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Given the inverse function theorem,

2-11 Theorem (Inverse Function Theorem). Suppose that $f: \mathbf{R}^{n} \rightarrow \mathbf{R}^{n}$ is continuously differentiable in an open set containing $a$, and $\operatorname{det} f^{\prime}(a) \neq 0$. Then there is an open set $V$ containing $a$ and an open set $W$ containing $f(a)$ such that $f: V \rightarrow W$ has a continuous inverse $f^{-1}: W \rightarrow V$ which is differentiable and for all $y \in W$ satisfies $$ \left(f^{-1}\right)^{\prime}(y)=\left[f^{\prime}\left(f^{-1}(y)\right)\right]^{-1} . $$

Is it necessarily true that the open set $W$ is contained in $f(V)$?

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  • $\begingroup$ What's $A$?${}{}$ $\endgroup$ Commented Feb 5, 2022 at 23:06
  • $\begingroup$ Edited its the open set containing $f(a)$ $\endgroup$ Commented Feb 5, 2022 at 23:16
  • $\begingroup$ Isn't $A$ already named $V$? And if $f\colon V\to W$ has an inverse $f^{-1}\colon W\to V$, then we know that $W=f(V)$ precisely. $\endgroup$ Commented Feb 5, 2022 at 23:21

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You have $f(V) = W$, otherwise it does not make sense to say that $f : V \to W$ has an inverse $f^{-1} : W \to V$.

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