Homework 1 Problem 7

Problem 7

An isomorphism between vector spaces is a linear bijection. Prove that if $f:M\rightarrow N$ is a diffeomorphism between manifolds then $T_{x}f:T_{x}M\rightarrow T_{f(x)}N$ is an isomorphism for each $x\in M.$


The following theorem is useful in proof.


Let $f$ and $g$ be a smooth maps of manifolds. If $h=g\circ f$, then

\begin{align} \[ T_{x}h=T_{x}(g\circ f)=T_{f(x)}g\circ T_{x}f. \] \end{align}

If $f$ is a diffeomorphism, then its inverse $f^{-1}$ is also a smooth function. If we put $g=f^{-1}$ in above theorem, we get;

\begin{align} \[ h=f^{-1}\circ f=I\Rightarrow T_{x}h=T_{x}(f^{-1}\circ f)=T_{f(x)}f^{-1}\circ T_{x}f \] \end{align}

Since, if $h$ is an identity map then $T_{x}h$ is also an identity map, the claim follows;

\begin{align} \[ T_{f(x)}f^{-1}\circ T_{x}f=I \] \end{align}

Last statement proves that $T_{x}f$ has an inverse map $(T_{x}f)^{-1}=T_{f(x)}f^{-1}$, therefore $T_{x}f$ is a bijection.

Lastly, linearity can be seen from the matrix of the linear transformation definition as follows;


For coordinates $\phi =(x^{1},...,x^{m})$ at $p\in M$, and $\psi=(y^{1},...,y^{n})$ at $f(p)\in N$, then;

\begin{align} \lefteqn\\ T_{x}f(\frac{\partial }{\partial x^{j}}) &=&\sum_{i=1}^{n}\frac{\partial y^{i}}{\partial x^{j}}\frac{\partial }{\partial y^{i}}\Rightarrow \\% T_{x}f\left( a_{j}\frac{\partial }{\partial x^{j}}+b_{k}\frac{\partial }{% \partial x^{k}}\right) &=&\sum_{i=1}^{n}a_{j}\frac{\partial y^{i}}{\partial x^{j}}\frac{\partial }{\partial y^{i}}+\sum_{i=1}^{n}b_{k}\frac{\partial y^{i}}{\partial x^{k}}\frac{\partial }{\partial y^{i}} \\ &=&a_{j}\sum_{i=1}^{n}\frac{\partial y^{i}}{\partial x^{j}}\frac{\partial }{% \partial y^{i}}+b_{k}\sum_{i=1}^{n}\frac{\partial y^{i}}{\partial x^{k}}% \frac{\partial }{\partial y^{i}} = a_{j}T_{x}f\left( \frac{\partial }{\partial x^{j}}\right) +b_{k}T_{x}f\left( \frac{\partial }{\partial x^{k}}\right) \end{align}

Thus, $T_{x}f$ is a linear bijection, i.e. isomorphism between vector spaces $T_{x}M$ and $T_{x}N.$

After correcting several typographical errors, Scott agrees!

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