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Let $U\in \mathbb R^n$ an open set. Thus we can take the identity map $\operatorname {Id}_U:U\to U$ to be a parametrization. My question is why this is the only parametrization of $U$. It seems easy but I couldn't prove it formally.

Definition of a parametrization:

A parametrization of class $C^k$ of a set $V\subset \mathbb R^n$ is an homeomorphism $\varphi:V_0\to V$ which is an immersion of class $C^k$ as well, defined in the open set $V_0\subset \mathbb R^m$.

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It's not. For a simple example, take $a \in \mathbb R^n$; then $f(x) = x + a$ is another one.

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    That's weird. I'm reading a book which says there is only one parametrization. – 2017-02-12
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    Does $x+a$ have to be in $U$? – 2017-02-12
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    @user42912, perhaps your book uses an uncommon meaning for the word "parametrization". Perhaps you could post (a screenshot of) an excerpt? – 2017-02-12
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    Please see my edit – 2017-02-12
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    @user42912, with that definition, my function (with the appropriate domain and codomain) is indeed a parametrization. I think there may be a small error in your book, or perhaps you misread about the identity being the *only* parametrization. – 2017-02-12