If I have a given continuous nonlinear map $T:\mathbb{R}^2\rightarrow \mathbb{R}^2$, and a region $D \subset \mathbb{R}^2$, is it necessarily true that $T(\partial D)=\partial T(D)$? That is, do boundary points of D get mapped to the boundary of the image of D after applying T?

I can see how this does not hold if $T$ is discontinous, but I can't think of a continuous $T$ where this does not hold. It also "feels right," but that's gotten people in trouble before!

I was attempting to prove this by looking at the effect on the open neighborhood around a boundary point after applying $T$... but couldn't make it very far.

Thanks for your help!

  • $\begingroup$ In addition to my answer, I recommend checking out Section "Properties" in the article Open and closed mappings. $\endgroup$ – avs Mar 20 at 22:16

Your instincts are right: this statement is not true.:) A couple of counterexamples:

  1. Let $D = {\mathbb R}^2$, and let $T$ be this projection: $$ T(x, y) = x^3. $$ Then $D$ has no boundary, while the image of $T$ is a line, hence is its own boundary.

  2. Let $D = \{ (x, y) \; : \; -\pi \leq x \leq \pi, \; 1 < y < 2 \}$, and let $$ T(x, y) = \left(\; y \cos(x), \; y \sin(x) \; \right). $$ This maps $D$ onto the open annulus centered at the origin and with radii 1, 2. However, the boundary points of $D$ with $|x|=\pi$ are mapped into interior points of the annulus.

  3. Let $D$ be the open strip $$ \{(x, y) \; : \; -\pi/2 < x < \pi/2, \; y \in {\mathbb R} \}. $$ Its boundary is the union of the two lines $|x| = \pi/2$. Now let $\tan$ be the principal branch of the tangent function and let--you guessed it!)-- $$ T(x, y) = (\tan(x), y). $$ Then $T(D)$ is all of ${\mathbb R}^2$, hence has no boundary at all.

  • $\begingroup$ These are good examples! I'm curious if it's possible to construct a polynomial map from a region to another region which does not have the property that $T(\partial D)=\partial T(D)$. In your first example you do provide a polynomial function (projection), but it's on the entirety of $\mathbb{R}^2$. $\endgroup$ – CuriousMathsStudent Mar 20 at 22:21
  • $\begingroup$ Oh, you want $D$ to be a proper subset of the plane, something like this? $D$ = the half-disc $|z| \leq 1, {\tt Re} z \geq 0$ in the complex plane, and $T(z) = z^2$. $\endgroup$ – avs Mar 20 at 22:40
  • $\begingroup$ Thank you very much! Good example. $\endgroup$ – CuriousMathsStudent Mar 20 at 22:55

It is not true.

Let's see $\mathbb{R}^2$ as $\mathbb{C}$, and consider the application $T : \mathbb{C} \rightarrow \mathbb{C}$ defined for all $z \in \mathbb{C}$ by $$T(z)=2z \text{ }\text{ if }\text{ } |z|\leq 1, \quad T(z)=(4-2|z|)z \text{ }\text{ if }\text{ } 1 < |z| \leq 2, \quad \text{and } T(z)=0 \text{ }\text{ if }\text{ } |z| > 2$$

You can check that $T$ is continuous.

Let $D = \lbrace z \in \mathbb{C} \text{ }|\text{ } |z| \leq 2\rbrace$. You can see that $T(D)=D$, but $T(\partial D)=\lbrace 0 \rbrace \neq \partial D$.

  • $\begingroup$ Ah, thank you! This is a great example. Thanks for the constructive response. As a followup, if we restrict our map $T$ to being polynomials (i.e. such that the function is smooth), does this hold? As an example, T(x,y)=<x^3+y,2x+y^2> would be a map fitting this description. $\endgroup$ – CuriousMathsStudent Mar 20 at 22:09

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