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Show that if $M subseteq Bbb R^n$ is a $k$-dimensional manifold, then there is no $x in M$ and $U_x$ a neighborhood of $x$ such that $M cap U_x$ has a good parametrization from $V_x subseteq Bbb R^m$ where $k neq m$.

The definition of "good parametrization" of $M$ is as follows: there exists a smooth, regular homeomorphism $r:V_x to Bbb R^n $ such that $r(V_x) =M$ where $V_x subseteq Bbb R^m$ is an open set.

I'm having a hard time approaching this one, any guidance will be appreciated

If locally arround $xin M$ we have two good parametrizations $phi$, $psi$ which are defined on open subsets of $mathbb R^k$ and $mathbb R^l$ respectively then after a restriction we get a homemorphism $tau=psi^{-1}circphi$ between open subsets of $mathbb R^k$ and $mathbb R^l$. Now one can use tools from algebraic topology to show that this imlpies $k=l$. However since we are in the differentiable setting we can also show this more directly:

Assume for a moment that we knew that $tau$ is in fact a diffeomorphism. Then by the chain rule for all $x$ in the domain of $tau$ we have

$$id_{mathbb R^k}=D(id)(x)=D(tau^{-1}circtau)(x)=D(tau^{-1})(tau (x))cdot D(tau)(x)$$
and in the same way

$$id_{mathbb R^n}=D(id)(tau (x))=D(taucirctau^{-1})(tau(x))=D(tau)( x)cdot D(tau^{-1})(tau (x))$$

so $D(tau)(x)$ is a linear isomorphism between $mathbb R^k$ and $mathbb R^l$ which implies $k=l$.

So it remains to show that $tau$ is a diffeomorphism. For this we can use the following lemma which is also of general interest:

$textbf{Lemma:}$

Assume $psi:Omega tomathbb R^n$ is a regular smooth map defined on an open subset of $mathbb R^k$. Then for all $xin Omega$ there are open neighbourhoods $Omega'$, $U$ of $x$, $psi(x)$ respectively and a smooth map $Gamma:UtoOmega'$ such that $psi_{|Omega'}$ is a homeomorphism onto it's image and $psi^{-1}_{|Omega'}=Gamma$ on $psi(Omega')cap U$.

So if $z$ is in the domain of $tau$ we can set $x=tau(z)$, find $Omega'$, $U$ and $Gamma$ as above and then $tau=Gammacircphi$ on $tau^{-1}(Omega')$ which is smooth. In the same way $tau^{-1}$ will then be smooth.

$textbf{proof of the Lemma:}$

Set $W=$im$(Dpsi(x))=Dpsi(x)(mathbb R^k)$ which is a $k$-dimensional subspace of $mathbb R^n$ and let $L:mathbb R^ntomathbb R^k$ be a linear map which maps $W$ onto $mathbb R^k$. Then by the chain rule

$$D(Lcircpsi)(x)=DL(psi(x))circ Dpsi(x)=Lcirc Dpsi(x)$$

is surjective and hence invertible. By the inverse function theorem there are open neigbhourhoods $Omega'subseteqOmega$, $Vsubseteqmathbb R^k$ of $x$ and $L(psi(x))$ such that $Lcircpsi:Omega'to V$ is a diffemorphism.

Now set $U=L^{-1}(V)$ and $Gamma=(Lcircpsi)^{-1}circ L$ which is defined on $U$. Then $Gamma$ is smooth and for all $yinpsi(Omega')cap U$ we have $Gamma(y)=Gamma(psi(psi^{-1}(y)))=psi^{-1}(y)$. Finally as $L$ is injective on $psi(Omega')$ for all $Wsubseteq Omega'$ open $psi(W)=L^{-1}(L(psi(W)))cap psi(Omega')$ which is an open subset of $psi(Omega')$ so $psi_{|Omega'}$ is a homeomorphism onto its image. $square$