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

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.

• In addition to my answer, I recommend checking out Section "Properties" in the article Open and closed mappings. – 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.

• 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 $\mathbb{R}^2$. – CuriousMathsStudent Mar 20 at 22:21
• 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 $T(z) = z^2$. – avs Mar 20 at 22:40
• Thank you very much! Good example. – 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 $$T(\partial D)=\lbrace 0 \rbrace \neq \partial D$$.

• 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. – CuriousMathsStudent Mar 20 at 22:09