Let $D \subseteq \mathbb{R}^2$ be the closed unit disk. Let $f:D \to D$ a smooth map with everywhere invertible differential. For each $t \in (0,1]$ let $D_t$ denote the closed disk with radius $t$.
Question: Suppose that $f$ is not a diffeomorphism on $D$. Does there exist a minimal $t \in (0,1]$ such that the restriction $f|_{D_t}$ is not a diffeomorphism onto its image?
Does the answer change if we replace $D_t$ by the open disk with radius $t$? (If not, does the critical value of $t$ change?)
Note that the set $A=\{ t \in (0,1] \, | \, f|_{D_t} \text{is a diffeomorphism onto its image} \}$ is clearly "closed down", i.e. $\,$ $(y \in A \, \text{ and }\, 0<x<y) \Rightarrow x \in A$. Writing $\sup A=s$, we then must have $A=(0,s)$ or $A=(0,s]$. An equivalent formulation of the question would then be to prove that $A$ is an open set, or $s \notin A$. (Then $s$ would be the required minimal $t$).
My immediate approach (assuming we are talking about open disks, actually) was the following:
Note that the restriction of $f$ to an open subset of $\text{int}(D)$ is a diffeomorphism onto its image if and only if it is injective (due to the inverse function theorem). Now, take $s_n \searrow s$; we know that $f|_{D_{s_n}}$ is not injective- so there exist $x_n \neq y_n \in D_{s_n}$ such that $f(x_n)=f(y_n)$. By compactness we can assume that $x_n \to x,y_n \to y$. If $x \neq y$ we are done since $x,y \in D_s$. But I don't see any reason to exclude the possibility $x=y$.
The inverse function theorem (applied at $x=y$) ensures that $f$ is one-to-one on a neighborhood of $x$.