# Continuous mapping $D^2 \to S^1$

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?

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$e^{i\pi|z|}{}$ – 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

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 $$(\textstyle\sqrt{1-y^2},y)\in S^1=\{(x,y)\in\mathbb{R}^2\mid x^2+y^2=1\}$$
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.