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There is the notion of Grothendieck topology on a category $\mathbf{C}$. This is a specific assignment of a collection $J(C)$ of sieves to each object $C\in \mathbf{C}$.

A Grothendieck pretopology or basis for a Grothendieck topology is a specific assignment of a collection $K(C)$ of covers for each object $C\in \mathbf{C}$. Such a basis generates a Grothendieck topology by $$ J(C)=\{S=\{C_i\to C\}\mid \exists R\in K(C):R\subseteq S\} $$ but there may exist many bases for one Grothendieck topology.

There is a maximal basis for a Grothendieck topology and I think this gives a one-to-one correspondence between Grothendieck topologies and maximal pretopologies.

Given a Grothendieck pretopology $K$, how can I define a maximal pretopology $\mathbf{K}$ without mentioning Grothendieck topologies and sieves?

I tried $$ \mathbf{K}(C)=\{\{D_k\to C\}_{k\in K}\mid \exists \{E_i\to C\}_{i\in I}\in K(C)\forall i\in I\exists k\in K:E_i\to D_k\to C\}, $$ is this right? Maybe this isn't a pretopology and I must intersect all pretopologies with the property after '$|$' instead.

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Let's be clear about what we mean. Let $\mathcal{C}$ be a category, not necessarily small or with pullbacks. A coverage on $\mathcal{C}$ is a collection of families $K (C)$ of sinks on $C$ for each object $C$ of our category $\mathcal{C}$ satisfying this condition:

  • For every sink $\mathfrak{U}$ in $K (C)$ and every morphism $g : D \to C$ in $\mathcal{C}$, there is some sink $\mathfrak{V}$ in $K (D)$ such that, for each morphism $h$ in $\mathfrak{V}$, the composite $g \circ h$ factors through some morphism in $\mathfrak{U}$. (More succinctly, the sink $\{ g \circ h : h \in \mathfrak{V} \}$ is subordinate to $\mathfrak{U}$.)

A saturated coverage on $\mathcal{C}$ is a coverage $K$ on $\mathcal{C}$ satisfying these additional conditions:

  • For each object $C$, the singleton sink $\{ \textrm{id}_C : C \to C \}$ is in $K (C)$.

  • Given any sink $\mathfrak{U}$ in $K (C)$ and any choice of sinks $\mathfrak{V}_f$ in $K (\operatorname{dom} f)$ for each $f$ in $\mathfrak{U}$, the family $\{ f \circ h : f \in \mathfrak{U}, h \in \mathfrak{V}_f \}$ is also in $K (C)$.

  • Given $\mathfrak{V}$ in $K (C)$ and any sink $\mathfrak{U}$ on $C$, if $\mathfrak{V}$ is subordinate to $\mathfrak{U}$ (i.e. each morphism in $\mathfrak{V}$ factors through some morphism in $\mathfrak{U}$), then $\mathfrak{U}$ is already in $K (C)$.

Given a Grothendieck topology $J$ on $\mathcal{C}$, one can define a saturated coverage $K$ by setting $$K (C) = \{ \mathfrak{U} : \mathfrak{U} \text{ is a sink on $C$ and generates a sieve in } J (C) \}$$ and conversely, given a saturated coverage $K$, one can define a Grothendieck topology $J$ by setting $$J (C) = \{ \mathfrak{U} : \mathfrak{U} \text{ is a sieve on $C$ and is in } K (C) \}$$ and these constructions are mutually inverse.

Your question is how to cut Grothendieck topologies out of the picture. This is simple enough: given any Grothendieck pretopology $T$, we can build a saturated coverage $K$ as follows: $$K (C) = \{ \mathfrak{U} : \mathfrak{U} \text{ is a sink on $C$ and, for some $\mathfrak{V}$ in $T (C)$, $\mathfrak{V}$ is subordinate to $\mathfrak{U}$ } \}$$ If I have not misread your symbolic formula, this is precisely the construction you have suggested. This is because the relation "$\mathfrak{V}$ is subordinate to $\mathfrak{U}$" is equivalent to "(the sieve generated by) $\mathfrak{V}$ is contained in the sieve generated by $\mathfrak{U}$", and it is well-known that the family of covering sieves in a Grothendieck topology is upward-closed.

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Thank you for this wonderful answer. If I get you right, my construction $\mathbf{K}$ is in fact a pretopology, right? –  McMurphy Nov 21 '12 at 8:16
    
It is when the category $\mathcal{C}$ has pullbacks. –  Zhen Lin Nov 21 '12 at 8:21
    
Is the pre-topology $\mathbf{K}$ (when $\mathcal{C}$ has pullbacks) also a Grothendieck topology, i.e. are the elements of $\mathbf{K}(C)$ actually sieves and satisfy the topology conditions? –  jeffrey Nov 6 at 21:34
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No, $K$ is almost never a Grothendieck topology in this case. –  Zhen Lin Nov 6 at 21:57
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Yes, $J$ is contained in $K$. –  Zhen Lin Nov 6 at 22:54

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