Is a smooth surjective map between disks which is injective a.e. everywhere injective?

Question

Let $0<\lambda$, and let $D_1,D_{\lambda} \subseteq \mathbb{R}^2$ be the closed Euclidean disks of radii $1$ and $\lambda$ respectively.

Suppose we have a smooth surjective map $f:D_1 \to D_{\lambda}$, and that for almost every $y \in D_{\lambda}$, the set $f^{-1}(y)$ is a singleton.

Is it true that $f$ is injective? i.e. can we deduce that $f^{-1}(y)$ is a singleton for every $y \in D_{\lambda}$, not just for almost every $y$?

Does the answer change if we assume in addition that $\det(df_x)=\lambda^2$ for every $x \in D_{1}$?

Answer 1

1. No: there exists a smooth surjective map $f: D_1\to D_\lambda$ (even diffeomorphic near the boundary) such that $f$ collapses a closed interval $\alpha$ in $int(D_1)$ to a point (the center $o\in D_\lambda$) and is 1-1 away from $\alpha$: $f:D_1 -\alpha \to D_\lambda -\{0\}$ is a diffeomorphism. You first construct a PL version of $f$ and then smooth it out. Thus, a smooth map between disks which is injective away from the preimage of a measure zero set, need not be injective.

Edit. It is technically easier to construct an example for (1) such that $f$ is constant on a subdisk $D'\subset int(D_1)$ of radius $\frac{1}{2}$ and injective on $D_1- D'$. Namely, let $\phi(r), r\ge 0$, be a $C^\infty$ function vanishing on $[0, \frac{1}{2}]$, strictly increasing on $[\frac{1}{2},1]$, and satisfying $\phi(1)=\lambda>0$. Then, in polar coordinates, set $f(r,\theta)= (\phi(r),\theta)$. The map $f$ will be $C^\infty$, constant on $D'$ and 1-1 on the annulus $D_1\setminus D'$, while $f(D_1)=D_\lambda$.

2. Yes: If you assume that $f: D_1\to D_\lambda$ is smooth with invertible derivative at every $x\in D_1$ then either (i) $f$ is injective or (ii) there exists an open subdisk $D'\subset int(D_1)$ such that $f^{-1}(f(D'))= D'\sqcup D''$, where $D''$ is another open subdisk in $int(D_1)$. In particular, in the case (ii), for a subset $A$ of positive measure in $D_\lambda$, $\forall a\in A$, $card(f^{-1}(a)\ge 2$. In particular, if you assume that $J_f$ is constant (and positive) then this observation applies.