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I was wondering what tools of algebraic topology are usually used to show that some things have the same homotopy type? Hatcher doesn't really talk about this in his book even though he defines the concept on page 3. Of course we can compute the homology or homotopy groups of a space, but just showing that they agree is not enough as far as I know.

For example, knowing that the Poincare conjecture is true, we know that every closed simply-connected 3-manifold is the 3-sphere. It follows that they must have the same homotopy type. Is this any easier to prove than Poincare itself? If so how? The reason I picked this example is that I know they are homotopy equivalent and I don't know an obvious map between the spaces.

EDIT: Dylan actually gave what's needed to finish off a proof. The map given by the generator of $\pi_3$ can easily be checked to induce isomorphisms on all homology groups. Now replace the $3$-manifold $M$ by a $2$-connected CW-model $Z$ by CW-approximation. Functoriality of CW-models then induces a map $f:S^3\to Z$ which induces isomorphisms on homology. The standard argument that replaces $Z$ by the mapping cylinder of $f$ and then applies Hurewicz on $H_n(M_f,S^3)$ shows that $\pi_n(M_f,S^3)=0$ for all $n$ on which implies that $M_f$ deformation retracts onto $S^3$ and they are homotopy equivalent. This gives the following chain of homotopy equivalences

$$S^3\simeq M_f\simeq Z\simeq M$$

so it follows that $M$ and $S^3$ has the same homotopy type.

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Yes, I know Whitehead, but for Whitehead you have to actually find a map between those spaces. This seems to be the tricky part unless there's a nice way to construct such things in general. Is there for example a canonical map between $S^3$ and a $3$-manifold? –  asdf Jan 14 '12 at 20:57
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Are you asking for a standard tool that gives the Poincare conjecture? Presumably such a thing doesn't exist or the Poincare conjecture would have been resolved much sooner... –  Qiaochu Yuan Jan 14 '12 at 21:09
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As for your comments about maps: In practice, we can usually cook these up, and if we can't then maybe the spaces don't have the same homotopy type! One way to get a map $S^3 \rightarrow M$ for a simply connected, closed and orientable 3-manifold is to notice that Poincare duality implies the vanishing of $H_2$ and Hurewicz implies that $H_3 \cong \pi_3$ so take your map to be the one representing the generator. –  Dylan Wilson Jan 14 '12 at 23:34
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Once you have a map and you know you wanna use Whitehead, usually you compute the homology and cohomology of everything in sight, hope that your spaces are simply connected or H-spaces, hope that your map induces the required isomorphisms, and then apply Whitehead. See Quillen's paper on the algebraic K-theory of finite fields for a wonderful example of this method. –  Dylan Wilson Jan 14 '12 at 23:38
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One basic tool for constructing homotopy equivalences is a gluing theorem for homotopy equivalences which I have advertised on this stackexchange answer. The proof uses no notions of homotopy groups, but the result was found by starting with the well known fact that a homotopy equivalence $f: Y \to Z$ of spaces induces an isomorphism of homotopy groups, and then generalising replacing $(S^n,x)$ by a pair $(X,A)$ with the Homotopy Extension Property. The theorem first appeared in the 1968 edition of the book now titled "Topology and Groupoids" . Its origin is also in special cases due to J.H.C. Whitehead. There is also a dual "cogluing theorem", for fibrations and pullbacks, rather than cofibrations and pushouts. The result has also been set up in various model categories, but an advantage of the original proof is that it gives control of the homotopies involved.

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It is generally difficult to show that two spaces are homotopy equivalent. Some of the better spaces are CW complexes, where the Whitehead theorem holds.

In this case, you only need that there be a map that induces an isomorphism on all of the homotopy groups in order to have a homotopy equivalence. As was mentioned in the comments, you still need a map that realizes this. It is not often the case that just knowing the isomorphism type of some (easily-computable) invariant gives you the homotopy type of a space.

Sometimes we luck out and we have very good invariants, e.g.

$$\{ \chi \text{ and orientability} \} \leftrightarrow \{ \text{homotopy types of closed surfaces} \}$$ but this is very rare.

A nice situation that is maybe worth mentioning is for rational spaces. Such spaces have a "super Whitehead theorem":

Theorem. Suppose that $X, Y$ are nilpotent spaces with homotopy groups that are finite dimensional rational vector spaces. Then the following are equivalent for a map $f: X \rightarrow Y$:

  1. $f$ is a homotopy equivalence;
  2. $f_*: H_* (X, \mathbb{Q}) \rightarrow H_*(Y, \mathbb{Q})$ is an isomorphism;
  3. $f_*: \pi_* (X) \rightarrow \pi_* (Y)$ is an isomorphism.

So, we see that in fact all one needs is a homology isomorphism in order to detect that a map is a homotopy equivalence, provided that the spaces are rational. Of course, you still need that this isomorphism is realized by a genuine map, as abstract isomorphism will not suffice to have homotopy equivalence.

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What you've written is not special to rational spaces: a map between spaces with the homotopy type of a CW-complex where the action of $\pi_1$ is trivial is a homotopy equivalence if and only if it induces an isomorphism on singular cohomology. –  Dylan Wilson Feb 5 '12 at 18:06
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I think what you do want is the more surprising fact: If $X$ is simply connected with finite-dimensional rational vector spaces as homotopy groups, then we can recover it up to homotopy from a certain commutative DGA (called the deRham complex). In other words, the surprising thing about rational spaces is: you don't need the map! –  Dylan Wilson Feb 5 '12 at 18:10
    
I was not actually aware of this, but after looking a bit this is the Eckmann-Hilton dual of the standard Whitehead theorem (c.f. May). This is kind of awesome. I think the issue is that I went too far with the assumption that $X$ be simply connected (therefore simple). If we just have that $X$ is nilpotent and rational, I believe that this Whitehead theorem still holds. I am quite familiar with the PL deRham theory that you mention, and it is indeed very surprising that there is such an algebraization of rational spaces. –  Matthew Pancia Feb 5 '12 at 18:51
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My point is that your answer is misleading. It makes it seem like what's important is not the nilpotence criterion, but the rational criterion. The point is that if you replace $\mathbb{Q}$ with $\mathbb{Z}$ then the claim holds without the rationality or finite-dimensionality hypotheses! If you're gonna state something special about rational spaces, then it should be the fact that the map is unnecessary, and that there is a purely algebraic model. That is, after all, the whole point of rational homotopy theory: everything reduces to algebra. –  Dylan Wilson Feb 5 '12 at 19:24
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