A question about why a atlas on a differentiable manifold consist of compatible charts

Question

As an Amazon Associate, I earn from qualifying purchases. For more practice problems like this, see Schaum’s Outline of Calculus, 7th Edition.

1. What the question is asking (in plain language)

We have a differentiable manifold (M) equipped with an atlas (\mathcal A={(U,\varphi)}).
Two charts ((U,\varphi)) and ((V,\psi)) are called (C^{m})-compatible when the change‑of‑coordinates map

[ \varphi\circ\psi^{-1}:\psi\bigl(U\cap V\bigr)\longrightarrow \varphi\bigl(U\cap V\bigr) ]

is a (C^{m}) function (i.e. has continuous partial derivatives up to order (m)).

A (C^{m}) function on the manifold is defined as follows:

(f:M\to\mathbb R) is of class (C^{m}) if for every point (p\in M) there exists a chart ((U,\varphi)\in\mathcal A) with (p\in U) such that the Euclidean function
[ f\circ\varphi^{-1}:\varphi(U)\longrightarrow\mathbb R ] is of class (C^{m}).

The books say that this definition does not depend on which chart we pick, because the charts in the atlas are compatible.

The student’s doubt is:

When we write [ f\circ\psi^{-1}= (f\circ\varphi^{-1})\;\circ\;(\varphi\circ\psi^{-1}), ] the second factor (\varphi\circ\psi^{-1}) is only defined on the overlap (\psi(U\cap V)), not on the whole set (\psi(V)). So we seem to obtain only that ((f\circ\psi^{-1})) restricted to the overlap is (C^{m}), not that (f\circ\psi^{-1}) itself is (C^{m}). How do we get the full statement?

In short: Why does compatibility of charts guarantee that the notion of a (C^{m}) function is independent of the chosen chart?

2. Detailed solution (all steps shown)

2.1. What we actually have to prove

Let ((U,\varphi)) and ((V,\psi)) be two charts in a (C^{m})-atlas (\mathcal A). Assume that for a point (p\in U\cap V) the Euclidean representation (f\circ\varphi^{-1}) is a (C^{m}) function on (\varphi(U)). We must show that the Euclidean representation using the other chart, (f\circ\psi^{-1}), is also (C^{m}) (at least on a neighbourhood of (p)). Doing this for every point of (V) will give the desired global independence.

2.2. The domains that appear

(\varphi:U\to\varphi(U)\subset\mathbb R^{n}) is a diffeomorphism onto its image.
(\psi:V\to\psi(V)\subset\mathbb R^{n}) is a diffeomorphism onto its image.
The overlap of the two coordinate neighbourhoods is (U\cap V). Its image in the (\psi)-coordinates is (\psi(U\cap V)\subset\psi(V));
its image in the (\varphi)-coordinates is (\varphi(U\cap V)\subset\varphi(U)).

The change‑of‑coordinates map [ \Phi:=\varphi\circ\psi^{-1}:\psi(U\cap V)\longrightarrow\varphi(U\cap V) ] is defined exactly on the overlap (\psi(U\cap V)); it is not defined outside that set.

2.3. The key observation

The definition of a (C^{m}) function on the manifold is local: to say that (f) is (C^{m}) at a point (p) we only need one chart that contains (p) and on which the Euclidean representation is (C^{m}) on a neighbourhood of the image of (p).

Therefore, to prove independence of the chart we do not have to show that the whole map (f\circ\psi^{-1}:\psi(V)\to\mathbb R) is (C^{m}); we only need to show that it is (C^{m}) on a neighbourhood of the point (\psi(p)). That neighbourhood is precisely (\psi(U\cap V)).

Thus the apparent “domain problem’’ is actually what we need.

2.4. Proof that the representation in the second chart is (C^{m})

Let (p\in U\cap V).
Assume that (f\circ\varphi^{-1}) is (C^{m}) on an open set (W\subset\varphi(U)) containing (\varphi(p)). (If it is (C^{m}) on the whole (\varphi(U)), we can simply take (W=\varphi(U)).)

