It's clear that a smooth chart on a manifold is a diffeomorphism. To me, the fact that smoothness of a manifold implies the smoothness of the transition function between the representation of two charts (whose domains overlap in M) should also imply that each plain old chart function is also a diffeomorphism. But is this correct?

  • $\begingroup$ What do you mean by a "plain old chart function"? $\endgroup$
    – bradhd
    Commented Dec 26, 2013 at 16:44

2 Answers 2


It is true that each chart map is a local diffeomorphism, but perhaps not for the reason you think.

When one defines manifolds, one starts with a topological space $X$. For a topological space it makes sense to talk about homeomorphisms and local homeomorphisms. Thus it makes sense to say that $X$ is covered by $U_i$ with each $U_i$ homeomorphic to $\mathbb{R}^n$ via chart map $\phi_i$. The notion of smoothness, however, does not make sense at this stage - $X$ (and also $U_i$'s) does not have any structure that would let one talk about smoothness and hence about diffeomorphisms. If the chart maps are such that transition functions are smooth (as maps between subsets of $\mathbb{R^n}$), then we say that the atlas of these charts is smooth and only then does $X$ obtain a smooth structure from the $\phi_i$'s. Now we can talk about smoothness of maps to or from $X$ - but the very definition is to compose with chart map (for maps to $X$) or to precompose with inverse of chart map (for maps from $X$).

In particular, smoothness of $\phi_i:U_i \to \mathbb{R}^n$ is by definition the same as the smoothness of the composition $\phi_i \cdot\phi_i^{-1}: \mathbb{R}^n \to \mathbb{R}^n$. Of course, that map is identity, so the maps $\phi_i$ are smooth. Similarly, by definition, smoothness of $\phi_i^{-1}$ is the same as smoothness of $\phi_i \cdot\phi_i^{-1}$. So indeed, after one makes $X$ into a smooth manifold by supplying a smoothly compatible collection of charts, the notion of local diffeomorphism makes sense for $X$, and the chart maps are diffeomorphisms.

  • $\begingroup$ Which notion of "local" diffeomorphism are you using? The problem I see is that, in general, coordinate maps as $\phi_i$ are not globally defined, w.r.t the manifold. Interpreting $(U,\phi)$, with $\phi \colon U \to V \subseteq \mathbb{R}^n$, as a global chart for the (sub)manifold $U$, $U$ automatically becomes a smooth manifold, using the maximal smooth atlas generated from this chart, and $\phi \colon U \to V$ can be interpreted as a MAP between the manifolds $U$ and $V$ with (trivially) smooth coordinate representation, as $id_V \circ \phi \circ \phi^{-1} = id_V$. $\endgroup$
    – el_tenedor
    Commented Mar 28, 2016 at 15:06
  • $\begingroup$ @el_tenedor I actually do not use local diffeomorphisms, as far as I can tell; but since you asked: A map $f$ is a local diffeomorphism if for every point $p$ there exists a neighbourhood $U$ of $p$ such that $f|_U$ is diffeomorphism onto its image. And yes, I implicitly assumed that "chart function is a diffeomorphism" means from its domain to its image (which I assume to be all of $\mathbb{R}^n$, and you assume to be $V$, a subset of $\mathbb{R}^n$, but that makes little difference). $\endgroup$
    – Max
    Commented Mar 29, 2016 at 12:06
  • $\begingroup$ Why does this question make sense ? The manifold $X$ is just a topological space. There is no structure to make sense of the notion of "differentiable" with $X$. Then why can you talk about "smoothness" of cart maps $\phi_i$ ? $\endgroup$
    – Hua
    Commented May 28, 2017 at 17:10
  • $\begingroup$ @Hua That's why Max is saying that one can talk of smoothness of the chart maps $\textit{only when}$ the chart maps are such that transition functions are smooth... $\endgroup$ Commented Nov 22, 2020 at 6:17
  • $\begingroup$ Why is the smoothness of $\phi_i$ the same as the smoothness of $\phi_i\circ\phi_i^{-1}$? $\endgroup$
    – kubo
    Commented Mar 11, 2022 at 17:15

Let's suppose we have an $n$-dimensional manifold $M$ equipped with a smooth atlas $\mathcal{A}$. Then the transition function for any pair of charts in $\mathcal{A}$ is indeed smooth, as a map between open subsets of $\mathbb{R}^n$. Now, you could consider a larger atlas (i.e. a collection of charts in the topological sense) $\mathcal{A}'$ containing $\mathcal{A}$, but there is no guarantee that the transition functions for pairs of charts drawn from $\mathcal{A}'\setminus\mathcal{A}$ are smooth. The charts in $\mathcal{A}'$ are homeomorphisms, but they need not be diffeomorphisms.

Here's a somewhat silly example. Let $M=\mathbb{R}$ with the usual smooth structure. Then the chart $(U,\phi_U)$ defined by $U=\mathbb{R}$ and $\phi_U(x) = x^3$ is a topological chart (i.e. a homeomorphism) but it is not a diffeomorphism, as its inverse $x\mapsto x^{\frac 13}$ is not smooth.

  • 5
    $\begingroup$ I don't understand why you added the last three lines. It is definitely true that $x \mapsto x^{\frac{1}{3}}$ is not smooth, but it seems pointless to differentiate regardless the fact that such map takes values on $\mathbb{R}$ seen as a manifold with a different structure. Indeed if the map was $\phi;(\mathbb{R},id) \to (\mathbb{R},id),x\mapsto x^{\frac{1}{3}}$ then it wouldn't be smooth, but our setting is $x:(\mathbb{R},id)\to (\mathbb{R},x^{\frac{1}{3}}), x \mapsto x^{\frac{1}{3}}$. $\endgroup$ Commented Jul 7, 2017 at 13:08
  • 3
    $\begingroup$ To check differentiability in the latter case we have to compute the derivative of $ x^{\frac{1}{3}} \circ (x^{\frac{1}{3}})^{-1} = x^{\frac{1}{3}} \circ x^3 = x $ which is indeed differentiable. $\endgroup$ Commented Jul 7, 2017 at 13:11
  • 1
    $\begingroup$ @TommasoSeneci I think he added the last three lines to suggest that $(\mathbb R, x^3)$ defines a smooth structure on $\mathbb R$, but $x^3$ is not a diffeomorphism. Hence, the coordinate maps need not be diffeomorphisms. $\endgroup$
    – user5826
    Commented Nov 11, 2019 at 18:56

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