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This is a somewhat vague question; I don't know how "soft" it is, and even if it makes sense.

[Edit: in the light of the comments, we can state my question in a formally precise way, that is: "Is the homotopy category of topological spaces a concrete category (in the sense, say, of Kurosh and Freyd)?". You may still want to read what follows, for a bit of motivation]

Historically, I'd bet people started to look at concrete metric spaces $(X,d)$ before exploring the utility of the abstraction given by general topological spaces $(X,\tau)$ ($\tau$ is a topology). The heuristic idea captured by the concept of a topological space is endowing a set $X$ with a "geometry" that forgets the rigidity of a hypothetical metric structure $(X,d)$, though retaining the "qualitative" aspects of the geometry given by the metric.

Of course there are non-metrizable topological spaces, but let stick to metrizable ones for the moment. I think it should be possible to see a topological space $(X,\tau)$ as an equivalence class of metric spaces: $(X,\[d\])$, where $\[d\]$ is the class of all metrics on the set $X$ that give rise to the same topology. So, an $(X,\tau)$ just has several "rigid" models $(X,d)$, and a morphism of topological spaces $f:(X,\tau) \rightarrow (Y,\tau')$ is given by taking any map $(X,d) \rightarrow (Y,d^{'})$ of "representatives" which is continuous according to the "metric ball" definition. [Please correct the above picture if it is imprecise or even just wrong!!]

The (perhaps naive) way I have always thought about homotopy is that it is an even "further step" in making the geometry more "qualitative" and less rigid: you can "collapse positive dimensional appendices" of a space as far as they are contractible, and so forth. When trying to make this formal, you consider "homotopy types", which are equivalence classes $\[(X,\tau)\]$ of topological spaces, where $(X,\tau) \sim (X',\tau')$ if there is a homotopy equivalence $\varphi:(X,\tau) \rightarrow (X',\tau')$. What are morphisms in the homotopy category? Just morphisms $f$ between "representatives", but now you have also to consider them up to homotopy, i.e. you take $\[f\]$ where $f \sim f'$ if there's a homotopy $\alpha: f \rightarrow f'$.

It's ugly to think of topological spaces as $\[(X,d)\]$ (or, rather, $(X,\[d\])$): it's better to use the simpler and more expressive abstraction $(X,\tau)$.

So, the question is:

  • Is there some kind of "homotopology" $h$ (whatever it is) that one can put on sets $S$ so that each homotopy type $\[(X,\tau)\]$ is fully described by a "homotopical space" (whatever it is) $(S,h)$ and homotopy classes of morphisms $\[f\]$ correspond to "$h$-compatible" (whatever it means) set-theoretic maps $F:S \rightarrow S'$?

(I don't even dare asking about the existence of "non-topologizable homotopical spaces" because the above question is already by far too vague!)

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Just out of curiosity: did the guy who voted to close the question regarded it as "no longer relevant" or as "offtopic"? –  Qfwfq Apr 17 '10 at 16:08
    
Well, in the light of the article linked by QiaochuYuan, I'd say it definitely is a question: "is the homotopy category of topological spaces a concrete category?". Perhaps I should edit to make it clear. –  Qfwfq Apr 18 '10 at 4:01
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My understanding is that it is preferable to say "concretizable" rather than "a concrete category": one should always think of a concrete category as a category equipped with a distinguished forgetful functor to Set, since the same category can be concretized in different ways. –  Qiaochu Yuan Apr 18 '10 at 4:34
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@Q.Y.: I think that when you ask "is $\mathcal{C}$ concrete?" you make a harmless abuse of language similar to the one you would commit asking "is the Banach space $X$ even a Hilbert space?". Every mathematician would understand it as "does $X$ admit a scalar product inducing the norm blah blah blah?" –  Qfwfq Apr 18 '10 at 19:39

1 Answer 1

up vote 40 down vote accepted

No.

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Perfect! I've accepted this answer. It's surprising that such a theorem ha sonly been proven in 2004. –  Qfwfq Apr 17 '10 at 15:35
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I've never seen such a short and effective answer. It does not just answer the question, but along the way makes it precise in terms of category theory. +1. By the way, I think it's more easy to understand the proof that the homotopy category of chain complexes is not concrete. –  Martin Brandenburg Apr 17 '10 at 15:41
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Ouh, I've read this paper some years ago, but I thought it's a well-known fact. Can anybody provide historical references who proved this first? –  Martin Brandenburg Apr 17 '10 at 15:45
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@unknown: in fact it was proven by 1970, and republished in 2004. –  Pete L. Clark Apr 17 '10 at 15:47
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If you click on the link, you see that it says: "Homotopy is not concrete. Peter Freyd. Originally published in: The Steenrod Algebra and its Applications, Springer Lecture Notes in Mathematics Vol. 168, © Springer-Verlag, 1970 and used by permission." So the conjecture that it was proved by Peter Freyd just before 1970 seems reasonable. When Pete says "republished", he means precisely that; he doesn't mean "reproved". –  Loop Space Apr 19 '10 at 7:17

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