Either this is horribly wrong, or it must be well-known. So I guess I’m asking for either a rebuttal or a reference.

Take a ‘smallish’ category $\mathbf{C}$. By this I mean that for every object $C$ the collection of all maps ending in $C$ must be a set. On this set, let’s call it $y(C)$ for Yoneda’s sake, we can define a pre-order $f \leq g$ if there is a commuting diagram

$\xymatrix{D \ar[rr]^f \ar[rd]_h & & C \\ & E \ar[ru]_g &}$

A *sieve* $S$ on $C$ is the same thing as a downset in $y(C)$ with respect to this pre-order. Composition with $h : D \rightarrow C$ gives a map $h : y(D) \rightarrow y(D)$ such that $h^{-1}(S)$ is a downset (or, sieve) in $y(D)$ whenever $S$ is a downset in $y(C)$.

A *Grothendieck topology* on $\mathbf{C}$ is a function $J$ which assigns to every object $C$ a collection $J(C)$ of sieves on $C$ satisfying:

- $y(C) \in J(C)$,
- if $S \in J(C)$ then $h^{-1}(S) \in J(D)$ for every morphism $h : D \rightarrow C$,
- a sieve $R$ on $C$ is in $J(C)$ if there is a sieve $S \in J(C)$ such that $h^{-1}(R) \in J(D)$ for all morphisms $h : D \rightarrow C$ in $S$.

From this it follows for all downsets $S$ and $T$ in $y(C)$ that if $S \subset T$ and $S \in J(C)$ then $T \in J(C)$ and if both $S,T \in J(C)$ then also $S \cap T \in J(C)$.

In other words, the collection $\mathcal{J}_C = \{ \emptyset \} \cup J(C)$ defines an ordinary topology on $y(C)$, and the second condition implies that we have a covariant functor

$\mathbf{J} : \mathbf{C} \rightarrow \mathbf{Top}$ sending $C \mapsto (y(C),\mathcal{J}_C)$

That is, one can view a Grothendieck topology as a functor to ordinary topological spaces.

Furher, the topos of sheaves on the site $(\mathbf{C},J)$ seems to fit in nicely. To a sheaf

$A : \mathbf{C}^{op} \rightarrow \mathbf{Sets}$

one associates a functor of flabby sheaves $\mathcal{A}(C)$ on $(y(C),\mathcal{J}_C)$ having as stalks

$\mathcal{A}(C)_h = Im(A(h))$ for all points $h : D \rightarrow C$ in $y(C)$

and as sections on the open set $S \subset y(C)$ all functions of the form

$s_a : S \rightarrow \bigsqcup_{h \in S} \mathcal{A}(C)_h$ where $s_a(h)=A(h)(a)$ for some $a \in A(C)$.

### Similar Posts:

- a noncommutative Grothendieck topology
- Connes-Consani for undergraduates (2)
- Connes-Consani for undergraduates (3)
- B for bricks
- a noncommutative topology 2
- A for aggregates
- noncommutative F_un geometry (2)
- How to dismantle scheme theory?
- noncommutative F_un geometry (1)
- Connes & Consani go categorical

## 3 Comments