A manifold is a homotopy type and _what_ extra structure? Motivation: Surfaces
Closed oriented 2-manifolds (surfaces) are "classified by their homotopy type". By this we mean that two closed oriented surfaces are diffeomorphic iff they're homotopy equivalent. This means that somehow the homotopy type of the surface contains essentially all information about the manifold.
Let's turn the question around. Let's choose a homotopy type $M$, and ask whether it specifies a manifold. I have to be more precise by what I mean here. First of all, let's fix a dimension, say 2 for now, although we can increase it later. Of course not every homotopy type will correspond to a surface. There a some restrictions on $M$, such as:


*

*It has to have cohomological dimension 2, i.e. isomorphisms $H^k(M) \cong 0 \quad \forall k > 2$.

*We have to specify an isomorphism $\phi\colon \mathbb{Z} \xrightarrow{\cong} H^2(M)$ which will in particular define the fundamental class $[M] := \phi(1)$, corresponding to orientation.

*Cohomology and homology have to exhibit Poincaré duality, which is to say that the cap product with the fundamental class is an isomorphism: $[M] \cap - \colon H_k(M) \xrightarrow{\cong} H^{2-k}(M)$


Now we're in better shape. Although I don't know a proof and haven't seen this statement anywhere, I'd venture the following conjecture, which should be easy to prove or disprove by anyone who knows more homotopy theory than me:
Conjecture Each homotopy type with the extra structure outlined in 1. - 3. corresponds to a closed, oriented surface. In particular, there is an equivalence between the category of homotopy types with extra structure and the category of closed, oriented surfaces.
Note also that the cap product is functorial, so a map of surfaces should be a map of the homotopy types preserving all of the structure.
The takeaway is this: I've come to believe that surfaces are essentially homotopy types with extra structure on cohomology and homology that comes from the manifold structure. Possibly I haven't captured all structure that is needed. But I guess one could amend the list in that case.
Question 1: Am I right so far?
Higher dimensions: Topological, PL and smooth structures
It gets hairier when we go up dimensions. There are closed 3-manifolds that are homotopy equivalent, but not homeomorphic. On the other hand, simply connected topological 4-manifolds are classified by their intersection form, so they can be completely recovered by the information in 1. - 3. ! For smooth structures, there is of course less luck, although the Kirby-Siebenmann class in 4th cohomology tells you whether there is a PL structure or not, so that sounds like a promising candidate for more extra structure along the lines of what we had so far.
Question 2: How far can we carry on the idea and classify higher dimensional (topological, PL, or smooth manifolds) by extra structure on the homotopy type, or its homology and cohomology?
Boundaries, noncompact manifolds
We could wonder whether it's possible to generalise the story to manifolds with boundaries, or noncompact manifolds. Then the homotopy type will certainly not be sufficient.
Already surfaces with boundary are not classified by their homotopy type. (Typical counterexample: The direct sum of two annuli, and the minimal 1-handlebody of a torus.) What is really relevant here is the homotopy type of the boundary inclusion $\partial M \hookrightarrow M$, and the corresponding relative cohomology.
(Similarly, for noncompact manifolds what seems to be relevant is compactly supported cohomology, which is related to the compactification of the manifold.)
How far can the idea be generalised here?
 A: Igor is giving a good reference to the topic. For completeness, my education, and satisfaction of other reader's laziness, I'm going to give a rough outline here.


*

*A Poincaré complex is, very similarly to what I've conjectured, a finite CW complex together with a chosen fundamental class that induces Poincaré duality.


*

*Such a complex of course does not correspond to a unique topological (let alone smooth) manifold, and there is a popular 5-dimensional counterexample.

*There are also "Poincaré pairs", corresponding to manifolds with boundary, essentially answering half of my last question.


*Another important construction that one can do with a manifold (indeed, any nice enough vector bundle) is the Thom-Pontryagin construction. It leads to something called the "Spivak normal fibration".


*

*Essentially, one takes a vector bundle $p\colon E \to M$, picks out the unit disk bundle and quotients by the unit sphere bundle (thus compactifying each fibre), yielding the Thom space $T(p)$.

*Now, one can also first embed $M$ (uniquely) into $\mathbb{R}^k$ for $k$ high enough, and compactify the complement of the stable normal bundle $\nu$. Since the 1-point-compactification of $\mathbb{R}^k$ is $S^k$, this gives a map $S^k \to T(\nu)$, the "normal invariant".

*One can generalise and axiomatise this construction for Poincaré duality spaces, which is then called a "Spivak normal fibration". It consists of a spherical fibration for the CW complex and something generalising the normal invariant.


*It is a theorem by Spivak that every Poincaré duality space has a Spivak normal fibration, unique up to homotopy equivalence. This is remarkable, since a Poincaré duality space will in general not even have a tangent or normal bundle. 


*

*When the spherical fibration actually comes from the stable normal bundle of a smooth manifold, it has a reduction to a $O(k)$ structure group. Alternatively, one can think about the classifying space $BG$ of circle bundles, and the classifying space $BO$ of vector bundles, which is mapped into $BG$ by means of the $J$-homomorphism. Only those spherical fibrations in its image can come from a smooth manifold.

*Similar arguments can be made for topological and $PL$-manifolds.


*For high dimensions, all this is already pretty close to a full answer. Dimensions 3 and 4 are the hardest, as usual.


*

*In 2 dimensions, oriented closed surfaces are classified by Poincaré complexes. (It seems this was proved as late as in the early '80s.) That seems to answer my first question.

*There is a theorem in $\geq 5$ dimensions by Browder that asserts that a simply connected Poincaré complex is homotopy equivalent to a manifold iff the normal fibration has a $TOP$-reduction.

*In 3 dimensions, Poincaré complexes were classified by Hendriks and Turaev. But I don't know which ones correspond to manifolds, and what extra data classifies the manifolds. (One could ask e.g. how to distinguish homotopy equivalent Lens spaces. Of course I'm not asking you to directly classify 3-manifolds.)

*In 4 dimensions, simply connected topological manifolds are classified by the intersection form and the Kirby-Siebenmann invariant. The intersection form is obviously defined for any 4-dimensional Poincaré complex. I think Kirby-Siebenmann does as well, but I'm not so sure about that.


*To really find a manifold from a Poincaré complex, one needs to study $L$-groups, which are a whole huge subject in itself. (See e.g. Wolfgang Lück's 2004 notes A Basic Introduction to Surgery Theory, Andrew Ranicki's Algebraic $L$-theory and topological manifolds (Cambridge Tracts in Mathematics 102 (1992)), or Wall's Surgery on compact manifolds.)


*

*The basic idea seems to be that one studies maps from a manifold into a Poincaré space, and then tries to change the manifold by surgery until the map is a homotopy equivalence. The $L$-groups give obstructions to this.


A: You are talking about the (much studied) Poincare duality spaces. For a survey, see the very nice one by John Klein: (seems to be unpublished, but dates to April 2010).
