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I'm wondering about the following:

Let $f:D \mapsto D$ be a continuous real-valued bijection from the open unit disc $\{(x,y): x^2 + y^2 <1\}$ to itself. Does f necessarily have a fixed point?

I am aware that without the bijective property, it is not necessarily true - indeed, I have constructed a counterexample without any trouble. However, I suspect with bijectivity it may be the case. I'm aware of the Brouwer Fixed Point Theorem and I imagine these two are intricately linked. However, i'm not certain where the bijectivity comes in - I believe we can argue something along the lines f now necessarily maps boundary to boundary - something about how if $x^2+y^2 \to 1$, $\|f(x,y)\| \to 1$ maybe. However, how does this help? Even if we could definitely define a limit to f(x,y) along the whole boundary and apply Brouwer, we can't guarantee the fixed points aren't all on the boundary anyway.

Conversely however, I still can't construct a counterexample. Could anyone help me finish this off please? Thanks!

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Hint: The open disk is homeomorphic to the plane. – Chris Eagle Jan 7 '11 at 4:37
Thanks Chris, I thought it would be true but considering that, it's quite easy to construct a counterexample! – Spyam Jan 8 '11 at 21:45
up vote 2 down vote accepted

Let $D$ be open disk with center $0$ of radius $1$ in complex plane. Consider holomorphic automorphism of $D$ given by formula $z \mapsto \frac{z+a}{1+\bar a z}$ with $a\in D$. Does it have fixed point if $a\neq 0$ ? [You must solve $z=\frac{z+a}{1+\bar a z}$]

You can also biholomorphically send $D$ to right half plane $H$ of $\mathbb C$ by $z \mapsto \frac{1+z}{1-z}$ and then apply Chris Eagle comment to $H$.

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Thank you very much, I figured out a way to construct such a map where the claim fails myself in the end! – Spyam Jan 8 '11 at 21:44

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