As title goes. A `strict 2-to-1 continuous map' refers to a continuous map $f$ such that for any point $x\in D^2$, the preimage $f^{-1}(x)$ has exactly 2 points.
Actually, there are three problems. Let $D^2=\{x\in \mathbb{R}^2: |x|\leq 1\}$.
- Is there any map $f: D^2\to D^2$ which is strict 2-to-1?
- Is there any map $f: D^2_0\to D^2_0$ which is strict 2-to-1? Where $D_0^2=\{x\in \mathbb{R}^2:|x|<1\}$ the interior of $D^2$. This is equivalent to find 2-to-1 continuous map $\mathbb{R}^2\to \mathbb{R}^2$.
- Is there any map $f: D^2\to D^2$ which is strict 2-to-1 with the restriction of $f|_{S^1}=[z\mapsto z^2]$?
Here are some remarks.
When $f$ is a covering, all the above are not true. Because there is no connected covering for a simple connected space.
But ``the map having same cardinality of fibre is covering map'' is not true, consider these four maps which are not covering maps.
When $f$ is holomorphism, then $f$ will map the boundary to boundary. $f$ has and only has two zeros, then $f$ will be a product of two Blahcke factors (Since such zero free function with $|f|=1$ on boundary is constant by maximal modulus principle) which is very easy to check.
When $f$ is an open map, we can construct a continous map $g: D^2\to S^1$ by the following way. Let $f^{-1}(x)=\{a_1,a_2\}$, let $b$ be the point where the line passing through $a_1$ and $a_2$ intersect with the circle $S^1$, and define $g(x)=b^2$. So when $f|_{S^1}=[z\mapsto z^2]$, since now $g|_{S^1}=\operatorname{id}_{S^1}$, it is not true by an argument of fundamental group.
(Proof that $g$ is continous.) Let $$h:\underbrace{\{(x,y)\in D^2\times D^2:x\neq y\}}_{D'} \to S^1$$ be the map of ``square of intersecting $S^1$'', so for any open subset $U\subseteq S^1$, $$g^{-1}(U)=\{x\in D^2: \exists (a_1,a_2)\in D', x=f(a_1)=f(a_2): h(a_1,a_2)\in U\}=(f\times f)(h^{-1}(U))$$ is open, so $g$ is continous. I think this is a right direction, but without the open map assumption, $g$ may not be continous.
If we change $D^2$ by $D^1$, it is easy --- a mathmatical analysis exercise using argument of extreme value.
It is very `topological', since whether there exists a 2-to-1 map are highly depends on the topological properties of space. And I guess the answer should not be true (at least the last one). But I do not figure out the answer yet. I do not know even under the assumption $f$ is smooth.