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Let $M$ be a differential manifold and let $f:M\to \mathbb{R}$ be a $C^{\infty}$ function such that there are exactly two points $x$ which satisfy $d_xf=0$. Let $p$ and $q$ those two points, and suppose that $f(p)=f(q)$. Prove that there is another $C^{\infty}$ function $g:M\to \mathbb{R}$ such that $p$ and $q$ are the only critical points but $g(p)\neq g(q)$.

I have been trying to solve this problem but I cannot come up with a solution. Clearly we can suppose $f(p)\neq 0$, so I intended to separate (by the $T2$ property) $p$ and $q$ with two open subsets $U$ and $V$, and letting $W$ an open set such that $p\in W\subseteq \overline{W}\subset U$ and $\overline{W}$ compact. Then I intended to make use either of a bump function or partition of unity, but I could not succeed.

How would you solve the problem?

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You're on the right track. Take $q\in W\subset\overline W\subset U$, and create a smooth function $\phi$ on $M$ that is $1$ on $\overline W$ of $q$ and $0$ outside $U$. We're going to consider the function $g=f+\phi$. Obviously, $p$ and $q$ are critical points of $g$ and $g(p)\ne g(q)$. But we may introduce other critical points, which necessarily must lie in $\overline U-W$.

So here's the trick. Let $C = \inf\limits_{x\in \overline U-W} \|df_x\|$ and let $C' = \sup\limits_{x\in\overline U-W} \|d\phi_x\|$. Now replace $\phi$ with $c\phi$, where $0<c<C/C'$. (You can compute these norms by working in a single chart centered at $q$. Just choose $U$ inside that chart.)

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  • $\begingroup$ But if the function is $0$ at a neighborhood of $p$, it would have an infinite number of critical points, and I want only two. $\endgroup$ Commented Jul 19, 2018 at 15:21
  • $\begingroup$ Ah, I intended you to add this function to the original function $f$. But we need to take some care to make sure that we don't introduce another critical point far away from $p$ and $q$. I'll edit slightly. $\endgroup$ Commented Jul 19, 2018 at 15:26
  • $\begingroup$ @solomeoparedes: Do you have any further questions? If not, please accept the answer so the question no longer appears as "unanswered." $\endgroup$ Commented Jul 21, 2018 at 16:57

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