The diffeomorphism of neighbourhood

$M,N$ are two smooth $n$-manifolds and $A$ is a subset of $M$. A smooth mapping $f$ from $M$ to $N$ has the property that $df$ is nonsingular and $f$ is injective on $A$. Is there a neighbourhood $U$ of $A$ such that $f$ is a diffeomorphism from $U$ to its image?

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Neighbourhood containing A, or neighbourhood contained in A? Actually, I think the answer either way is no... – Zhen Lin Mar 20 '12 at 7:58

If $A$ is allowed to be arbitrary, the answer is no. Here's a counterexample: let $M=\mathbb R$, $N=\mathbb R^2$, $A = [0,2\pi)$, and $f(t) = (\cos t,\sin t)$. Then any smooth extension of $f$ to a neighborhood of $A$ will take a small neighborhood of $0$ onto a neighborhood of $(0,0)$, and thus cannot be injective.
If $A$ is compact, then the conjecture is true. Here's a sketch of a proof. First, let $F$ be any smooth extension of $f$ to an open neighborhood $U$ of $A$. After shrinking $U$ if necessary, we can arrange that $dF_x$ is nonsingular for all $x\in U$, so $F$ is a local diffeomorphism. Let $g$ be a Riemannian metric on $N$, and give $U$ the pullback metric $F^*g$, so $F$ is a local isometry.
For each $x\in A$, choose $\epsilon_x>0$ small enough that $F$ is injective on $B_{3\epsilon}(x)$, and such that the distance from $F(x)$ to the compact set $F(A)\smallsetminus B_{2\epsilon_x}(F(x))$ is greater than $\epsilon_x$. Then the restriction of $F$ to the open set $\bigcup_{x\in A} B_{\epsilon_x}(x)$ is an injective local diffeomorphism, hence a diffeomorphism.
Let $M:={\mathbb R}$ and $N:={\mathbb R}/{\mathbb Z}$ and consider the quotient map $f:\ t\mapsto [t]$. Then $df={\rm id}$ $\ (t\in{\mathbb R})$, and $f$ is injective on $A:=[0,1[\$, but there is no neighborhood $U$ of $A$ such that $f$ can be extended to a diffeomorphism $\tilde f:\ U\to N$.