Differentiability of a function on a manifold

["Don't panic!"]

I am currently working through Nakahara's book, "Geometry, Topology and Physics", and have reached the stage at looking at calculus on manifolds. In the book he states that "The differentiability of a function $f:M\rightarrow N$ is independent of the coordinate chart that we use". He shows this is true in $M$ and then leaves it as an exercise to show that it is also true in $N$. Here is my attempt, please could someone tell me if what I've done is correct, and if not, what the correct way is?

Let $f:M\rightarrow N$ be a map from an $m$-dimensional manifold to an $n$-dimensional manifold $N$. A point $p\in M$ is mapped to a point $f(p)\in N$, namely $f:p\mapsto f(p)$. Take a chart $(U,\phi)$ on $M$ and $(V,\psi)$ on $N$, where $p\in U$ and $f(p)\in V$. Then $f$ has the following coordinate presentation: $$\psi\circ f\circ\phi^{-1}:\mathbb{R}^{m}\rightarrow\mathbb{R}^{n}$$
If we write $\phi (p)=\lbrace x^{\mu}\rbrace =x$ and $\psi(f(p))=\lbrace y^{\alpha}\rbrace = y$, then $y=\psi\circ f\circ \phi^{-1} (x)$ is a vector-valued function of $m$ variables.The differentiability of $f$ is independent of the coordinate chart in $N$. Indeed, let $(V_{1},\psi_{1})$ and $(V_{2},\psi_{2})$ be two overlapping charts in $N$. Take a point $q=f(p)\in V_{1}\cap V_{2}$, whose coordinates by $\psi_{1}$ are $\lbrace y_{1}^{\mu}\rbrace$, while those $\psi_{2}$ are $\lbrace y_{2}^{\nu}\rbrace$. When expressed in terms of $\lbrace y_{1}^{\mu}\rbrace$, $f$ takes the form $$y_{1} =\psi_{1}\circ f\circ\phi^{-1}$$ while in $\lbrace y_{2}^{\nu}\rbrace$, it takes the form $$y_{2} =\psi_{2}\circ f\circ\phi^{-1} =\psi_{2}\circ\left(\psi^{-1}_{1}\circ\psi_{1}\right)\circ f\circ\phi^{-1}= \left(\psi_{2}\circ\psi^{-1}_{1}\right)\circ\psi_{1}\circ f\circ\phi^{-1}$$ Now, by definition, $\psi_{2}\circ\psi^{-1}_{1}$ is $C^{\infty}$, and therefore if $\psi_{1}\circ f\circ\phi^{-1}$ is $C^{\infty}$ with respect to $y_{1}^{\mu}$ and $\psi_{2}\circ\psi^{-1}_{1}$ is $C^{\infty}$ with respect to $y_{2}^{\nu}$, then $\psi_{2}\circ f\circ\phi^{-1}=\left(\psi_{2}\circ\psi^{-1}_{1}\right)\circ\psi_{1}\circ f\circ\phi^{-1}$ is also $C^{\infty}$ with respect to $y_{2}^{\nu}$.

Would this be correct?

 

[WWGD]

Yes, that looks correct, but remember that not all manifolds are $C^{\infty}$

 

[Fredrik]

There are some inaccuracies, but you seem to have the right idea. Your post should be talking about an f that's "differentiable at p", and not necessarily "differentiable" (i.e. differentiable at p for all p).

If x and y are coordinate systems with p in their domains, then $f\circ x^{-1}$ is typically not equal to $f\circ y^{-1}\circ y\circ x^{-1}$, because $y^{-1}\circ y$ is the identity map on the domain of $y$, which typically doesn't include the entire domain of $x$. So $f\circ y^{-1}\circ y\circ x^{-1}$ is a restriction of $f\circ x^{-1}$ to a smaller domain. This is irrelevant when we consider the partial derivatives of these functions, but it's a bit of a pain to prove that. (I don't think someone who's learning differential geometry should spend too much time on these subtleties, because they don't contribute much to your understanding).

 

["Don't panic!"]

Ah, ok. Thanks for your help Fredrik, there's so much to take in, but I think it's slowly starting to become clearer!

 

[blue_raver22]

Yes, your explanation is correct. By using the coordinate charts, you have shown that the function f is differentiable in both charts, and the differentiability is independent of the choice of chart. This is because the composition of two differentiable functions is also differentiable. Therefore, the differentiability of f on the manifold is well-defined and independent of the coordinate chart used.