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$A\subset \mathbb{R}^m$ is an open and bounded set, $f:A\longrightarrow \mathbb{R}^n$, $m\leq n$, is injective, continuously differentiable and its Jacobian matrix has full rank on $A$. Does this suffice for $f$ to be a $C^1$ diffeomorphism onto its image?

Maybe an additional assumption that $f^{-1}$ is continuous would make this enough?

Also is boundedness of $A$ of any help or could it be done away with?

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Indeed, these conditions together with the assumption that $f^{-1}$ is continuous are enough, see https://en.wikipedia.org/wiki/Embedding

The first conditions guarantee that $f$ is a local diffeomorphism, but without the latter condition that $f^{-1}$ is continuous, i.e. that $f$ is a homeomorphism onto its image, you can have global problems. The typical example is $A = (0,1)$, $n = 2$ and $f$ is a curve that converges upon itself, for example $$ \lim_{t \to 1} f(t) = f(1/2). $$ Then a neighborhood of the point $f(1/2) \in \mathbb{R}^2$ contains both a neighborhood of $1/2$ and $1$ in $(0,1)$.

Boundedness of $A$ is not relevant for this.

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  • $\begingroup$ Are these assumptions also necessary for $f$ to be a diffeomorphsim? Could they be used as a definition? $\endgroup$ – Tom Jun 27 '16 at 17:13
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    $\begingroup$ Yes, by definition a $C^1$ diffeomorphism $f$ is a one-to-one map such that both $f$ and $f^{-1}$ are $C^1$. This implies all these assumptions. Note that the assumption $f^{-1}$ is necessary when you equip $f(A)$ with the subspace topology induced by $\mathbb{R}^n$ (the normal choice); if you equip it with the topology induced by $f$, then continuity of $f^{-1}$ automatically follows from the inverse function theorem. $\endgroup$ – Jaap Eldering Jun 27 '16 at 19:24

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