When can we recover a manifold when we attach a $2n$-cell to $S^n$? I have a question related to this one.  In my answer I was going to try and say something about the possible manifolds that might arise in this way, i.e. as mapping cones of elements of $\pi_{2n-1}(S^n)$.  Certainly not all of them will be manifolds.  At first I felt like there should be a reason why we'd need to be using a torsion-free homotopy generator if we want a manifold (in which case we could appeal to Serre's theorem that the only non-torsion part of the homotopy groups of spheres is $\pi_n(S^n)=\mathbb{Z}$ and $\pi_{4n-1}(S^{2n})=\mathbb{Z}$), but then I realized I couldn't think of any reason why that should be true.  Also I was hoping the Hopf invariant would enter into the picture too, since that's a pretty obvious tool at our disposal, but I couldn't get anywhere with that either...and I think at this point I'm more or less all out of tricks.  Does anyone have any ideas?
 A: I believe this is related to the Hopf invariant one problem.  Let $X = S^n \cup_f B^{2n}$.
The attaching map from $B^{2n}$ is a map $f$ from $S^{2n-1}$ to $S^n$.  I think one way of thinking about the Hopf invariant of this map is as follows.
Let $x\in H^n(X)$.  Then $x^2 = k y$ where $y$ generates $H^{2n}(X)$.  The number $k$ is the Hopf invariant.
In order for $X$ to be a manifold, it must satisfy Poincare duality (so long as $n>1$ - if $n=1$ different stuff can happen), which implies that $k = \pm 1$.  But Adams has shown you can only have Hopf invariant 1 when $n = 2, 4, 8$, getting you something homotopy equivalent to $\mathbb{C}P^2$, $\mathbb{H} P^2$ or $\mathbb{O}P^2$.
Edit:  As pointed out by Denis Gorodkov, (communicated by G. Sassatelli), this last sentence is false.  In the case of $\mathbb{C}P^n$, it is true as any manifold with the cohomology ring of $\mathbb{C}P^n$ is homotopy equivalent to it.  But the cohomology rings of $\mathbb{H}P^2$ and $\mathbb{O}P^2$ do not determine the homotopy type - see the comments.
End edit
The $n=1$ case, you're attaching a disc to a circle and getting a manifold.  Here, since $\pi_1(X)$ may be nontrivial, one cannot neccesarily use Poincare duality.  However, by a simple analysis, the only manifold this can give rise to is $\mathbb{R}P^2$.
Edit  If one wants to avoid cases ($n = 1$ vs $n > 1$) then one can work with cohomology with $\mathbb{Z}/2\mathbb{Z}$ coefficients instead.
