Invertibility theorem on the boundary for a function between two closed 2D manifolds

Question

Assume a function $f:\mathbb{R}^2\to\mathbb{R}^2$ on a simply connected, closed domain $D\subset\mathbb{R}^2$ including its boundary $\partial D$.

I am interested in the local invertibility of $f$ in a point $x\in D$. For inner points, I know that it is related to the Jacobian determinant at $x$. However, I wonder if a similar condition is necessary for points on the boundary $x\in\partial D$.

I am sorry if this is basic analysis, but I could not find anything about it. I know that for 1D functions $f:\mathbb{R}^1\to\mathbb{R}^1$ with $D$ being an interval, no condition is needed on the boundary points. However, in the 2D case I would expect the derivative along the boundary's tangent to play a role.

Apart from the case where the associated manifolds $D$ and $f(D)$ have smooth boundaries, I am also interested in the case, where they have (the same number of) corners. For simplicity, one could assume $D$ being a unit square or circle.

Answer 1

For interior points the local inversion theorem says that the mapping is a local diffeo at the point if the jacobian det is not zero. For boundary points one has to assume differentiability at the point, which means the function has some differentiable extension to a nbhd of the point in $\mathbb R^2$. Then again if the jacobian det is not zero, you get a local diffeomorphism at the point. However, I think the point here is what happens for mappings $f:D\to D$, that is local inversion theorem for manifolds with boundary. The difficulty is that the inverse doesn't exit the domain! Here one needs more than just jacobian determinant not zero: the additional condition is that the map preserves the boundary. This can be seen in any book discussing manifolds with boundary (e.g. this ). Now, the square $[0,1]^2$ is not a manifold with boundary, but a manifold with corners. Well there is also a local inversion theorem for these objects, and the condition is the preservation of the corners.

Answer 2

Manifolds with corners. A manifold with corners is a space locally homeomorphic to open sets of quadrants $x_1\ge0,\dots,x_k\ge0$ in $\mathbb R^n \ (k\le n)$. Two such local homeomorphisms (charts) $\psi,\varphi$ must be compatible in the usual sense: $\psi^{-1}\circ\varphi$ must be a diffeomorphism (where defined, that is, among quadrants). Then they form an atlas and the usual machinery follows: differentiable mappings, tangent spaces, derivatives.

The notion of a differentiable mapping on arbitrary subsets of $\mathbb R^n$ (as quadrants) is that it has local differentiable extensions to open sets in $\mathbb R^n$. Then a bijection of two arbitrary sets which is differentiable as well as its inverse is a diffeomorphism. Here one can extend the inverse, but this extension needn't be the inverse of anything. Concerning derivatives, they are well defined through extension when one deals with regularly closed sets, that is, sets whose interior is dense in them, as quadrants are.

Boundary and corners. In a manifold with corners $M$ the boundary $\partial M$ consists of all points that are mapped into some face of a quadrant $x_1\ge0,\dots,x_k\ge0$. But there are differences here: points mapped into $x_1=0,\dots,x_k=0$ have index $k$, those into $x_1=0,\dots,x_{k-1}=0,x_k>0$ have index $k-1$, and so on. The interior points have index $0$, so to say. This index is invariant by diffeomorphisms, which makes things consistent. This index stratifies the boundary into pieces, and corners needn't be finitely many. But note that points of index equal the dimension of $M$ (the maximum possible index, not necessarily existing) are isolated points (hence at most countable).

A maybe unexpected fact is that $\{x\ge0\}\cup\{y\ge0\}$ is not a manifold with corners. (See Is there any diffeomorphism from A to B that $f(A)=B$?)

Local inversion theorem. Let $f:M\to N$ be a differentiable mapping of manifolds with corners. Then $f$ is a local diffeomorphism at $p\in M$ if and only if (i) $d_pf:T_pM\to T_{f(p)}N$ is a linear isomorphism and (ii) there is an open nbhd $U$ of $a$ in $M$ such that $f(U\cap\partial M)\subset\partial N$. If that is the case, $f$ preserves indices: $\text{ind}(x)=\text{ind}(f(x))$ in a nbhd near $a$.

This last remark is interesting. For instance, it follows that one cannot deform differentiably a halfplane onto a quadrant: the corner of the quadrant is the obstruction.

Concerning the previous answer, if one has a mapping $\ f:D\to D$ and a point $a\in\partial D$ at which the jacobian det is $\ne0$, then $f$ is a local diffeo from an open nbhd $U$ of $a$ in $\mathbb R^2$ onto an open nbhd $V$ of $f(a)$ in $\mathbb R^2$. If $f$ doesn't preserve the boundary, there will be points in $V\cap D$ such that $f^{-1}(x)\in U\setminus D$. For instance, one can map $D$ onto a smaller disc $E\subset D$ tangent at $f(a)$ and check what happens.

The only full reference for all of this "cornered" stuff I can recommend is Margalef-Outerelo's Differential Topology, North Holland 1992. The book is a bible of the topic.