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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 $(a,b)$?

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    Yes and no. Think of two disjoint spheres one of which has its "bottom" point at the halfway point of the other. If you pick a Morse function that is just the "height", then you will get critical values in the interior of the interval. On the other hand, you can alter critical values and get an equivalent Morse function such that $f(X)=[a_1, b_1]\cup \cdots \cup [a_m, b_m]$ so that the index $0$ happen at the $a_i$ and index $n$ happen at the $b_i$.2012-08-31
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    Are you saying the answer to the first question is in the affirmative? Please prove it. And as far as the second question goes, I meant *connected* manifold.2012-08-31
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    Woops. I meant "yes and no" to the second question. The point is that for the purposes of Morse theory you can alter the Morse function so that all critical points have distinct critical values. Then the previous argument shows that if they have index $0$ and $n$ and $X$ is connected, then there are only 2 critical points and the correspond to the max and min, so they must happen at the endpoints. The answer to the first question is yes, but I only know a complicated proof by attaching handlebodies.2012-08-31
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    Actually. I'm being somewhat circular. Let me think about this more. I guess I'm using that if you only have index $n$ and $0$, then you keep attaching $0$ and $n$ handlebodies and hence get a bunch of spheres. Once you know they are spheres you know you can spread out the critical values enough to get disjoint intervals. A priori you can only spread out the critical values a small amount.2012-08-31

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