Homotopy type of a space 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.
 A: 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 $f: X \rightarrow Y$: 


*

*$f$ is a homotopy equivalence;

*$f_*: H_* (X, \mathbb{Q}) \rightarrow H_*(Y, \mathbb{Q})$ is an isomorphism;

*$f_*: \pi_* (X) \rightarrow \pi_* (Y)$ 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.
A: I agree with the point of the question that one does want to know how to construct homotopy equivalences. 
One basic  tool for this is a gluing theorem for homotopy equivalences  which I have advertised on this stackexchange answer. It gives conditions for a pushout of homotopy equivalences to be a homotopy equivalence. 
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. 
