• Title: Regular Value Theorem in $\mathbb{R}^n$

  • Series: Manifolds

  • YouTube-Title: Manifolds 15 | Regular Value Theorem in ℝⁿ

  • Bright video: https://youtu.be/SENcpcFg5sA

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  • Original video for YT-Members (bright): https://youtu.be/tm4FeIz4Cas

  • Original video for YT-Members (dark): https://youtu.be/I9cXeqaOniM

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  • Quiz Content

    Q1: Let $f: \mathbb{R} \rightarrow \mathbb{R}^2$ be a $C^1$-function. Which claim is correct?

    A1: All the points $x \in \mathbb{R}$ are critical points of $f$.

    A2: All the points $x \in \mathbb{R}^2$ are critical points of $f$.

    A3: There are no critical points of $f$.

    A4: There is only one critical point of $f$.

    Q2: Let $f: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ be the function given by $f(x) = x$. Which claim is correct?

    A1: There are no critical points of $f$.

    A2: All the points $x \in \mathbb{R}^2$ are critical points of $f$.

    A3: There are no regular values of $f$.

    A4: There is only one critical point of $f$.

    Q3: Let $f: \mathbb{R}^n \rightarrow \mathbb{R}^m$ be a $C^\infty$-function. What is correct formulation of the regular value theorem?

    A1: If $c$ is a regular value of $f$, then $f^{-1}[{ c } ]$ is an $(n-m)$-dimensional submanifold of $\mathbb{R}^n$.

    A2: If $f^{-1}[{ c } ]$ is an $(n-m)$-dimensional submanifold of $\mathbb{R}^n$, then $c$ is a regular value of $f$.

    A3: If $f^{-1}[{ c } ]$ is a critical point of $f$, then $c$ is a regular value of $f$.

    Q4: Let $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ be given by $f(x,y) = x^2 + y^2 - 1$. Is $f^{-1}[{ 0 } ]$ a submanifold?

    A1: Yes, it is by the regular value theorem.

    A2: No, 0 is not a regular value.

    A3: No, the function is not well-defined.

    A4: One needs more information.

  • Last update: 2024-10

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