Proving global injectivity (and perhaps surjectivity) of these smooth maps In this article, in sect. 4 (p. 13), Léna introduces a family of functions, and says that, for $\mathbf t$ sufficiently close to zero, «it is easy to check that $\Phi_t$ is a smooth diffeomorphism that sends $\Omega$ onto itself. Here is the definition of $\Phi_t$:

The $X_i$ are fixed points (the poles of an Aharonov-Bohm operator) in an open (perhaps simply connected) domain $\Omega$. So this is essentially a "smooth junction" of a translation in each $B_i'$ and the identity outside all the $B_i$'s.
Now, I have used IFT and proved that, for $\|\mathbf t\|$ small enough, the Jacobian determinant of $\Phi_t$ has modulus at least $\frac12$, and in particular is nonzero, making this a locally invertible map. To make it a diffeomorphism, I need smoothness and global injectivity. Smoothness is evident from the definition. I have, btw, also proved $\Phi_t(\Omega)\subseteq\Omega$, but not vice versa.
But this is not enough. No-one guarantees this map will be globally injective. For example, the complex exponential, viewed as a map from the plane into itself, is not injective, yet the determinant is greater than zero: $\det J\operatorname{exp}(z)=e^{2\operatorname{Re}z}$. If we take any subset $\operatorname{Re}z\geq K$, we will make said determinant greater than a constant. For example, if $K=\frac12\log\frac12$, we will have $\det\geq\frac12$, just like in our $\Phi_t$ case, but still the function won't be globally injective.
I honestly have no clue how to prove this $\Phi_t$ is globally injective on $\Omega$. Any suggestions?
Also, is it perhaps also surjective? I know the image is open because for each point in the image I can find a neighborhood of the point and one of the preimage (any preimage) such that $\Phi_t$ maps one neighborhood to the other diffeomorphically. $\Omega$ should be connected, so to prove surjectivity we merely need the image to be closed. Suppose $x\in(\Phi_t(\Omega))'$. Then we have a sequence $\Phi_t(y_n)=x_n\to x$. Since all the components of $J\Phi_t$ (the Jacobian) are bounded, we can say that $\|\Phi_t(x)-\Phi_t(y)\|\leq K\|x-y\|$, but that is only local, because it involves Taylor, or it only holds for $x,y$ such that $tx+(1-t)x\in\Omega$ for all $t$ (and I don't think $\Omega$ is assumed to be convex), and besides, we would want the reverse inequality to conclude $y_n$ is Cauchy, hence converges, and then conclude by continuity of $\Phi$. Naturally, without injectivity, I have no way to prove this reverse inequality, because injectivity follows from it. So how can I either prove injectivity first and use it to deduce this reverse inequality and surjectivity, or prove this reverse inequality directly and conclude everything from it?
 A: We need to show that for every $i$, there is a $c_i > 0$ such that for $\lVert t\rVert < c_i$, the restriction of $\Phi_t$ is a bijection of $\overline{B_i}$ with itself.
Once that is done, since there are only finitely many $B_i$ to consider, we see that $\Phi_t$ is a global bijection for $\lVert t\rVert < c:= \min \{ c_i : 1 \leqslant i \leqslant N\}$, and since $\Phi_t$ is a local diffeomorphism for $\lVert t\rVert < \delta$, it follows that $\Phi_t$ is a diffeomorphism for $\lVert t\rVert < \min \{c, \delta\}$. Since none of the $\partial B_i$ intersects $\partial \Omega$, each $B_i$ is either contained in $\Omega$ or in the complement of $\overline{\Omega}$, and since $\Phi_t$ fixes all points not contained in any $B_i$, and maps $B_i$ to itself, it follows that $\Phi_t(\Omega) = \Omega$ for these $t$.
So let's look at an arbitrary but fixed $B_i$. To simplify notation, we can assume that the centre $X_i$ of $B_i$ is $0$. Also, since $\chi_j(x) = 0$ for $x\in B_i$ and $j \neq i$, we can ignore the $\chi_j$ for $j\neq i$. Effectively, we can assume that $N = 1$ and drop the indices.
As a first step, we want to see that for all sufficiently small $\lVert t\rVert$, we have $\Phi_t(B) \subset B$. Since
$$\lVert\Phi_t(x) - x\rVert \leqslant \chi(x)\cdot \lVert t\rVert$$
(I'm not reducing $t$ to only the two components used for this ball, we'd have equality if I did), a sufficient condition is that
$$\chi(x)\lVert t\rVert < r - \lVert x\rVert$$
for all $x\in B$. By the mean value theorem, since $\chi$ vanishes on $\partial B$, we have
$$\chi(x) \leqslant M\cdot \operatorname{dist}(x,\partial B) = M\cdot (r - \lVert x\rVert),$$
where $M$ is a bound for the norm of the derivative of $\chi$. It follows that $\Phi_t(B) \subset B$ if $\lVert t\rVert < 1/M$. We consider only such $t$ from here on.
Since $\chi$ vanishes on $\partial B$, we have $\Phi_t(x) = x$ for all $x\in \partial B$, regardless of $t$. If $\Phi_t\lvert_{\overline{B}} \colon \overline{B} \to \overline{B}$ were not surjective, it would induce a retraction $\overline{B} \to \partial B$, but such a retraction doesn't exist, since balls have trivial reduced homology while spheres have nontrivial reduced homology (or argue with homotopy groups or whatever you are familiar with). Thus $\Phi_t\lvert_{\overline{B}}(\overline{B}) = \overline{B}$ for $\lVert t\rVert < 1/M$.
It remains to see that $\Phi_t$ is injective on $\overline{B}$ for small enough $\lVert t\rVert$. If $\Phi_t(x) = \Phi_t(y)$ for $x,y \in \overline{B}$, then
$$x - y = \bigl(\chi(y) - \chi(x)\bigr)\cdot v,\tag{$\ast$}$$
where $v = t_{2i-1}e_1 + t_{2i}e_2$. But since $\overline{B}$ is convex, we have
$$\lvert \chi(y) - \chi(x)\rvert \leqslant M\lVert y-x\rVert,$$
and since $\lVert v\rVert \leqslant \lVert t\rVert$, $(\ast)$ can only happen for $x = y$ when $\lVert t\rVert < 1/M$.
Thus, $\Phi_t$ is a (global) bijection for
$$\lVert t\rVert < \frac{1}{\max\:\{\lVert D\chi_i(x)\rVert : 1 \leqslant i \leqslant N,\; x \in \mathbb{R}^n\}}.$$
