Smooth maps between manifolds are continuous The comments in this question Smooth maps (between manifolds) are continuous (comment in Barrett O'Neill's textbook) seem to suggest that a smooth map "in charts", id est satisfying the definition
A mapping $\phi:M\rightarrow N$ 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.
might not be continuous.
Here, I'm thinking every meaning every coordinate system in the maximal atlas.
I am not sure if O'Neil uses maximal atlases but I would like to know the answer for maximal atlases all the same. The definition of maximal atlas can be found here: https://en.wikipedia.org/wiki/Smooth_structure.
Since in this definition, I am not guaranteed that the domain of $\eta\circ\phi\circ\xi^{-1}$ is open let's say that here smooth means that it is the restriction of a smooth map defined on an open set. Does there exist a counterexample of a map satisfying this property that is not continuous?
 A: The remark in the accepted answer in that post refer to the following equivalent definitions of smoothness of a map $f : M \to N$ between manifolds (that automatically implies continuity)

*

*$\color{blue}{(\text{Avoid dealing with domain})}$ $f$ is smooth if for every $p \in M$, there exists smooth charts $(U, \varphi)$ containing $p$ and $(V,\psi)$ containing $f(p)$ such that $f(U) \subseteq V$ and the composite map $\psi \circ f \circ \varphi^{-1}$ is smooth from $\varphi(U)$ to $\psi(V)$.


*$\color{blue}{(\text{Directly impose openness of domain})}$ $f$ smooth if for every $p \in M$, there exists smooth charts $(U, \varphi)$ containing $p$ and $(V,\psi)$ containing $f(p)$ such that $U \cap f^{-1}(V)$ is open in $M$ and the composite map $\psi \circ f \circ \varphi^{-1}$ is smooth from $\varphi(U \cap f^{-1}(V))$ to $\psi(V)$.


*$\color{blue}{(\text{Impose openness by continuity})}$ $f$ is smooth if $f$ is continous and there exists smooth atlases $\{(U_{\alpha},\varphi_{\alpha})\}$ and $\{(V_{\beta},\psi_{\beta})\}$ for $M$ and $N$ respectively, such that for each $\alpha$ and $\beta$, $\psi_{\beta} \circ f \circ \varphi_{\alpha}^{-1}$ is a smooth map from $\varphi_{\alpha}(U_{\alpha} \cap f^{-1}(V_{\beta}))$ to $\psi_{\beta}(V_{\beta})$.
An exercise in Lee's $\textit{Introduction to Smooth Manifolds, 2nd ed}$ illustrate what can go wrong if we drop the requirement that the domain should be open in the second definition : The following function $f : \mathbb{R} \to \mathbb{R}$ (with $\mathbb{R}$ with standard maximal atlas) defined as
$$
  f(x) =
  \begin{cases}
                                   1 & \text{if $x \geq 0$} \\
                                   0 & \text{if $x<0$}.
  \end{cases}
$$
is clearly not continous but it is smooth in the sense of second definition minus openness of domain : as you can see, the composite map $\text{Id}_{(1/2,3/2)} \circ f \circ \text{Id}_{(-1,1)}^{-1}$ is a smooth map from $\text{Id}_{(-1,1)} \big( (-1,1) \cap f^{-1}(\frac{1}{2},\frac{3}{2})  \big) = [0,1)$ to $(1/2,3/2)$.
Here is how one can modifiy example above to find a counterexample for the third definition (minus continuity of course) : Let $f : \mathbb{R} \to \mathbb{R}$ be the same function as above, with domain $\mathbb{R}$ equipped with atlas contain only identity chart $(U,\varphi)=(\mathbb{R},\text{Id}_{\mathbb{R}})$, but atlas of the codomain $\mathbb{R}$ consist of two charts $(V_1,\psi_1)=(\frac{1}{4},\infty), \text{Id}_{(\frac{1}{4},\infty)}$ and $(V_2,\psi_2) = (-\infty,\frac{3}{4}), \text{Id}_{(-\infty,\frac{3}{4})}$. So that now $f$ is not continous but for each $\beta$, $\psi_{\beta} \circ f \circ \varphi^{-1}$ is a smooth map from $\varphi(U \cap f^{-1}(V_{\beta}))$ (which is either $[0,\infty)$ or $(\infty,0)$) to $\psi_{\beta}(V_{\beta})$. In short, we just throw away charts where the representation of $f$ in these charts not smooth/continous but still have enough charts to cover the manifolds.
