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Consider a function $f:X\rightarrow \mathbb{R}^+$, where $X=[-M,M]\subseteq \mathbb{R}$. $f(x)=\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{x^2}{2\sigma^2}}$ The paper goes on to define a cover of $\mathbb{R}^+$ by $\{[0,5),(4,10),(9,15),(14,\infty)\}$.

They say that $f^{-1}([0,5))$ consists of a single connected component and $f^{-1}([4,10))$,$f^{-1}([9,15))$,$f^{-1}([14,\infty))$ have two distinct connected component on the positive and negative half-line respectively. They have not provided a value of $\sigma$ and $M$, so I assume that they say it is true for any of them. This is not clear to me as $f$ goes to $0$ only at $\pm \infty$, so without knowing them how is it possible to claim this?

Reference: https://research.math.osu.edu/tgda/mapperPBG.pdf Page 4, example 2.4

cookiemonster
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1 Answers1

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It is false for all values of $M$ and $\sigma$.

Note that

  1. $f$ is an even function with positive values.

  2. $f$ is strictly decreasing on $[0,M]$ and strictly increasing on $[-M,0]$.

  3. If $f(M) \le a \le f(0)$, then there exists a unique $r_a \in [0,M]$ such that $f(r_a) = a$. This follows from the intermediate value theorem and 2.

Consider a subinterval $J$ of $\mathbb R^+$ which is open in $\mathbb R^+$. It has the form $(a,b)$ or $[0,b)$ with $a \ge 0$ and $b \le +\infty$.

Case 1. $J = (a,b)$ with $a \ge f(0)$. Then $f^{-1}(J) = \emptyset$ which certainly does not have two components. I would say that $\emptyset$ has $0$ components (see Empty space connected? If so, then what are its components?.

Case 2. $J = (a,b)$ with $a < f(0) <b$. Then $f^{-1}(J)$ has one component. In fact, if $f(M) > a$, then $f^{-1}(J) = X$, and if $f(M) \le a$, then $f^{-1}(J) =(-r_a,r_a)$.

Case 3. $J = (a,b)$ with $b \le f(0)$. If $f(M) \ge b$, then $f^{-1}(J) = \emptyset$. If $f(M) < b$, then $f^{-1}(J) = [-M,-r_b) \cup (r_b,M]$ provided $f(M) > a$ and $f^{-1}(J) = (-r_a,-r_b) \cup (r_b,r_a)$ provided $f(M) \le a$. In both cases $f^{-1}(J)$ has two components.

Case 4. $J = [0,b)$ with $f(0) < b$. Then $f^{-1}(J) = X$ which has one component.

Case 5. $J = [0,b)$ with $b \le f(0)$. If $f(M) \ge b$, then $f^{-1}(J) = \emptyset$. If $f(M) < b$, then $f^{-1}(J) = [-M,-r_b) \cup (r_b,M]$ which has two components.

It is said that $f^{-1}([0,5))$ consists of a single connected component. Due to the above considerations this requires $f(0) < 5$ which implies $f(X) \subset (0,5)$. Thus $f^{-1}((9,15))$,$f^{-1}((14,\infty))$ are empty and certainly do not have two components, and $f^{-1}((4,10))$ has one component.

Paul Frost
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