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What does the inverse function say when $\det f'(x)$ doesn't equal $0$?

I know that if a function is one-to-one, than it has an inverse. However, I'm confused about the question above. Would appreciate if someone could get to me!

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I'm assuming you're talking about the inverse function theorem. Basically, if you have a function $f: U \to V$ where $U$ and $V$ are open sets in Euclidean space, and furthermore, if $det(Df(p))$ where $p \in U$ is non-zero, then there is an inverse differentiable function from a neighbourhood of $f(p)$ to a neighbourhood of $p$. Furthermore, the derivative of $f^{-1}$ at $f(p)$ is $(Df(p))^{-1}$. Intuitively, this says that $f$ is $locally$ invertible (or more interestingly, a local differentiable homeomorphism).

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