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Disclaimer: I know practically nothing about category theory, so sorry if the answer is "obvious".

Consider a category C, a subcategory D and a "closure" (i.e. idempotent) functor F : C → D. Then, given an object X ∈ D, what is the "smallest" Y ∈ C such that F(Y) = X?

It's not clear that "smallest" is well-defined for all such categories and idempotent functors, but here are two motivational examples from algebra where the categories are concrete (so objects can be ordered by isomorphic inclusion) and the functors are a form of set extension:

C the category of fields, D the category of algebraically closed fields, F algebraic closure.

C the category of (integral) domains, D the category of fields, F "field of fractions".

There are more examples of the sort of "closure functors" I'm talking about in the answers to this question.

Edit: a followup question - if the smallest such Y is X itself, i.e. X has no proper subobjects Y ∈ C such that F(Y) = X, what does this imply about X?

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Another example might be the (poset) category of subsets of a topological space, with the functor being "closure." Was this what motivated your question? I also assume you mean C to equal D, as otherwise idempotency ($F\circ F=F$) does not make sense, as one cannot compose $F$ with $F$. In any case, one might try to take the limit of e.g. the diagram given by all objects that map to $X$ and all arrows that map to the identity. Or all objects that map to something isomorphic to $X$ and all arrows that map to an isomorphism. –  Daniel Litt Jun 28 '10 at 21:48
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Do you mean F(Y) = Y and Y contains X? Otherwise it doesn't seem to match your examples. –  Kiochi Jun 28 '10 at 22:26
    
Kiochi: For my first example, given an algebraically closed field X [∈ D], what is the smallest field Y [∈ C] whose algebraic closure F(Y) = X? For my second example, given a field X [∈ D], what is the smallest ring Y [∈ C] whose field of fractions F(Y) = X? Since F is a closure operator, I should really have stipulated that not only is F idempotent, but in fact F(X) = X ∀ X ∈ D. So the smallest Y containing X such that F(Y) = Y would be X itself. Daniel: I do not mean C to equal D; D is a subcategory of C, so F can certainly be composed with itself. –  Robin Saunders Jun 28 '10 at 23:16
    
And yes, closure of subsets of a topological space would be an example of the sort of functor I'm talking about, although in general there wouldn't be a "minimal" subset Y whose closure was a given closed subset X. –  Robin Saunders Jun 28 '10 at 23:18
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Hmm, how is the «algebraic closure» a functor? Suppose $f:K\to L$ is an inclusion of fields: there are in general several maps $\overline K\to\overline L$, even several which extend $f$. –  Mariano Suárez-Alvarez Jun 29 '10 at 14:13
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3 Answers 3

One reasonable notion of “smallest” here could be “initial”?

So for an object $W$ of $\mathbf{D}$, we would look at the iso-comma category $(F \downarrow_{\cong} W)$, where an object is an object $X \in \mathbf{C}$ together with an isomorphism $i_X \colon F(X) \cong W$, and a map $f: (X,i_X) \to (Y,i_Y)$ is a map $f : X \to Y$ with $i_Y \cdot F(f) = i_X$. Then we could define a “minimal $\mathbf{C}$-object generating $W$” as an initial object of $(F \downarrow_{\cong} W)$. (Or possibly weakly initial?)

Idempotent functors of the kind you're describing often fall into one of two classes: reflections or co-reflections. Reflections are more usual: they're functors that come from an adjuntion $\mathbf{C} \leftrightarrow \mathbf{D}$ where the left adjoint goes from $\mathbf{C}$ to $\mathbf{D}$, the free way to $\mathbf{D}$-ify a $\mathbf{C}$-object. The abelianisation of a group, the field of fractions of a ring, the Stone-Cech compactification of a space, sheafification of a presheaf, are all examples of this. The unit of the adjunction gives a natural map $X \rightarrow F(X)$.

I can't think of many examples of reflections where $(F \downarrow_{\cong} W)$ will have an initial object for all (or even most) $W \in \mathbf{D}$. In the “field of fractions” case, for instance, $\mathbb{Z}$ is an initial object in $(\mathit{Frac} \downarrow_\cong \mathbb{Q})$, but fields of fractions $k(x)$ will have no minimal generating ring. (High-falutin' explanation: $(F \downarrow_{\cong} W)$ has all connected colimits that $\mathbf{C}$ does, but not other limits/colimits in general.)

