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Let me begin by noting that I know quite little about category theory. So forgive me if the title is too vague, if the question is trivial, and if the question is written poorly.

Let $\mathcal{C}$ and $\mathcal{D}$ be categories. Say that a functor $T: \mathcal{C} \to \mathcal{D}$ has property X (maybe there is a real name for this property?) if a morphism $f: A \to B$ between objects of $\mathcal{C}$ is an isomorphism whenever $T(f): T(A) \to T(B)$ is an isomorphism. For example, the obvious forgetful functor $CH \to Set$ where $CH$ is the category of compact Hausdorff spaces has property X because a continuous bijection from a compact space to a Hausdorff space is automatically a homeomorphism.

Here is my question. Is there a (nontrivial) functor $T: LCH \to \mathcal{D}$ from the category of locally compact Hausdorff spaces to some category $\mathcal{D}$ with property X? Even better, can we assume that $LCH$ is a subcategory of $\mathcal{D}$ and that $T$ is a forgetful functor?

I don't care to specify what I mean by "nontrivial", except that the "identity" functor from $LCH$ to itself doesn't count. I want it to be genuinely easier to decide whether or not a morphism is an isomorphism in $\mathcal{D}$. If there happen to be lots of ways to do this, perhaps it will help to know that my interest comes from some problems in analysis.

Thanks in advance!

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  • $\begingroup$ Since your interest comes from analysis, can we assume that the spaces of interest are not just locally compact Hausdorff but have other properties as well, for example first-countability? $\endgroup$
    – Todd Trimble
    Commented Jun 7, 2010 at 20:48
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    $\begingroup$ This isn't an answer, but it's still worth mentioning: your property X is called 'conservativity' (as well as 'reflecting isos', as in D.C.'s answer below). Given a monad T on a category C, the forgetful functor $G^T \colon C^T \to C$ from the category of T-algebras is always conservative. This includes your first example, because compact Hausdorff spaces are the algebras for the ultrafilter monad on Set. I don't know if LCH is monadic, but Top is monadic over Rel, via the ultrafilter monad again. $\endgroup$
    – Finn Lawler
    Commented Jun 7, 2010 at 21:04
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    $\begingroup$ @Finn, this is a good point about $CH$ and I should have caught this. However, be careful about saying $Top$ is monadic- the ultrafilter monad doesn't extend to a strict monad on (the bicategory) $Rel$- the unit transformation is only oplax. Moreover, in order to really view $Top$ as algebras for the ultrafilter monad, you'd have to view $Top$ as a bicategory with "continuous relations" as arrows (or to REALLY make things natural, you'd have to view it as double category, and $Rel$ as well). $\endgroup$
    – David Carchedi
    Commented Jun 7, 2010 at 21:10
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    $\begingroup$ Oh... I almost forgot- I am not sure about the monadicity of locally compact Hausdorff spaces, but, the category of locally compact LOCALES is monadic. See: tac.mta.ca/tac/volumes/10/13/10-13abs.html $\endgroup$
    – David Carchedi
    Commented Jun 7, 2010 at 21:22
  • $\begingroup$ @David: Yes, you're right. Forget that bit. $\endgroup$
    – Finn Lawler
    Commented Jun 7, 2010 at 21:51

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The property $X$, as you call it, is well-known. A functor with this propery is said to "reflect isomorphisms". Another example of such a functor is the geometric realization functor from simplicial sets to compactly generated Hausdorff spaces. There are all sorts of ways of building a category $D$ and a functor $T$ with the properties you want, however, depending on what you want to do, different answers can be more suiting. For example, you can let $D$ be the category $Sh(CH)$ of sheaves on the site of compact Hausdorff spaces and $T$, be "Yoneda": if $Y$ is a $LCH$ space, then $T(Y)$ is the sheaf that assigns each compact Hausdorff space $X$ the set $Hom(X,Y)$. Then, it is a simple exercise to verify that $T$ is fully-faithful and that $T(f)$ is an isomorphism implies that $f$ is (SINCE every locally compact Hausdorff space is compactly generated).

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  • $\begingroup$ If you elaborate on your interests, perhaps I can think of an example that fits you better. $\endgroup$
    – David Carchedi
    Commented Jun 7, 2010 at 20:28
  • $\begingroup$ I am glad to know that this can be done. I'll try to elaborate a little. I am trying to pursue a possible mechanism for constructing isomorphisms in C*-algebra K-theory which works fine for unital commutative C*-algebras precisely because as a category they are equivalent to CH spaces and continuous bijections between CH spaces are automatically homeomorphisms. However, to get anything interesting (e.g. Bott periodicity), the mechanism needs to work for nonunital C*-algebras (i.e. LCH spaces) and my question is about the place where it breaks down. $\endgroup$
    – Paul Siegel
    Commented Jun 8, 2010 at 13:23
  • $\begingroup$ So my hope was that I could associate (possibly noncommutative) C*-algebras to some category which "reflects isomorphisms" between LCH spaces. I was thinking along the lines of convolution algebras for an appropriate category of groupoids, for instance. Also, for applications it would be ideal if the desired functor enjoyed some sort of compatibility with group actions, so that if a group acts on a LCH space then it also acts on its "image" under $T$ and $T$ preserves the group action in the appropriate sense. $\endgroup$
    – Paul Siegel
    Commented Jun 8, 2010 at 13:34
  • $\begingroup$ If this is the case, you should probably pass to the bicategory of etale topological stacks. Of course, the inclusion of $LCH$ into this bicategory "reflects isos", in the appropriate 2-categorical sense. Moreover, by "Equivalence between the Morita categories of etale Lie groupoids and of locally grouplike Hopf algebroids" by Kalisnik and MrCun, you can conclude that etale differentiable stacks are equivalent to the Morita-bicategory of locally grouplike hopf algebroids. I would imagine a similar statement can be made in the topological setting.... $\endgroup$
    – David Carchedi
    Commented Jun 8, 2010 at 13:55
  • $\begingroup$ @Paul: there is not so much space to explain more, but if you have more questions, feel free to email me. $\endgroup$
    – David Carchedi
    Commented Jun 8, 2010 at 14:05

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