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Let $U\subset \mathbb R^m$ and $V\subset \mathbb R^n$ be open sets and $f:U\to \mathbb R^n$ and $g:V\to \mathbb R^m$ differentiable functions. Suppose $g(f(x))=x$ for every $x\in U$.

I want to prove the image of the linear transformations $f'(x):\mathbb R^m\to \mathbb R^n$ and $g'(f(x)):\mathbb R^n\to \mathbb R^m$ has the same dimension.

I've proved using the chain rule:

$$g'(f(a))\cdot f'(a)=(g\circ f)'(a)=Id_U'=Id_{\mathbb R^m}$$

So using this fact which I proved above

$g'(f(a))\cdot f'(a)=Id_{\mathbb R^m}$

does give me the image of these linear transformations have the same dimension?

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    How can they have the same range when they are parts of different spaces? One is in $\Bbb R^n$, the other in $\Bbb R^m$. – 2017-01-07
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    @RossMillikan The range is the dimension of the image – 2017-01-07
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    @user251257 As I said to Ross, range is the dimension of the image – 2017-01-07
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    @RossMillikan I've just edited the question – 2017-01-07
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    @user251257 I'm sorry, you're right – 2017-01-07
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    Do you have a guess what the rank might be? – 2017-01-07
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    @user251257 rank is the dimension of the range, no? – 2017-01-07
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    @user42912 I meant here in particular. You asked to show the ranks are equal. Do you have any guesses? What can you tell about the rank of the matrices $A$ and $B$ if $AB=Id$? – 2017-01-07
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    @user251257 essentially that's my question, I don't know – 2017-01-07
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    @user42912 could the rank be less than $n$? – 2017-01-07
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    @user251257 The linear transformation $T_A$ represented by $A$ is surjective, because it has a right inverse, then $A$ has rank $n$? $AB$ has rank $n$, no? – 2017-01-07
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    @user251257 $AB$ can't have rank less than $n$. – 2017-01-07
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    @user42912 yes $AB=I$ has rank $n$, so does $A$. What about $B$? – 2017-01-07
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    @user251257 $B$ has rank $n$, because the linear transformation reprasented by $T_B$ is injective, then its kernel is the trivial one, then it's rank is $n$ – 2017-01-07
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    @user42912 the only matrix with zero rank is the zero matrix... – 2017-01-07
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    @user251257 I've just edited my last comment, am I right? – 2017-01-07
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    @user42912 on I made a mistake, i switched $n$ for $m$. Sorry. But in principle your argument is right. – 2017-01-07
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    @user251257 Thank you very much. Fell free to convert these comments into an answer – 2017-01-07

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