1

I was asked whether a closed unit ball is a retract of the euclidean space $\mathbb R^2$. I think the answer is yes and the retraction might be defined as follows: for all the points in $\mathbb R^2$ join them with the origin by a straight line, then wherever it will cut the ball first that would be the image of that point under the function.

This is just an idea but I want a rigorous proof. Can somebody tell me? Any other function is also welcome.

R. H. Vellstra
  • 3,263
akansha
  • 1,913
  • 1
    Hint: Let $w$ be a vector such that $|w|>1$ and $v$ the corresponding unit vector. Consider the retraction given by $(1-t)w+tv$ for $t\in[0,1]$. For points inside the disk, they should be fixed by the retraction. – Michael Burr Aug 18 '15 at 13:19
  • 3
    The function you have defined is simply $$x\mapsto \begin{cases} x &: |x| \leq 1 \ \frac{x}{|x|} &: |x| > 1\end{cases}$$ where $|(x,y)| = \sqrt{x^2+y^2}$. You now have to check that this is continuous (use the fact that $x\mapsto |x|$ is continuous by the triange inequality) – Prahlad Vaidyanathan Aug 18 '15 at 13:31
  • 1
    @MichaelBurr that's a deformation retract, which is more than is needed. – Dan Rust Aug 18 '15 at 14:22

1 Answers1

1

To expand slightly on the comments, let $D^n = \{ x \in \mathbb R^n \mid \lVert x \rVert \leq 1\}$ and define $r:\mathbb R^n \to D^n$ piecewise by $$r(x) = \begin{cases} x, & \lVert x \rVert \leq 1 \\ \frac{x}{\lVert x \rVert}, & \lVert x \rVert \geq 1 \end{cases}.$$

Note that if $\lVert x \rVert =1$ then the two pieces agree.

As $x \mapsto \lVert x \rVert$ is continuous (see, e.g., Why are norms continuous?), then $r$ is continuous on $\{ x \mid \lVert x \rVert \geq 1\}$, and hence continuous on $\mathbb R^n$, e.g., by the Pasting Lemma.

If $i : D^n \hookrightarrow \mathbb R^n$ denotes the inclusion, then it is also evident that $r \circ i = \operatorname{id}_{D^n}$, hence $r$ is a retraction.

R. H. Vellstra
  • 3,263