Nonlinear transformation of region from $\mathbb R^2\to\mathbb R^2$

Question

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!

Answer 1

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.

Answer 2

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 $T(\partial D)=\lbrace 0 \rbrace \neq \partial D$.