivalent:

  • $S \in K (a)$.
  • For all $a' \le a$, $a'$ is a supremum of $\{ a'' \wedge a' : a'' \in S \}$.  ◼

Corollary. Let $S \in \Omega (a)$. Assuming $A$ is a complete Heyting algebra, the following are equivalent:

  • $S \in K (a)$.
  • $\sup S = a$.  ◼

Thus, $K$ is the same thing as the sup topology when $A$ is a complete Heyting algebra. (In general, the sup "topology" fails to be a Grothendieck topology at all!) I claim that $K$ is the canonical topology on $A$. That is:

Theorem. The following are equivalent:

  • $J$ is subcanonical.
  • For every $a \in A$, $J (a) \subseteq K (a)$.

Proof. First, we have to show that $K$ itself is subcanonical. For this, we can check that each ${\downarrow} (a)$ is $K$-closed. This is clear: if ${\downarrow} (a) \cap {\downarrow} (a') \in K (a')$, that means $a'$ is a supremum of a subset of ${\downarrow} (a)$, so we must have $a' \le a$. Hence, any topology smaller than $K$ is also subcanonical.

Now, suppose $J$ is subcanonical. We have to show that $J$ is smaller than $K$. Let $S \in J (a)$. Suppose $a' \in A$ is an upper bound for $S$, i.e. $S \subseteq {\downarrow} (a')$. By hypothesis, ${\downarrow} (a')$ is $J$-closed, so we must have $a \in {\downarrow} (a')$, i.e. $a \le a'$. Hence, $a$ is a supremum of $S$. By the same argument, for every $a' \le a$, $a'$ is a supremum of $S \cap {\downarrow} (a')$, which is an element of $J (a')$ because $J$ is a Grothendieck topology. Therefore $S \in K (a)$, as required.  ◼

Corollary. The following are equivalent:

  • $J$ is not subcanonical.
  • There exist $a \in A$ and $S \in J (a)$ such that $a$ is not a supremum of $S$.  ◼

So it should now be clear how to construct a non-subcanonical Grothendieck topology on any non-trivial preorder. (In any case, it should be clear that such things exist. Otherwise $\Omega$, which is a Grothendieck topology, would be subcanonical, but $S \subseteq A$ is $\Omega$-closed if and only if $S = A$!)

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