# Consequence of Brouwer / No Retraction

I am trying to prove a consequence of the Brouwer Theorem and/or the No Retraction Theorem and I think it must be simple I am just not seeing the point. The result is: If $f:D^2\rightarrow D^2$ is continuous and satisfies $f|_{S^1}=Id_{S^1}$, then it is surjective.

So far, this is what I have. Let $x\in D^2$. If $x\in S^1$, then $f(x)=x$. Now I don't know how to argue if $x\in D^2-S^1$.

Many thanks!

• If $x\in D^2 - S^1$, you can project from $x$ to $S^1$. Call this projection $p$. $p\circ f$ is then a retraction from $D^2$ to $S^1$. – Callus - Reinstate Monica Apr 18 '16 at 23:30
• Reason by contradiction: If it failed to be surjective, you could give a retraction of the disk to its boundary. – Ted Shifrin Apr 18 '16 at 23:31
• It's not really clear to me. It seems to me that I could do that to all $x$, even if they are image of something – user194469 Apr 18 '16 at 23:38
• Oops, I should have said $x\in D^2$ but not in $\text{Im}(f)$. Note that $p$ is not defined at $x$. – Callus - Reinstate Monica Apr 19 '16 at 4:25
• The intuition is simply that if $f : D^2 \to D^2$, $f|_{\partial D^2} = \text{id}$ and $f$ failed to be surjective (say, the image doesn't hit the origin), then you could push $f$ away from the origin to get the image fit into $\partial D^2$ using the retract $r : D^2 - 0 \to \partial D^2$. But $f$ leaves $\partial D^2$ identity by hypothesis, so composition of the two would also leave $\partial D^2$ identity. That's a retract. Maybe trying to visualize it with $[0, 1]$ replaced by $D^2$ would help. – Balarka Sen Apr 19 '16 at 9:19

Suppose $f$ is not surjective then we can find an $x_0\in D^2\setminus \operatorname{Im}f$ and this point is not on $S^1$.
Define a map $r:\operatorname{Im}f\to S^1$ as follows - For every $x\in \operatorname{Im}f\subseteq D^2$ consider the ray beginning at $x_0$ and ending at $x$. Extend this ray till it meets the boundary $S^1$ of $D^2$. Define $r(x)$ to be the point on $S^1$ that this ray hits. Then $r$ is clearly well-defined and continuous. So $r\circ f:D^2\to S^1$ gives a retraction of $D^2$ onto $S^1$ which is a contradiction to the no retraction theorem.