Suppose we have a manifold $X$ embedded in $\mathbb{R}^n$. Define the vector field $v_u(p) : X \rightarrow TX$ by taking the point $u \in \mathbb{R}^n$ to its natural (orthogonal) projection onto $TX_p$. How can we show that the set of $u \in \mathbb{R}^n$ such that every zero of $v_u$ is nondegenerate is dense in $\mathbb{R}^n$?
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Your question doesn't seem to make any sense. For example, is $M$ supposed to be $X$? If the vector field is only defined on $X$, how can we end up with a dense subset of $\mathbb{R}^n$? – Rhys May 17 '13 at 13:29
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Fixed. $M$ was supposed to be an $X$. – Vladimir Chernousov May 17 '13 at 13:33
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Oh, and ignore my second question; I misunderstood. Sorry! :-) – Rhys May 17 '13 at 13:39
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It looks like the Morse lemma could be useful. – Hans Engler Jul 29 '13 at 06:33