Continuous bijections from the open unit disc to itself - existence of fixed points 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!

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 $H$. 
