Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

(Needless to say, I'm a total newbie in differential geometry so I apologize if this seems rather too obvious to many of you).

As a comment on his definition of smooth mapping, Barrett O'Neill in his Semi-Riemannian Geometry book states that smooth mappings are continuous. I've been thinking how to prove it, to no avail.

His definition is: "A mapping $\phi:M\longrightarrow N$ is smooth provided that for every coordinate system $\xi$ in M and $\eta$ in N the coordinate expression $\eta\circ\phi\circ\xi^{-1}$ is Euclidean smooth (and defined on an open set of $\mathbb{R}^m$ [which I assume to be $\xi(U)$ where $U$ is the domain of $\xi$] )."

In his definition of smooth manifolds the set $M$ has a topology, and the rest I assume you know (atlas, smooth overlap, etc.).

Thank you very much.

share|improve this question
To prove that $\phi$ is continuous it suffices to show that $\phi^{-1}(U)$ is open for a basis of open sets of $N$. Since $N$ is locally Euclidean... –  Qiaochu Yuan Aug 9 '11 at 16:04
Let's say the charts are $(\xi, U)$ and $(\eta, V)$. My hope is that his parenthetical remark is meant to imply that the largest possible domain of definition $\xi(\phi^{-1}(V) \cap U) \subset \mathbf{R}^m$ for that composition is open. –  Dylan Moreland Aug 9 '11 at 16:39
@Dylan: that confuses me. Is he requiring that set in $\mathbb{R}^m$ to be open, or is it open as a result of the continuity of $\phi$? –  Weltschmerz Aug 9 '11 at 16:50
@Weltschmerz I'm hoping that it's the first; otherwise, I'm a little worried about this, although I don't have a counterexample. –  Dylan Moreland Aug 9 '11 at 17:36

1 Answer 1

you have to know that your coordinate charts are smooth which follows essentially from the def. of a manifold (locally homeomorphic to open euclidean sets). Then represent your map in question simply as $\eta^{-1} \eta \Phi \xi^{-1} \xi$. The middle term is cont. by def. and the outer twos by what I just said.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.