# Question about theorem 2.2.1 in Wald's General Relativity

1. Sep 9, 2012

### CJ2116

Hi everyone, first of all I have been a lurker here for years and have benefited greatly from many of the discussions in the math and physics sections. Thanks, I have received a lot of helpful information from these forums!

I have been working through Wald's General Relativity book and I am having trouble following the reasoning behind one part of a theorem. From page 15, the theorem and part of the proof is (For those who don't have the book):

Let M be an n-dimensional manifold. Let $p \in M$ and let $V_p$ denote the tangent space at p. Then dim $V_p=n$

Proof We shall show that dim $V_p=n$ by constructing a basis of $V_p$, i.e. by finding n linearly independent tangent vectors that span $V_p$. Let $\psi : O \rightarrow U\subset R^n$ be a chart with $p\in O$. If $f\in \mathfrak{F}$, then by definition $f\circ \psi^{-1}:U\rightarrow R$ is $C^{\infty}$. For $\mu=1,...,n$ define $X_{\mu}:\mathfrak{F}\rightarrow R$ by
$$X_{\mu}(f)=\frac{\partial}{\partial x^{\mu}}(f\circ \psi^{-1})\bigg|_{\psi (p)}$$
$$\vdots$$

I can't seem to figure out how the term $\frac{\partial}{\partial x^{\mu}}(f\circ \psi^{-1})\bigg|_{\psi (p)}$ is a mapping from $\mathfrak{F}\rightarrow R$. $f\circ \psi^{-1}$ was defined to be a mapping from $U\rightarrow R$. In other words, I don't see why these last two terms should be equal. I think I am missing something obvious here. Is there maybe some sort of chain rule argument?

Thanks, any pointer in the right direction would be greatly appreciated!

2. Sep 10, 2012

### haushofer

You have $f o \ \psi^{-1}(\psi(p)) = f(p)$.

Last edited: Sep 10, 2012
3. Sep 10, 2012

### CJ2116

Wow, that was more embarrassingly obvious than I thought!