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Let $C$ be a small category. Consider the class of all Grothendieck topologies on $C$, it is a preorder with the relation "finer". Does this preorder has all infima and suprema?

(For example, how do you prove that there is always the canonical topology?)

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up vote 6 down vote accepted

Yes. In fact, Grothendieck topologies on any small category constitute a locale. See Proposition 3.2.13 in Borceux's Handbook of Categorical Algebra 3.

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An intuitively attractive way of packaging it is that Grothendieck topologies on a small category $C$ are in bijective correspondence with those operators $j: \Omega \to \Omega$ (where $\Omega$ is the truth value object in the topos $Set^{C^{op}}$) that satisfy the conditions $1_\Omega \leq j$, $j j = j$, and $j \circ \wedge = \wedge \circ (j \times j)$, (where $\wedge: \Omega \times \Omega \to \Omega$ is internal intersection). This is discussed in Mac Lane and Moerdijk for instance. Such meet-preserving closure operators form an inf-lattice, in fact a sub-inf-lattice of $\hom(\Omega, \Omega)$. The fact that $\hom(\Omega, \Omega)$ is an inf-lattice is easy because it is assembled as a limit of inf-lattices $\hom(h_c, \Omega) \cong \Omega(c)$ (the poset of sieves = subfunctors of the representable $h_c$), where infs are ordinary set-theoretic intersections. If one writes out the details of this, one is led to the proof in Borceux's book.

Added later: I suppose it is well-known that inf-lattices are automatically sup-lattices: the sup of a family is the inf of the set of upper bounds.

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This follows from the lemma that the intersection of any family of Grothendieck topologies is a Grothendieck topology. We define the supremum to be the intersection of all topologies containing the union (the topology spanned by a family), and we can define the infimum of a family to be the intersection over the family.

Recall that a Grothendieck topology is a function $J$ assigning to each object $x$ of $C$ a set of subfunctors (called covering sieves) of $h_x$ satisfying the following closure properties:

  • If $f:x\to y$ is a morphism, and $s\in J(x)$, then $f^*s\in J(y)$
  • If $s\in J(x)$, and $t\subseteq h_x$ such that for all $y\in Ob(C)$ and all $f\in s(Y)$, $f^*t\in J(Y)$, then $t\in J(x)$
  • For any object $x\in Ob(C)$, $h_x\in J(x)$.

We define the intersection of two (or any family of) topologies $J, K$ to be the pointwise intersection $J(x)\cap K(x)$, viewed as subsets of the set of subfunctors of $h_x$.

To avoid appearing too pedantic, we will refer to Grothendieck topologies as topologies unless there is reason to believe that the reader will be confused.

Lemma: The intersection of two topologies is a topology.

Proof. We check the conditions:

  • If $f:x\to y$ is a morphism, and $s\in K \cap J(x)$, then $f^*s\in K\cap J(y)$ follows immediately.
  • If $s\in K\cap J(x)$, and $t\subseteq h_x$ such that for all $y\in Ob(C)$ and all $f\in s(Y)$, $f^*t\in K\cap J(Y)$, then $t\in K\cap J(x)$ follows again by intersection
  • That all representables are present in the intersection is immediate.

We can see immediately how this generalizes to any family of topologies. $\Box$

I note that we are ignoring set-theoretic considerations here, since in reality, our proof relies on the fact that the class of subfunctors is a set, but this can be rectified using universes (because I am lazy).

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So that's the inf of a set of topologies, and for the sup of a set of topologies you have the inf (min in this case) of the set of topologies containing this set. – Tom Goodwillie Sep 20 '10 at 3:55
@Tom: Is that incorrect? – Harry Gindi Sep 20 '10 at 5:03

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