Morse function with indices of only $0$ and $n$ Q1: If a Morse function on a smooth closed $n$-manifold $X$ has critical points of only index $0$ and $n$, does it follow that $X\approx \mathbb{S}^n\coprod\ldots\coprod\mathbb{S}^n$?
I think the following question is essential in regard to the one above:
Q2: If $f$ is a Morse function on a closed connected smooth $n$-manifold $X$ that has critical points of only index $0$ and $n$ and $f(X)\!=\![a,b]$, can a critical point of index $0$ or $n$ be mapped into $(a,b)$?
 A: As Matt mentioned, a Morse function gives you a handle decomposition.
In the connected case when $n>1$, Poincare duality forces you to have exactly two critical points. The handle decomposition then means that you can form your smooth manifold by taking two $n$-discs and gluing them together along their boundary.
As a result, the actual answer to your question depends on what you mean by $\approx$.
If you mean homeomorphism, the answer is yes.  To pull out a heavy hammer, the Poincare conjecture tells you that the resulting topological manifold has to be the $n$-sphere $S^n$, because it's $(n-1)$-connected.
If you mean diffeomorphism, then I suspect that the answer is no.  After all, this procedure of taking two discs and gluing them together along a diffeomorphism of their boundaries, which is what the handle decomposition gives you, is also how Milnor originally produced exotic spheres.
A: Say $M$ has $j$ connected components and you only have critical points of index 0 and $n$, then you must have $j$ of each index.  You can probably also get this from the Morse inequalities but a very easy way to see this is because the homology of $M$ can be computed from the complex whose $k$th degree group is the free $\mathbb Z$ module generated by the critical points of index $k$.  Then since the chain complex is zero in degrees $n-1$ or $n+1$, the $n$th homology must be $\mathbb Z^m$ where $m$ is the number of critical points of index $n$.  But an $n$-manifold with $j$ components must have $\mathbb Z^j$ for $n$th homology.  Similarly using Poincare duality (i.e. replacing $f$ by $-f$), shows that there must be $j$ critical points of index 0 as well. 
Now $f$ restricted to each connected component has exactly two critical points and it is a theorem of Reeb that any such manifold must be homeomorphic to $S^n$.  An easy proof of this is in Milnor's Morse Theory. 
