Why inner variations are diffeomorphisms? Let $\Omega \subset \mathbb{R}^n $ be an open, bounded, connected domain with smooth boundary.
Let $\eta \in C_c^{\infty}(\Omega,\mathbb{R}^n)$ be a smooth compactly supported $n$-tuple of real valued functions.
Define $\phi(x)=x+\epsilon\, \eta(x)$.
How to prove that for sufficiently small $\epsilon$, $\phi$ is a diffeomorphism $\Omega \to \Omega$?
I can see why $\phi(\Omega)\subseteq \Omega$ for small $\epsilon$, but I don't see why $\phi|_{\Omega}$ is injective, or why $\phi(\Omega)\supseteq \Omega$. I do see why $d\phi$ is invertible for small $\epsilon$.
Edit:
Here is some progress: (Is there a more elementary way, which do not use the fact the Jacobian is null-Lagrangian?).
Assume $\phi$ is injective (see comment below).
First, we have
$$
\phi(\bar \Omega) \subseteq \overline{\phi( \Omega)} \subseteq \bar \Omega.
$$
The injectivity of $\phi$ implies
$$
\text{Vol}(\phi(\bar \Omega))=\int_{\bar \Omega} J\phi\stackrel{(1)}{=}\int_{\bar \Omega} \text{Id}=\text{Vol}(\bar \Omega),
$$
where equality $(1)$ follows, since the Jacobian is null-Lagrangian.
Thus, we have
$$
\phi(\bar \Omega) \subseteq \bar \Omega, \,\,\, \text{Vol}(\phi(\bar \Omega))=\text{Vol}(\bar \Omega).
$$
This implies that $\phi(\bar \Omega)=\bar \Omega$. Thus, $\phi:\bar \Omega \to \bar \Omega$ is bijective, hence by the inverse function theorem, it is a diffeomorphism.
Is there a direct way to see $\phi:\bar \Omega \to \bar \Omega$ is surjective?
 A: So, if we prove that $\phi$ is a diffeomorphism on $\mathbb{R}^n$, we are done, since it maps the complement of $\Omega$ and of $\bar\Omega$ in themselves.
As you and previous comments suggest, injectivity follows for small $\varepsilon$ as well as the invertibility of the differential. This actually holds in the whole $\mathbb{R}$.
To prove surjectivity, we just need to show that the image is open and closed (and non-void) since $\mathbb{R}^n$ is connected.
The openness of $\phi(\mathbb{R}^n)$ follows from the inverse function theorem, as for a given point $x$ of the image, the gradient calculated in its preimage is invertible (thus exists a whole neighborhood of $x$ that is in the image).
The closure follows by a compactness argument: Assume $x_n$ is a sequence of vectors in the image of $\phi$ that converges to $x\in\mathbb{R}^n$. Then, the preimage of the subsequence (let's call it $y_n$) is bounded. Taking a convergent subsequence $y_{n_k}\to y$, by the continuity of $\phi$, we must have $\phi(y)=x$.
I guess there are some theorems concerning bijections whose gradient is everywhere surjective, but I am not an expert in that.
