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

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. $\square$

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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.