Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
13  
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
add comment

1 Answer

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 $H$.

share|improve this answer
    
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
add comment

Your Answer

 
discard

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.