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James R. Munkres "Analysis on Manifolds"
p.60

Theorem 7.4.

Let $A$ be open in $\mathbb{R}^n$; let $f : A \rightarrow \mathbb{R}^n$; let $f(a) = b$. Suppose that $g$ maps a neighborhood of $b$ into $\mathbb{R}^n$, that $g(b) = a$, and

$g(f(x)) = x$

for all $x$ in a neighborhood of $a$. If $f$ is differentiable at $a$ and if $g$ is differentiable at $b$, then

$Dg(b) = [Df(a)]^{-1}$.

I cannot understand these statements very easily.

Why didn't he write as follows?

Let $A$ be open in $\mathbb{R}^n$.
Let $f : A \rightarrow \mathbb{R}^n$.
Let $B$ be open in $\mathbb{R}^n$.
Let $g : B \rightarrow \mathbb{R}^n$.
Let $a \in A$.
Let $b \in B$.
Let $f(a) = b$.
Let $f$ be differentiable at $a$.
Let $g$ be differentiable at $b$.
Let $g(f(x)) = x$ for all $x$ in a neighborhood of $a$.

Then,

$Dg(b) = [Df(a)]^{-1}$.

Are these statements equivalent to Munkres' statements?

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It is just pedagogical purposes;

The author just wants to emphasise some of the points. Plus, of course there are things that is just a matter of style.

For example, in your statement you do not explicitly state that $g(b) = a$, but of course this can be seen from $g(f(x)) = x$ for some neighbourhood of $a$, but this is one of the things that we all (I think) love Munkres' books; they are not terse.

Moreover, what another reason (which is I love about) is that it lets you form an image in your mind while reading it, whereas in your statement, you just list the hypothesis that you have in a unrelated order; which is not something that is helpful to the reader.

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  • $\begingroup$ @tchappyha Not a problem (see also my last edit). Sorry, by the way, for the first comment; I just misunderstood the question. $\endgroup$ – onurcanbektas Aug 15 '18 at 5:14

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