Past exam of the short options course of the University of Oxford / 2021 / 4 (Introduction to manifolds)

Take $\Omega ={\mathbb{R}}^2$ andIts Jacobian matrix and determinant areThus its total derivative is invertible everywhere, and the inverse function theorem makes it locally invertible. But $F(s,t+2\pi )=F(s,t)$, so it is not injective and has no global inverse on its image. The domain is open and connected as required.

No. On $\Omega ={\mathbb{R}}^2$ letIt is a bijection onto ${\mathbb{R}}^2$, with inverse $(u,v)\mapsto (\sqrt[3]{u},v)$. Nevertheless $D{F}_{(s,t)}=\mathrm{diag}(3s^2,1)$ is singular whenever $s=0$. Set-theoretic invertibility, even together with a continuous inverse, does not force invertibility of the total derivative. If the inverse were assumed differentiable, the chain rule applied to $F^{-1}\circ F$ would force both derivatives to be invertible; that stronger assumption is absent here.

Fix $x\in U$. Since $U$ is open, choose an open ball $B(x,\epsilon )\subset U$. For any $y$ in this ball the segment $x+t(y-x)$ remains in it. For each coordinate $G_j$, the chain rule givesThe ordinary mean value theorem therefore gives $G_j(y)=G_j(x)$ for every coordinate, hence $G(y)=G(x)$. Thus $G$ is a locally constant function.

For each $z\in {\mathbb{R}}^m$, its preimage is a union of such open balls, so $G^{-1}(\{z\})$ is open in ${\mathbb{R}}^n$. Its complement in $U$ is the union of the other fibers and is also open. If $U$ is nonempty and connected, the nonempty fiber through any chosen point is both open and closed in $U$, and must be all of $U$. ThereforeThe empty-domain case is vacuous. Notice that continuity of $DG$ was never needed.

Write $a={\partial }_1{f}_1={\partial }_2{f}_2$ and $b={\partial }_2{f}_1=-{\partial }_1{f}_2$. The Cauchy-Riemann equations giveConsequentlyEquivalently, it is singular exactly when all four partial derivatives vanish. The sum of squares is over $\mathbb{R}$, so no cancellation is possible.

Suppose $D{F}_x$ were invertible at some $x\in \Omega$. The inverse function theorem supplies a neighborhood $V$ of $x$ for which $F(V)$ is a nonempty open subset of ${\mathbb{R}}^2$. But $F(V)\subset M$, and a one-dimensional embedded submanifold of ${\mathbb{R}}^2$ has empty interior. To see this last assertion, a slice chart for an embedded submanifold takes $M$ locally into a coordinate line; a coordinate line cannot contain an open plane neighborhood, and a chart preserves openness. This is a contradiction.

Thus $D{F}_x$ is singular everywhere. Part (c)(i) strengthens singularity to $D{F}_x=0$ everywhere. By (b) and connectedness of $\Omega$,The special form of the Jacobian matrix is essential: an arbitrary differentiable map can have nonconstant image in a curve with derivative of rank one.