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(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.

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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.

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