Smooth maps (between manifolds) are continuous (comment in Barrett O'Neill's textbook) (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.
 A: 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. 
A: 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.
