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?


1 Answer 1


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.

  • $\begingroup$ What about the definition of my question that requires that the map is smooth when read through all charts? This example looks not to be a counterexample in this case as I can take the whole real line as a chart. $\endgroup$ Commented Jul 6, 2020 at 15:04
  • $\begingroup$ @NicolòCavalleri $f$ is smooth at $x=0$ because the representation map is the restriction of the constant map $F\equiv 1$ on $\mathbb{R}$. There's no need to insist the chart must be all of $\mathbb{R}$ since the definition require us to find only one. $\endgroup$ Commented Jul 6, 2020 at 15:14
  • $\begingroup$ What definition? Your definition does require only one chart but mine asks that the representation map is smooth for every choice of charts in the maximal atlases of the two manifolds. Under normal circumstances (ie the map is continuous) the two are equivalent, but are they now? $\endgroup$ Commented Jul 6, 2020 at 16:14
  • $\begingroup$ @NicolòCavalleri Ok. Now i have a feeling that the counterexample that you want is for the third definition above which is the same as Barrett O'neils definition. I have one example in mind. Let me add this to my answer. $\endgroup$ Commented Jul 6, 2020 at 17:22
  • $\begingroup$ @NicolòCavalleri I think it's ok now. $\endgroup$ Commented Jul 6, 2020 at 18:00

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