Composite of an immersion with the inverse map of another immersion is a diffeomorphism

Question

Let $U\subset \mathbb{R}^k$ be an open set, $n>k$ and $\varphi_1,\varphi_2 : U\to \mathbb{R}^n$ be immersions, meaning continuously differentiable such that the differential taken in any point of $U$ is injective. Also, let $\varphi_1(U)=\varphi_2(U)=:U'$ and $\varphi_1$ and $\varphi_2$ be injective (and homeomorphisms $U\to U'$).

Then $\varphi_1^{-1} \circ \varphi_2 : U\to U$ is a diffeomorphism.

My problem is that $\varphi_1^{-1}$ need not be differentiable in the usual analysis sense, because $U'$ is not an open subset of $\mathbb{R}^n$. I am guessing that $d (\varphi_1^{-1}\circ \varphi_2)(x)$ should be $d(\varphi_1)(\varphi_1^{-1}(\varphi_2(x)))^{-1}\cdot d(\varphi_2)(x)$, because this is what happens if $k=n$ and it is still defined for $k<n$. However, I would need a way to apply the theorem on inverse functions to $\varphi_1$.

Answer 1

Your maps $\phi_1, \phi_2$ are usually called embeddings (= injective proper immersions).

Are you familiar with abstract manifolds? This is in some sense the appropriate language to discribe the map $\phi^{-1}: U' \rightarrow U$ and its differential $d \phi^{-1}(p)$ (this would be a map from the tangent space of $U'$ at $p$ onto $\mathbb{R}^k$).

For an elementary treatment, try to extend the map $\phi_1$ to a map $$\tilde{\phi}_1 : U \times \mathbb{R}^{n-k},\,\,\,\, \tilde{\phi}_1(x,y) = \phi_1(x) + A(x)y$$ where $A(x) \in \mathbb{R}^{(n-k)\times k}$ is a smooth map of matrices, such that $d\tilde{\phi_1}(x,0) = d\phi_1(x) + A(x)$ is an isomorphism for every $x \in U$. Do the same for $\phi_2$ and then argue by the inverse function theorem.

Answer 2

I managed to solve this after all now (which is in part thanks to Sam's post, but I use a different idea).

Let $x\in U$ and $a:=\varphi_1(x)\in U'$. Choose $y\in U$ with $\varphi_2(y)=a$. It is enough to show that there is an open neighbourhood $V$ of $a$ in $U'$ such that $\varphi_1^{-1} \circ \varphi_2 : \varphi_2^{-1}(V)\to U$ is differentiable (by symmetry and because differentiability is a local property).

Without loss of generality, let the first $k$ rows of $d\varphi_1(x)$ be independent. I want to use the fact that for suitably small $V$ as above, $V=W\cap U'$ with $W$ open in $\mathbb{R}^n$, there is a diffeomorphism $\Phi:W\to W'\subseteq \mathbb{R}^n$ such that $\Phi (V) = W'\cap \{ (x_1,..,x_n)\in W'\, |\, x_{k+1},..,x_n = 0 \}$.

In fact, one can be chosen as $\Phi: (x_1,..,x_k,x_{k+1},..,x_n)\mapsto (x_1,..,x_k,x_{k+1}-\varphi_{1,k+1}(\psi(x_1,..,x_k)),..,x_{n}-\varphi_{1,n}(\psi(x_1,..,x_k)))$

where, using the inverse function theorem, $\psi$ is the locally defined inverse function of $(\varphi_{1,1},..,\varphi_{1,k})$. (The differential of $\Phi$ is everywhere lower triangular with determinant 1.)

Let $p:\mathbb{R}^n\to\mathbb{R}^k$ be the projection to the first $k$ coordinates. The map $f:=p\circ \Phi \circ \varphi_1$ (defined on a suitable open set in $U$ containing $x$ and mapping to $\mathbb{R}^k$) is differentiable and injective.

For any $x'$ close to $x$, $df(x') = dp(\Phi(\varphi_1 (x'))) \cdot d\Phi (\varphi_1 (x')) \cdot d\varphi_1 (x')$. Now the first $k$ rows of $d\varphi_1(x')$ are independent (since $x'$ is close to $x$ and this is the case for $d\varphi_1(x)$) and for $1\leq i\leq k$, the $i$th row of $d \Phi$ taken at any point is simply a $1$ at the $i$th coordinate and zeroes otherwise. The same goes for $dp$. So this composition of three matrices still has its first $k$ rows independent, so it is an invertible $k\times k$ matrix. It follows that $f$ is a diffeomorphism onto its image.

$\varphi_1^{-1}\circ \varphi_2$ equals the composition $f^{-1} \circ p \circ \Phi \circ \varphi_2$ on a suitably small neighbourhood of $y$ and all these maps are differentiable.