(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.

  • 2
    $\begingroup$ 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... $\endgroup$ – Qiaochu Yuan Aug 9 '11 at 16:04
  • $\begingroup$ 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. $\endgroup$ – Dylan Moreland Aug 9 '11 at 16:39
  • $\begingroup$ @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 $\phi$? $\endgroup$ – Weltschmerz Aug 9 '11 at 16:50
  • $\begingroup$ @Weltschmerz I'm hoping that it's the first; otherwise, I'm a little worried about this, although I don't have a counterexample. $\endgroup$ – Dylan Moreland Aug 9 '11 at 17:36

Let $M$ and $N$ be smooth manifolds of dimensions $m$ and $n$, respectively, and let $\phi : M \to N$ be a smooth map. It is sufficient to show that $\phi$ is locally continuous, i.e., that every point $x \in M$ has a neighborhood $U_x$ such that $\phi\left|_{U_x}\right.$ is continuous.

Thus, let $x \in M$. Since $M$ and $N$ are smooth manifolds, there exist local coordinate systems $(U,\xi)$ at $x$ and $(V,\chi)$ at $\phi(x)$, and since $\phi$ is smooth, the coordinate expression $$\Phi = \chi \circ \phi \circ \xi^{-1} : \xi(U\cap \phi^{-1}(V)) \to \mathbb{R}^n$$ is smooth. Choose $U_x = U \cap \phi^{-1}(V)$. Then $U_x$ is a neighborhood of $x$. Since $U_x \subset \phi^{-1}(V)$, it follows that $\phi(U_x) \subset \phi(\phi^{-1}(V)) \subset V$ and therefore $$\phi\left|_{U_x}\right. = \chi^{-1} \circ \Phi \circ \xi = \chi^{-1} \circ \left(\chi \circ \phi \circ \xi^{-1}\right) \circ \xi.$$ The maps $\chi^{-1}$ and $\xi$ are homeomorphisms and $\Phi$ is a smooth map between Euclidean spaces, so these maps are all continuous. Therefore $\phi\left|_{U_x}\right.$ is the composition of continuous maps, and so is continuous. $\square$

Remarks: Usually smoothness is only defined this way for continuous maps $\phi$, so that the requirement that $U \cap \phi^{-1}(V)$ be open is automatically satisfied. More recent books such as John M. Lee's Introduction to Smooth Manifolds use the following definition of smoothness:

A map $\phi : M \to N$ between smooth manifolds is smooth if for every $x \in M$ there exist charts $(U,\xi)$ at $x$ and $(V,\chi)$ at $\phi(x)$ such that $\phi(U) \subset V$ and the coordinate expression $\Phi = \chi \circ \phi \circ \xi^{-1}$ is smooth.

Using this definition slightly shortens the proof given above, since you don't have to worry about the domains. For continuous maps, this definition implies the condition used by O'Neill.

  • $\begingroup$ Why is local continuity enough? $\endgroup$ – YoTengoUnLCD Jun 1 '17 at 4:12

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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