Change of coordinates is (C^{m}).
By the hypothesis that the atlas is a (C^{m})-atlas, the map
[ \Phi=\varphi\circ\psi^{-1}:\psi(U\cap V)\longrightarrow\varphi(U\cap V) ] is a (C^{m}) diffeomorphism. In particular, its inverse (\Phi^{-1}=\psi\circ\varphi^{-1}) is also (C^{m}).
Compose the two (C^{m}) maps.
Consider the composition [ (f\circ\psi^{-1})\big|_{\psi(U\cap V)} \;=\; (f\circ\varphi^{-1})\;\circ\;\Phi . ] Here:
- (\Phi) is (C^{m}) on the open set (\psi(U\cap V));
- (f\circ\varphi^{-1}) is (C^{m}) on the open set (W) that contains (\Phi\bigl(\psi(U\cap V)\bigr)=\varphi(U\cap V)).
By the chain rule (or the standard theorem that the composition of two (C^{m}) maps is (C^{m})), the composition ((f\circ\varphi^{-1})\circ\Phi) is a (C^{m}) function on (\psi(U\cap V)).
Conclusion for the point (p).
Since (\psi(p)\in\psi(U\cap V)), we have shown that (f\circ\psi^{-1}) is (C^{m}) on a neighbourhood of (\psi(p)). Hence (f) is (C^{m}) at (p) when we use the chart ((V,\psi)).

Because the argument works for any point (p\in V) (choose a chart ((U,\varphi)) that contains (p) and on which (f\circ\varphi^{-1}) is (C^{m}); such a chart exists by the definition of a (C^{m}) function), we obtain that the property “(f) is (C^{m}) at (p)” does not depend on the particular chart that contains (p).

2.5. Why we do not need the whole (\psi(V))

The definition of a (C^{m}) function on a manifold never asks for a single chart to work everywhere on (M). It only asks that locally around each point there is some chart giving a (C^{m}) Euclidean representation. Consequently, when we change from one chart to another we only have to compare the representations on the overlap of the two charts, which is exactly what the compatibility condition guarantees.

If we did want a statement of the form “(f\circ\psi^{-1}) is (C^{m}) on all of (\psi(V))”, it would be false in general: the function might not be (C^{m}) on parts of (V) where we have not yet verified the condition. What matters is that every point of (V) belongs to some overlap with a chart where we already know (C^{m}), and then the above argument shows the desired regularity at that point.

3. Final answer

The compatibility of charts in a (C^{m})-atlas guarantees that the notion of a (C^{m}) function on the manifold is independent of the chosen chart:

If (f\circ\varphi^{-1}) is (C^{m}) on a chart ((U,\varphi)) and ((V,\psi)) is any other chart containing the same point, then on the overlap (\psi(U\cap V)) we have

[ f\circ\psi^{-1} = (f\circ\varphi^{-1})\;\circ\;(\varphi\circ\psi^{-1}), ]

a composition of two (C^{m}) maps; therefore it is (C^{m}) on that overlap. Hence (f) is (C^{m}) at the point when expressed in the (\psi)-coordinates.
Doing this for every point of (V) shows that the definition of a (C^{m}) function does not depend on which chart from the atlas we use.

Thus the requirement that the charts of an atlas be pairwise (C^{m})‑compatible is exactly what makes the definition of differentiability on a manifold well‑posed.

4. Common mistakes

Mistake	Why it’s wrong	How to avoid it
Thinking we must prove (f\circ\psi^{-1}) is (C^{m}) on the whole (\psi(V)).	The definition of a (C^{m}) function is local; we only need (C^{m}) on a neighbourhood of each point.	Remember that the statement “(f) is (C^{m}) at (p)” requires a chart containing (p) with a (C^{m}) representation near (p).
Ignoring the domain of the change‑of‑coordinates map.	The map (\varphi\circ\psi^{-1}) is defined only on the overlap; trying to compose it with a function defined on a larger set leads to a meaningless expression.	Explicitly restrict all compositions to the overlap (\psi(U\cap V)).
Assuming compatibility of two charts automatically gives a global (C^{m}) structure.	Compatibility ensures smooth transition maps, but a single chart may not cover the whole manifold.	Use the atlas: cover the manifold with charts and apply the local argument at each point.
Confusing “(C^{m}) on an open set” with “(C^{m}) on its closure”.	Differentiability is defined on open sets; extending to the boundary can fail.	Work only with open neighbourhoods (e.g., (\psi(U\cap V)) is open in (\psi(V))).
Forgetting to use the chain rule for compositions of maps between Euclidean spaces.	The regularity of the composition is not automatic without invoking the chain rule.	State explicitly: “Since both (\Phi) and (f\circ\varphi^{-1}) are (C^{m}), their composition is (C^{m}) by the chain rule.”

By keeping these points in mind, the proof that chart compatibility makes the definition of a (C^{m}) function well‑defined becomes clear and rigorous.

Original question: A question about why a atlas on a differentiable manifold consist of compatible charts on Mathematics Stack Exchange, licensed CC BY-SA.