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Consider a smooth $n$-dimensional submanifold $A$ in $\mathbb{R}^{n+1} \times \mathbb{R}$ and the projection $f:\mathbb{R}^{n+1} \times \mathbb{R}\rightarrow \mathbb{R}$ onto the second factor. Is it possible to isotope $A$ such that $f$ is a Morse function with respect to the 'new' (isotoped) submanifold?

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up vote 3 down vote accepted

The answer is yes. A variant of this is proven in Milnor's "Morse Theory" text. I don't have it here with me, but what he does is instead of looking at linear functions on the ambient space, he looks at the function that gives the distance from a point in the ambient Euclidean space. He proves this function is Morse on $M$ for a generic choice of point off the manifold. You can do the same for linear functions on the ambient space, and the proof is much the same.

You may worry, that you don't care about linear functions but you have a prescribed linear function (coordinate projection). The idea is that if projection onto some direction is Morse, then you can simply rotate your manifold so that "some direction" becomes the coordinate direction.

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So, the projection to the last factor can be made Morse not just by an ambient isotopy, but by a rigid motion. Neat! – Neal Nov 15 '12 at 6:14

If $A$ is a submanifold of $M$ and you already have a Morse function $f$ on $M$. Then by definition $f$ must not have non-degenerate critical points on $A$. If $f$ is not a Morse function on $M$, $f$ could still be a Morse function on $A$. However it is not clear to me if this holds for arbitrary submanifold on $M$. You may need some embedding theorem like Hoef embedding theorem.

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But the $f$ in the question is very far from being a Morse function. Notice the question is somewhat different from what you answered: it asks if one can move the submanifold so that the fixed function $f$, restricted to it, is Morse. – Mariano Suárez-Alvarez Nov 13 '12 at 16:11
I should have undeleted my deleted parts, which asserts $f$ is not a Morse function. – Bombyx mori Nov 13 '12 at 18:38
In any case, it is not true that the restriction of a Morse function on $M$ to a submanifold (even if the latter is even an embedded one —a closed one, say) For example, restricting to a level surface of $f$ is not going to work! – Mariano Suárez-Alvarez Nov 13 '12 at 19:01
Yes....I did written down all these but I deleted them.... – Bombyx mori Nov 13 '12 at 19:05

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