Is an Immersion which is also a homeomorphism always a diffeomorphism? In my differential geometry script we defined a Immersion the following way:
Let $f \colon U \rightarrow \mathbb{R}^3$ be such that $\forall x \in U: f^{\prime}(x)$ is injective, where $U \subset \mathbb{R}^2$ is open. Then $f$ is called an immersion.
My question is, whether an injective immersion which is also a homeomorphism onto its image has got a differentiable inverse $f^{-1} \colon \operatorname{im}(f) \rightarrow U$.
I would suggest that the answer is no, but it is a blind guess and all the theorems I know only work for $\operatorname{dim}(U)=3$, but I am interested in $\operatorname{dim}(U)=2$.
If necessary one can assume that $f \in C^{\infty}$, but I am mostly excited about results for $C^1$ or just differentiable.
 A: The answer is yes, once you add the phrase "onto its image" everywhere and assume enough regularity. More precisely:

Theorem.
Let $M,N$ be smooth manifolds (of dimensions $m,n$ repsectively), $f:M\to N$ a smooth map. If

*

*$f$ is an immersion (meaning for each $x\in M$, the tangent mapping $T_xf:T_xM\to T_{f(x)}N$ is injective)

*$f$ is a homeomorphism of $M$ onto $f(M)$, equipped with the subspace topology from $N$,

then $f(M)$ is an embedded submanifold of $N$, and the restricted mapping (on the target) $f:M\to f(M)$ is a diffeomorphism.

First, some remarks.
Notice that I'm not saying $f:M\to N$ is a diffeomorphism; this would be absurd (consider $M=\Bbb{R},N=\Bbb{R}^2$, $f(t)=(t,0)$). Also, the theorem above and proof below holds if we replace "smooth" by $C^1$ everywhere. The idea of the proof is pretty easy once you know the basic consequences of the inverse/implicit function theorems (particularly the local canonical form for immersions). The annoying part is making sure all the open sets are small enough so that things match up.
Regarding terminology, a mapping $f$ which satisfies the hypotheses of the theorem (an immersion which is a homeomorphism onto its image) is known as an embedding. So, the theorem can be stated as "the image of an embedding is an embedded submanifold" (where 'embedded submanifold' means it satisfies the 'local slice-chart' condition).

TL;DR of Proof
The crux of the proof is the first statement, that $f(M)$ is an embedded submanifold of $N$ (of the same dimension as $M$). Once you do this, it is basic stuff that $f:M\to N$ being smooth implies $f:M\to f(M)$ is smooth, so being an immersion and bijective easily prove that the latter is a diffeomorphism (usual inverse function theorem argument).

Proof of the Theorem.
To show $f(M)$ is an embedded submanifold, we need to show the existence of a "slice chart" around any point. So, take an arbitrary point $f(x)\in f(M)$. Since $f$ is an immersion, we can:

*

*find a chart $(U,\phi)$ of $M$ around $x$ with $\phi(x)=0\in\Bbb{R}^m$,

*find a chart $(V_0,\psi)$ of $N$ around $f(x)$ such that $\psi(f(x))=0\in\Bbb{R}^m$

*$f$ maps $U$ into $V_0$ (we can always arrange this WLOG by replacing $U$ with $U\cap f^{-1}(V_0)$ if necessary) and the local representative $F:=\psi\circ f\circ \phi^{-1}:\phi(U)\to \psi(V_0)$ is given by $F(\xi^1,\dots, \xi^m)=(\xi^1\dots, \xi^m,0,\dots 0)$.

i.e the local form of $f$ is (the restriction of) the canonical injection of a lower dimensional vector space into a higher dimensional one (this stuff with inverse function theorem and charts is the non-linear analogue of basic linear algebra results regarding row-reduction of matrices corresponding to injective linear transformations). Now, we shall have to choose small enough open sets.

*

*First of all, since $f:M\to f(M)$ is a homeomorphism, it is an open mapping, so $U$ being open in $M$ implies $f(U)$ is open in $f(M)$; by definition this means there exists some open $V_1\subset N$ such that $f(U)=V_1\cap f(M)$.

*Next, let us define $V_2:=\psi^{-1}\left(\psi(V_0)\times \left(\phi(U)\times \Bbb{R}^{n-m}\right)\right)\subset V_0$.

*Finally, define $V:= V_0\cap V_1\cap V_2=V_1\cap V_2$ (since $V_2$ is already contained in $V_0$). In particular $V$ is open in $N$ since it is the intersection of three open sets in $N$.

Then, we have $f(U)\subset V_0$ (by definition), and next by definition we have $f(U)=V_1\cap f(M)$, and we also have $\psi(f(U))=F(\phi(U))=\phi(U)\times \{0_{\Bbb{R}^{n-m}}\}$ (see the explicit formula for $F$), so that $f(U)\subset V_2$. These three statements imply that $f(U)= V\cap f(M)$, and a little more set-theoretic unwinding of these definitions shows that
\begin{align}
\psi(V\cap f(M))=\psi(V)\cap \left(\Bbb{R}^{m}\times\{0_{\Bbb{R}^{n-m}}\}\right).
\end{align}
So, what I have shown is that for each point in $f(M)$, there is a chart (namely the restricted $(V,\psi|_V)$ above) such that $\psi(V\cap f(M))=\psi(V)\times (\Bbb{R}^m\times \{0\})$ looks like a flattened $m$-dimensional piece sitting inside of $\Bbb{R}^n$. This is precisely the slice-chart we're looking for so the definition of $f(M)$ being an ($m$-dimensional) embedded submanifold of $N$ is satisfied.
It is now clear that $f:M\to f(M)$ is a diffeomorphism, because $f:M\to f(M)$ is a bijective mapping between two smooth manifolds of the same dimension ($m$) and has injective derivative, so by the inverse function theorem is a local diffeomorphism. Coupled with bijectivity, this implies $f$ and $f^{-1}$ are smooth.
Alternatively, we can avoid usage of the inverse function theorem as in the previous paragraph to finish things off. We already used it once to obtain the charts $(U,\phi), (V,\psi)$ above. With these charts, the local representative of $f$ is $F: (\xi^1,\dots,\xi^m)\mapsto (\xi^1,\dots, \xi^m,0,\dots, 0)$. Now recall that a chart for the submanifold is obtained by ignoring the last few zero coordinates, so $(V\cap f(M), \tilde{\psi})$ is a chart for $f(M)$, and now the local representation is just $\tilde{F}:(\xi^1\dots, \xi^m)\mapsto (\xi^1,\dots, \xi^m)$, i.e it locally looks like the identity mapping (which is actually exactly what the inverse function theorem told us in the previous paragraph). So, $f:M\to f(M)$ is a local diffeomorphism, and due to bijectivity, it follows $f:M\to f(M)$ and $f^{-1}:f(M)\to M$ are both smooth functions (since smoothness is a local property). Hence $f:M\to f(M)$ is a diffeomorphism.
