# Continuous surjective functions from the unit disk to itself that agree nowhere

Do there exist two continuous surjective functions $f,g:D \to D$ such that $f(z) \neq g(z)$ for all $z \in D$, where $D$ is the closed unit disk?

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Since you are new, I want to give some advice about the site: To get the best possible answers, you should explain what your thoughts on the problem are so far. That way, people won't tell you things you already know, and they can write answers at an appropriate level; also, people are much more willing to help you if you show that you've tried the problem yourself. If this is homework, please add the [homework] tag; people will still help, so don't worry. Also, many would consider your post rude because it simply demands an answer, instead of requesting help, so please consider rewriting. –  Zev Chonoles Feb 18 '13 at 16:18
Closed disk or open disk? –  Jonas Meyer Feb 18 '13 at 20:59

Note: $D$ is now known to be the closed unit disk... I'll edit when I can.

I'll assume you mean $D$ to be the open unit disk.

Take $$f(z)=z$$ and $$g(z)=\frac{z+1/2}{1+z/2}=\frac{2z+1}{z+2}.$$

The latter is a biholomorphism of $D$ onto itself.

And it is easy to check that it has no fixed point in $D$.

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@user62832 Do you agree with this example? Also, let me know if you want more details. –  1015 Feb 19 '13 at 2:21
$D$ is closed unit disk –  user62832 Feb 19 '13 at 2:53
@user62832 This makes a big difference. You should have made this clear from the beginning. –  1015 Feb 19 '13 at 2:56
sorry for the mistake. could you give me some hints on how to proceed? –  user62832 Feb 19 '13 at 5:41
@user62832 I'm still thinking about it on and off. I know the answer is no if additionnally require one of them to be bijective. That's all I have so far. –  1015 Feb 19 '13 at 5:43

Here is a example. Let $D$ be the closed unit disk of $\mathbb{C}$ and $D_L:=\{ z \in D | Re(z) \geq 0 \}$, $D_R:=\{ z \in D | Re(z) \leq 0 \}$ be the right half and the left half unit disk.

Let $f_1 : D \rightarrow D_L$ be an homeomorphism such that $f_1([-1,1])=[0,1]$ and such that $f_1$ sends the upper (resp. lower) half disk of $D$ homeomorphically to the upper (resp. lower) quater disk of $D_L$. Define an homeomorphism $g : D \rightarrow D_R$ in the same way.

Define $$f_2 : D_L \rightarrow D, z=r e^{i\theta} \mapsto z^2=r^2 e^{i2\theta},$$ $$g_2 : D_R \rightarrow D, z=r e^{i(\pi-\theta)} \mapsto r^2 e^{i(\pi+2\theta)}.$$ You should think $f_1$ and $g_1$ as Horseshoe maps. Note $g_2$ is a horseshoe map composed with a symmetry (Make a picture !).

Now define $f,g : D \rightarrow D$ by $f = f_2 \circ f_1$ and $g = g_2 \circ g_1$.

It is obvious that $f$ and $g$ are continuous and onto. The functions $f$ and $g$ disagree everywhere because :

• $f$ sends the upper (resp. lower) half disk onto the upper (resp. lower) half disk.
• $g$ sends the upper (resp. lower) half disk onto the lower (resp. upper) half disk.
• $f$ sends $[-1,1]$ to $[0,1]$ with $f(-1)=0$.
• $g$ sends $[-1,1]$ to $[-1,0]$ with $g(1)=0$.
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Hint: Look up the classification of elements of $SL(2;\mathbb{R})$. The answer will depend (but only slightly) on whether you are considering the open or closed unit disk.

Further hint: An element of $SL(2;\mathbb{R})$ gives an isometry of hyperbolic space. (Isometries are in particular continuous and surjective.) Hyperbolic space can be identified with the unit disk. Take one of your maps to be the identity. For the other map, it will suffice to find an isometry of hyperbolic space which fixes no points. This is where the classification of elements of $SL(2;\mathbb{R})$ is useful.

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Please be more specific. Beyond superficial connection with isomorphisms of the unit disk I don't see any relevance of your answer to the OP. –  Marek Feb 18 '13 at 19:20
@Marek I have clarified. Unless I am unable to read today (always a possibility), I think this indicates the answer to the OP. –  Neal Feb 18 '13 at 20:38
@Neal :sir can i take f to be the translation map and g=identity map? –  user62832 Feb 19 '13 at 3:52
@user62832 Does a translation map the unit disk onto itself? –  Neal Feb 19 '13 at 3:56
@Neal any matrix $A \in SL_2(\mathbb R)$ with $tr(A) > 2$ has fixed points on the real line which is not a part of hyperplane –  user62832 Feb 19 '13 at 5:31