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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1$\begingroup$ 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? $\endgroup$– Steven GubkinCommented Oct 28, 2010 at 14:27
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$\begingroup$ 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.) $\endgroup$– KConradCommented Oct 28, 2010 at 14:38
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2$\begingroup$ 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. $\endgroup$– Qiaochu YuanCommented Oct 28, 2010 at 15:57
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$\begingroup$ 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. $\endgroup$– Johannes EbertCommented Oct 28, 2010 at 17:55
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1 Answer
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Corollary 4.1.(i) in Johnstone's book Stone Spaces 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.
answered Oct 28, 2010 at 15:53
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$\begingroup$ For reference: en.wikipedia.org/wiki/Realcompact_space $\endgroup$– David Roberts ♦Commented Oct 28, 2010 at 20:13
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$\begingroup$ Thank you. Paracompactness does not suffice here, right? $\endgroup$– roger123Commented Nov 1, 2010 at 12:48
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$\begingroup$ @roger123 Every paracompact Hausdorff space admits a complete uniformity (see Kelley's General Topology Chapter 6, exercise L (d)), so a paracompact Hausdorff space is realcompact iff it has no discrete subset of (2-valued) measurable cardinality by Shirota's theorem. (See, for instance Gillman and Jerison's Rings of Continuous Functions Theorem 15.20). Therefore it is consistent that every paracompact Hausdorff space is realcompact. $\endgroup$ Commented Oct 30, 2018 at 16:33