Alternative characterization of homotopy equivalence - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T01:58:48Z http://mathoverflow.net/feeds/question/86016 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/86016/alternative-characterization-of-homotopy-equivalence Alternative characterization of homotopy equivalence Giorgio Mossa 2012-01-18T17:15:20Z 2012-06-11T12:51:59Z <p>Using the formalism of model categories its possible define the concept of homotopy as done <a href="http://ncatlab.org/nlab/show/homotopy" rel="nofollow">here</a>.</p> <p>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.</p> <p>From this point of view seems that weak equivalence are what really matter, so here's my question:</p> <blockquote> <p>Is there any way to characterize homotopy equivalence (in $\mathbf{Top}$) without using the concept of homotopy? </p> </blockquote> <p>I'm wondering if there's a way to discriminate homotopy equivalence <em>without using the concept of homotopy at all</em>, 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 $1_Y$.</p> http://mathoverflow.net/questions/86016/alternative-characterization-of-homotopy-equivalence/86021#86021 Answer by Karol Szumiło for Alternative characterization of homotopy equivalence Karol Szumiło 2012-01-18T18:35:04Z 2012-06-11T12:51:59Z <p>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 <code>$X = \{ 1, 2, 3, \ldots, \infty \}$</code> with discrete topology and <code>$Y = \{1, \frac{1}{2}, \frac{1}{3}, \ldots, 0 \}$</code> with the topology inherited from <code>$\mathbb{R}$</code>. Let <code>$f \colon X \to Y$</code> be a map given by <code>$f(m) = \frac{1}{m}$</code> (and $f(\infty) = 0$). Then <code>$f$</code> 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.</p> <p>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.</p> <hr> <p>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.</p> <p>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 $Y^I \times_{Y \times Y} (X \times *)$.</p> <p>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".</p> http://mathoverflow.net/questions/86016/alternative-characterization-of-homotopy-equivalence/86032#86032 Answer by David White for Alternative characterization of homotopy equivalence David White 2012-01-18T20:08:37Z 2012-01-19T01:29:21Z <p>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.</p> <p>Here is a nice article on <a href="http://ncatlab.org/nlab/show/localization" rel="nofollow">localization</a>. 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.</p> <p>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!</p>