Using the formalism of model categories its possible define the concept of homotopy as done here.

If we take as model category $\mathbf{Top}$ having homotopy-equivalence as weak-equivalence and fibration and co-fibration defined in the standard topological way, these type of homotopies are just homotopies as defined in basic courses of algebraic topology.

From this point of view seems that weak equivalence are what really matter, so here's my question:

Is there any way to characterize homotopy equivalence (in $\mathbf{Top}$) without using the concept of homotopy?

I'm wondering if there's a way to discriminate homotopy equivalence without using the concept of homotopy at all, meaning that I'm looking for a criteria which enable to say that a certain continuous map $f \colon X \to Y$ is an homotopy equivalence without looking for a morphism $g \colon Y \to X$ and continuous maps $\mathcal F \colon X \times I \to X$ and $G \colon Y \times I \to Y$ which are indeed respectively homotopies of $g \circ f$ with $1_X$ and $f \circ g$ with $1_Y$.

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    $\begingroup$ Both weak equivalences and homotopy equivalences are important in abstract homotopy theory. A usual way of constructing the homotopy category of a model category is: first, divide out the homotopy relation (perhapes restricting to fibrant or cofibrant objects), second, calculus of fractions with respect to weak equivalences. $\endgroup$ Jan 18 '12 at 22:29
  • $\begingroup$ Have you looked at the MO question mathoverflow.net/questions/51091/computing-homotopies ? Answers there may or may not tell you something that you want to know. $\endgroup$ Jan 18 '12 at 22:35
  • $\begingroup$ It appears that you really mean to consider $Top$ as a model category with the homotopy equivalences (not the weak homotopy equivalences) as the weak equivalences. Is that right? $\endgroup$ Jan 19 '12 at 2:49
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    $\begingroup$ @TomGoodwillie: yeah you're right. The point is, because homotopy equivalences, fibrations and cofibrations seem to be the only things you need to rebuild the whole homotopy theory, I was wondering if there is any why to present this model category structure of $\mathbf{Top}$ without using the concept of homotopy. $\endgroup$ Jan 19 '12 at 12:41

EDIT: Now that the OP has edited his question to make clearer what he wants as an answer, I'm removing speculation about what he wanted. The answer is: yes, you can characterize homotopy equivalences as the maps which become isomorphisms after applying the localization functor to invert the weak equivalences. This answer doesn't require you to "use the notion of homotopy" because it's part of a much more general framework.

Here is a nice article on localization. One of the best reasons for studying model categories is that they let you get your hands on the maps (weak equivalences) which build homotopy equivalences. Many of the axioms for a model category are there to get around set-theoretic issues that arise when you try to localize a category with respect to an arbitrary class of morphisms. It turns out you need to localize at a class of morphisms which admits a calculus of fractions, and the class $W$ of weak equivalences does. If you read the article, you'll see how to construct $C[W^{-1}]$ in general, and you can then specialize to the case where $C=Top_*$ and $W$ is the class of weak equivalences.

Unfortunately, for computation, this highly general approach is not as good as just using the Path and Cylinder objects mentioned in the nLab article the OP links to. That's why most people who study model categories use those instead of this general framework: because a model category is more than just a category with a nice class $W$--it's a category where you can really get your hands on everything and compute!

  • $\begingroup$ I'm sorry but what about a map that is a weak equivalence without being a homotopy equivalence. Remember you are not assuming that the spaces are CW complexes. The Warsaw circle has trivial homotopy groups but is by no way contractible. $\endgroup$
    – Tim Porter
    Jan 19 '12 at 15:59
  • $\begingroup$ @Tim Porter: I'm not claiming that all homotopy equivalences come from weak equivalences, just that the process of localization builds the homotopy equivalences directly from the cofibrations, fibrations, and weak equivalences without reference to path objects, cylinder objects, or the word "homotopy." This is because the homotopy equivalences are the isomorphisms in a category which is obtained by this general theory of localization. You need the cofibrations and fibrations in order to get $W$ to admit the calculus of functors, but you don't need any distinguished objects $\endgroup$ Jan 19 '12 at 19:01
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    $\begingroup$ I'm afraid I still don't understand what the question is about and how this answer addresses it. In any model category a map is a weak equivalence if and only if it becomes an isomorphism in the localization. If we start with the standard model structure on $\mathrm{Top}$ with weak homotopy equivalences as weak equivalences, then we don't recover actual homotopy equivalences this way. $\endgroup$ Jan 19 '12 at 20:02
  • $\begingroup$ One more thing -- it is not true in general that weak equivalences in a model category satisfy the calculus of fractions. It only becomes true after dividing by the homotopy relation, for which you need to use cylinders or path objects. Think about the classical construction of the derived category of an abelian category. You have to divide by the chain homotopy relation before you can prove that quasi-isomorphisms satisfy the calculus of fractions. $\endgroup$ Jan 19 '12 at 20:15
  • $\begingroup$ @Karol. Your second objection is something I wasn't aware of. I guess this means even using the general theory of localization you can't escape from using cylinder objects. I read the question as seeking more information about how to pass from a model category to the homotopy category, and in particular how to obtain the homotopy equivalences in a different way than is done in classical algebraic topology. That's why I phrased my answer the way I did and pitched it at the level I did. $\endgroup$ Jan 19 '12 at 21:12

EDIT: This is an old question, but I have stumbled upon it by accident and realized that my answer is wrong. It turns out that genuine homotopy equivalences cannot be characterized in terms of their homotopy fibers. Here's a counterexample. Let $X = \{ 1, 2, 3, \ldots, \infty \}$ with discrete topology and $Y = \{1, \frac{1}{2}, \frac{1}{3}, \ldots, 0 \}$ with the topology inherited from $\mathbb{R}$. Let $f \colon X \to Y$ be a map given by $f(m) = \frac{1}{m}$ (and $f(\infty) = 0$). Then $f$ has contractible homotopy fibers (in fact they are all one-point spaces), but it is not a homotopy equivalence since the only candidate for a homotopy inverse is not continuous.

It is well-known that a map with all homotopy fibers weakly contractible is a weak equivalence and my mistake was to assume that there is a similar result for homotopy equivalences.

I have to say I don't understand the motivation behind your question. Why exactly do you want to characterize homotopy equivalences without using homotopies? Moreover I'm not sure the question is well-posed, the answers would probably refer to homotopies implicitly and it would be a matter of taste whether they count as "not mentioning homotopies". To illustrate my point, here's an attempt at an answer.

First, define the space $X$ to be contractible if it is a retract of its cone (more precisely if the canonical inclusion $X \to C X$ admits a retraction). Then define a map $X \to Y$ to be a homotopy equivalence if its homotopy fiber at every point $y \in Y$ is contractible. The homotopy fiber can be defined as a mapping cocylinder i.e. the pullback $Y^I \times_{Y \times Y} (X \times *)$.

Now, I didn't use the word "homotopy" (except in "homotopy fiber", but I explained how to "go around it"). However, for example the retraction $C X \to X$ is nothing else but a homotopy from $\mathrm{id}_X$ to some constant map. I suspect that you won't be satisfied with this kind of hiding the homotopies backstage. If this is the case you should explain precisely what counts as "not mentioning homotopies".

  • $\begingroup$ I've edited the question and added some specification, I hope I made myself more clear. Thanks. $\endgroup$ Jan 18 '12 at 21:36

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