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The contravariant functor $C(-)$ given by $$ \hom_{Top}(-,\mathbb{R}):cCW\to Rng $$ where $cCW$ is the category of compact CW complexes is injective on objects. What is known about surjectivity, faithfulness and fullness of this functor?

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Surjective on objects: Definitely not, how do you get mathbb{Z} or worse a noncommutative ring? Full: How do you induce the zero map between two rings of continuous functions with a continuous function between the spaces? Faithful: This is the only interesting one. I am guessing that it is faithful. Examining the proof that it is injective on objects (looking at the MaxSpec construction), should point you in the right direction I think. Maybe your question is more interesting if you restrict your attention to R-modules? –  Steven Gubkin Oct 28 '10 at 14:27
    
Steven: the zero map shouldn't really count: it's not a ring homomorphism here. (I assume rings have identity and the identity is preserved, a standard convention for commutative rings.) –  KConrad Oct 28 '10 at 14:38
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Rng is a strange choice of target category. You want at least commutative R-algebras and you actually get a commutative Banach algebra or, even better, a commutative C*-algebra over R with trivial involution. –  Qiaochu Yuan Oct 28 '10 at 15:57
    
The Gelfand-Naimark-Theorem gives an answer. But it does not tell you how to see whether a space is a CW by looking at its function algebra. –  Johannes Ebert Oct 28 '10 at 17:55

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up vote 5 down vote accepted

Corollary 4.1.(i) in Johnstone's book Stone Spaces (electronic version: http://gen.lib.rus.ec/get?nametype=orig&md5=C26F62F69C32101307213F1960F85BA3) states that the category of realcompact spaces is dual to the full subcategory of the category of commutative rings consisting of rings of the form C(X). The functor C implements the duality.

The category of compact CW-complexes embeds into the category of realcompact spaces as a full subcategory, hence the functor C is fully faithful.

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For reference: en.wikipedia.org/wiki/Realcompact_space –  David Roberts Oct 28 '10 at 20:13
    
Thank you. Paracompactness does not suffice here, right? –  roger123 Nov 1 '10 at 12:48
    
I think there are non-homeomorphic paracompact spaces with isomorphic algebras of continuous functions. This is plausible because not all paracompact spaces are realcompact. –  Dmitri Pavlov Nov 1 '10 at 15:50

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