Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

It is a well known result in algebraic topology that there is no retraction of $D^2$ onto $S^1$.

Does anyone know any continuous maps $D^2 \to S^1$ which are not constant?

share|cite|improve this question
$e^{i\pi|z|}{}$ – t.b. May 13 '11 at 2:08
There are no continuous maps of $D^2$ to $S^1$ that are the identity on $S^1\subset D^2$. – Thomas Andrews May 13 '11 at 2:49
In particular, a map $f:S^1\rightarrow S^1$ can be extended to all of $D^2$ if and only if $f$ is homotopy-equivalent to a constant. – Thomas Andrews May 13 '11 at 2:53
I have to ask: how does one get to the point where one knows what a retraction is and, well, that algebraic topology is something, and is not able to construct a non-const. cont. function like that? :/ – Mariano Suárez-Alvarez May 13 '11 at 3:50
Since Mariano's comment was up voted so many times I feel compelled to respond! I was trying to come up with a map from D^2 to S^1 which when restricted to the boundary has degree 2. I couldn't think of one which is why I asked the question. I probably should have been more explicit. Thomas answered this question though... – DBr May 14 '11 at 4:44
up vote 3 down vote accepted

How about the map that sends $$(x,y)\in D^2=\{(x,y)\in\mathbb{R}^2\mid x^2+y^2\leq 1\}$$ to $$(\textstyle\sqrt{1-y^2},y)\in S^1=\{(x,y)\in\mathbb{R}^2\mid x^2+y^2=1\}$$

share|cite|improve this answer

If you can think of a path $\gamma\colon I \to S^1$, then you get a continuous map $D^2 \to S^1$ by first collapsing $D^2$ to the interval $I$ and then composing with your path.

If your path $\gamma$ was not constant, then the resulting map $D^2 \to S^1$ will not be constant.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.