What about when $F$ (let's call it $G$ now) is a co-reflection, i.e. comes from a right adjoint to the inclusion of $\mathbf{D}$? This situation typically looks a little different: the “group core” of invertible elements in a monoid, for instance; in this case the natural map goes the other way, $G(X) \to X$.

In this case, there'll always be an initial element of $(G \downarrow_{\cong} W)$: just $W$ itself! (This is dual to how in the reflection case, W is always a “maximal generating object” for itself, i.e. a terminal object of $(F \downarrow_{\cong} W)$.) On the other hand, here perhaps a terminal object is a better sense of “minimal” (“maximally quotiented”, or something): taking the sheaf of germs of a bundle is a co-reflection where this might be an interestingly non-trivial question?

Some examples, of course, aren't quite either reflections or co-reflection, like “algebraic closure”, which is nearly a reflection but not quite, because of the automorphisms. This case will look, I suspect, similar to the reflection case…

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By the way, I realise I was misusing terminology here. The (co-)reflection is used for the left (resp. right) adjoint to the inclusion CD. The resulting functor from C to itself is then an idempotent monad, i.e. a monad whose multiplication μ: FF^2 is an isomorphism. –  Peter LeFanu Lumsdaine Jun 30 '10 at 10:24
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Old answer, retrieved from the rollback dump and wikified in response to @supercooldave's lament

Don't really know much about this, but stumbled upon these a while back, they're called the Kuratowski Closure axioms and they concern a generalised 'closure' operation cl:

1) $A\subseteq cl(A)$

2) $cl(cl(A))=cl(A)$

3) $cl(\emptyset)= \emptyset$

4) $cl(A \cup B)= cl(A) \cup cl(B)$

They allow you to define a topology by declaring the closure of every set to be closed, giving continuity a pretty little characterisation of: $f$ is cts iff $f(cl(A)) \subseteq \hat{cl}(f(A))$ for all subsets $A$ (where $\hat{cl} $ is the closure operation in the target space).

The problem, of course, for taking this into a more general category would be axiom 4- straight up, old fashioned union doesn't work for, say, rings- but I'm sure with some adjoint construction in place of the union, this could describe all of your examples.

I think the reason this construction isn't exactly famous, though, is why would anyone want to put a topology on, say, the category of integral domains??? What would its homotopy type tell you??!! (Disclaimer: there may be, despite my skepticism, an incredibly interesting and useful answer to these questions...)

Edit: Just to clarify, this is on the potential 'applications' side of the question. One would hope, with the particular adjoint condition used to define a generalised 'union', that the above axioms would be satisfied by your definition and hence give a topology to a category for which such a construction is possible.

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Final Edit: Hadn't read above answer properly as I assumed it would be too technical, in fact it contained the main part of my previous edit. The only thing worth saving, therefore, is deemed the name of/nlab link to the general construction: completion.

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One should probably say: «the left adjoint, when it exists, provides...» :) –  Mariano Suárez-Alvarez Jun 29 '10 at 14:08
    
Pity you deleted it. In some recent work I was working with a closure operator which satisfied 3 of the 4 properties you mentioned (not the $\cup$ one). It turns out (conjecture!) that this was related to a pullback in a certain coslice category, which is quite related to what Peter LeFanu Lumsdaine is saying (though perhaps dual). Unfortunately, I cannot say anything in general, perhaps except that pullbacks in coslice categories correspond to taking certain maximums and pushouts in slice categories correspond to taking certain minimums. –  supercooldave Jun 29 '10 at 14:10
    
@ Mariano: correct, and corrected :) –  Tom Boardman Jun 29 '10 at 14:14
    
@ Supercooldave- intriguing enough to warrant retrieval, feel free to play with below if you're up for expanding vvvv :) –  Tom Boardman Jun 29 '10 at 14:28
    
I am an idiot. :( –  Tom Boardman Jun 29 '10 at 15:05
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