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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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    http://en.wikipedia.org/wiki/Whitehead_theorem2012-01-14
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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?2012-01-14
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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...2012-01-14
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    Nope, I'm looking for homotopy type, not homeomorphism, which is stronger. I just picked this particular example, since at least we know a priori that they are homotopy equivalent (because of Poincare).2012-01-14
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    You pretty much need a map. For example, suppose you were so awesome that you computed all of the homotopy groups of both spaces you were interested in (this pretty much never happens)... you still would not know that the two spaces had the same homotopy type unless you had a map between them (and they were of the homotopy type of a CW complex). Really the only finer invariant of a space we know of aside from its homotopy groups would be... its homotopy type.2012-01-14
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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.2012-01-14
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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.2012-01-14

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