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Question 1.7.5 (Differential Topology - Guillemin and Pollack) Exhibit a smooth map $f : \mathbb{R} \to \mathbb{R}$ whose set of critical values is dense.

From [Exercise 1.1.18], there is a function $g : \mathbb{R} \to \mathbb{R}$ such that $g(x) = 1$ if $|x| \leq 1/4 $ and $g(x) = 0$ if $|x| \geq 1/2 $. Now, write $ \mathbb{Q} = \{q_1, q_2,...\}$, and then for $i \in \mathbb{N}$, define $g_i :\mathbb{R} \to \mathbb{R}$ by $g_i(x) = q_ig(x-i)$. Now define $f := \sum g_i$, then all the rationals are critical values for $f$ and dense in R.

I understand roughly the solution, except for one point. How can I get the function $g$ from the exercise 1.1.18(a). I think we would have applied a composition of an unknown function or played with parameters of the function provided in the exercise.

Book of Guillemin and Pollack : http://math.ucr.edu/~res/math260s10/old/difftopGP.pdf

  • The question is not clear. If you're looking for clarification on the exercise with the bump function (1.1.18), why focus the attention on a different exercise (1.7.5)? – Amitai Yuval Jul 23 '15 at 18:59
  • This exercise 1.7.5 is realized from the exercise 1.1.18. The solution provided the Answer Key from a web. I just need details on how they arrive at this result. –  Jul 23 '15 at 19:12
  • There exist a Answer Key from the website https://chriseurmath.files.wordpress.com/2014/07/eur_difftopgp_soln.pdf –  Jul 23 '15 at 19:14
  • No, I'd need precision on the solution of the exercise 1.7.5. –  Jul 23 '15 at 20:33

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Let $f(x) = \exp (-1/(x(1-x)), x\in (0,1), f(x) = 0$ elsewhere. Then $f\in C^\infty(\mathbb {R}),$ and $f(1/2) = e^{-4}$ is a critical value. Let $q_1,q_2, \dots $ be the rationals. Then

$$F(x)=\sum_{n=-\infty}^{\infty}q_nf(x-n)$$

is $C^\infty$ on $\mathbb {R},$ and the critical values of $F$ include $q_1e^{-4},q_2e^{-4},\dots ,$ which comprise a dense subset of $\mathbb {R}.$

zhw.
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  • How do you get $f(x)$? A part of the function in the exercise is $exp(-1/x^2)$ and none $exp(-1/x)$. –  Jul 24 '15 at 02:00
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    You can let $f(x) = \exp (-1/(x^2(1-x)^2), x\in (0,1)$ if you like. It doesn't really matter. Any $f$ smooth on $\mathbb R$ with $f=0$ off of $(0,1)$ and $f\ne 0$ somewhere in $(0,1)$ would work as well. – zhw. Jul 24 '15 at 13:58