$f: X \rightarrow Y$ is a homotopy equivalence $\iff$ $X$, $Y$ are both homeomorphic to a deformation retract of a space $Z$.So what?

Question

I have started reading some algebraic topology.In this thread there is the following result:

$f: X \rightarrow Y$ is a homotopy equivalence $\iff$ $X$, $Y$ are both homeomorphic to a deformation retract of a space $Z$

What is the use of the above result? I have heard that the above result give more intuitive idea of homo topic spaces.Can some one explain me what is the use(or better way to think) of above result?

Thank you.

A very important corollary of this is that given a CW pair $(X, A)$ and homotopic maps $f, g: A \to Y$, the adjuction spaces $X \cup_f Y$ and $X \cup_g Y$ are homotopy equivalent (see eg prop. 0.18 in Hatcher). This says that homotopy type of CW complexes only depend on the attaching maps. — Balarka Sen, Sep 30 '16 at 07:01

score 2 · Answer 1 · answered Sep 29 '16 at 07:35

2

This is a great example of the result and why you should care!

answered Sep 29 '16 at 07:35

Faraad Armwood

8,938

Pedro · Accepted Answer · 2016-09-29T08:10:43.143

Suppose I take a space $X$ and deform it in a reasonable way. I get another space $X_1$, and by reasonable I mean that $X_1$ and $X$ have the same homotopy type. If I perform another deformation on $X$ to obtain $X_2$, it is manifest that $X_1$ and $X_2$ have the same homotopy type: they are deformations of the same original space. Thus, for two spaces to have the same homotopy type, it is sufficient they are deformations of a common space. The result you cite says this is a necessary condition, too.

score 1 · Answer 3 · edited Nov 23 '19 at 13:15

As noted in two other answers, the evident direction is also a useful way to argue that two spaces are of the same homotopy type. I will try to address the question of how is one a better way of understanding the notion. Well, I would not say that one is better, but certainly that each is giving us a rather different point of view on the notion.

Firstly, the notion of a homotopy equivalence is a way of saying that two spaces are in some sense the same. That particular sense is in the existence of a homotopy equivalence between them. In more detail, we want to declare $X$ and $Y$ as essentially the same if there exists a function $f\colon X\to Y$ which is a homotopy equivalence. What this in fact is saying is that we would like it if $f$ were an isomorphism. So, we have our category $\bf Top$ of topological spaces and continuous functions, and a bunch of morphisms, the homotopy equivalences, and we'd like all of them to have inverses. More generally, one can start with any category and a bunch of morphisms in it and wish to invert those morphisms. This is called localizations. So, the definition of homotopy equivalence, and the resulting view of certain spaces which are far from homeomorphic as still being essentially the same, is a special case of localization in a category - a very morphism-oriented approach (not that there is anything wrong with that).

Then comes the second definition. It replaces the auxiliary function by an auxiliary space. The claim that two spaces are essentially the same if they are each obtained as a deformation of single space is (arguably) a more geometrically appealing notion then that of a localization (which is rather algebraic). Regardless, the interesting part of this result says that if two spaces have the same homotopy type, then you can think of these spaces as essentially being the same space $Z$, but not in the strict sense, but rather that each is homeomorphic to a deformation of $Z$. That is comforting. Of, course, preservation of difficulty implies that understanding deformations is just as hard as understanding homotopy equivalence.

$f: X \rightarrow Y$ is a homotopy equivalence $\iff$ $X$, $Y$ are both homeomorphic to a deformation retract of a space $Z$.So what?

3 Answers3