Suppose that there is indeed an diffeomorphism $f: \Bbb R^m \to \Bbb R^n$, and $f(a) = 0$. Then $f \circ f^{-1} = identity$. In class my teacher asserts that $Df^{-1}_a \cdot Df_a = identity$ and $Df_a \cdot Df_a^{-1} = identity$. Why this is true and how does it imply that $f$ cannot exist?
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8This is a perfect example of why the foundations of differential topology are easier than those of algebraic topology. Proving that there is no diffeomorphism between them is just linear algebra. Proving that there is no homeomorphism between them is much more difficult. – Dustan Levenstein Sep 12 '16 at 18:20
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The matrix $M=Df_a$ is of size $m\times n$ and $N=Df_a^{-1}$ of size $n\times m$. If $f$ is a diffeomorphism between the two spaces then the rule for taking derivative of compositions implies that $M N= 1_m$ (identity in ${\Bbb R}^m$) and $NM=1_n$. The rank, however, can not be bigger than the minimum value of $n$ and $m$ so $n=m$.
H. H. Rugh
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This would imply $\mathbb{R}^n$ and $\mathbb{R}^m$ are isomorphic as vector spaces, which is of course a contradiction! The fact that $Df_a^{-1} \circ Df_a = Df_a \circ Df_a^{-1} = Id$ is just the chain rule.
Zestylemonzi
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What maps are you considering between linear spaces $R^m$ and $R^n$? The left multiplication by $Df_a$ and $Df_a^{-1}$? – Keith Sep 12 '16 at 20:20
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The derivative/Jacobian $Df_a$ is a linear map. That's the whole idea surrounding derivatives - the idea of approximating a function (at a given point i.e at $a$) as a linear map. – Zestylemonzi Sep 12 '16 at 20:24