# Non-smooth internal function of composition which happens to be smooth

This is a problem which addresses the dilemma of what can happen of we restrict the conditions for the definition of a smooth function between manifolds.

Firstly, the definition of a smooth function between manifolds is that if we have a function $F:M \rightarrow N$, with $(U, \varphi)$ and $(V, \psi)$ being smooth charts containing $p \in M$ and $F(p) \in N$ respectively, then we say that $F$ is smooth if and only if the composition $\psi \circ F \circ \varphi^{-1}$ is smooth with respect to ordinary calculus.

Now, if we have the function $f: \mathbb{R} \rightarrow \mathbb{R},\ \text{where} \ f(x) = \begin{cases} 1 & x\geq 0 \\ 0 & x \lt 0 \end{cases}$ and $(U,\varphi)$ and $(V, \psi)$ are both smooth charts for the domain and codomain respectively, how do we choose the coordinate maps for smooth charts of both copies of $\mathbb{R}$ such that the composition $\psi \circ f \circ \varphi^{-1}: \varphi(U \cap f^{-1}(V)) \rightarrow \psi(V)$ is smooth even though the internal function $f$ is clearly not smooth.

Now, I have been thinking about this problem and have come to the conclusion that if we have the standard smooth chart for the first copy of $\mathbb{R}$, where $\varphi$ is the identity function we are left to choose some smooth coordinate map of the second chart that smoothly maps $\{0,1\}$ into $\mathbb{R}$. If we choose the atlas of the codomain $\mathbb{R}$ to have $\psi$ as a constant function (being non-zero) is this sufficient? I ask about the sufficiency of such a map because the composition becomes smooth, $f$ is not smooth, however I am in doubt if the chart for the codomain is actually a smooth chart, recalling that a smooth coordinate chart is given as any chart that is housed in the maximal smooth atlas.

• I should state that the condition that is being discarded is $f(U) \subseteq V$ – D.S. Dec 8 '16 at 6:56
• how can $\psi$ be constant function? it must be homeomorphism and presereves dimension. – Paladin Dec 8 '16 at 8:03

What you are trying to do is not possible. Let us say you have picked smooth structures $\mathcal{A}_i$ for $i = 1,2$ on $\mathbb{R}$ (defined using atlases). A function $f \colon \mathbb{R} \rightarrow \mathbb{R}$ that is smooth as a function $f \colon (\mathbb{R}, \mathcal{A}_1) \rightarrow (\mathbb{R}, \mathcal{A}_2)$ between the manifolds $(\mathbb{R}, \mathcal{A}_i)$ will always be continuous. To see why, let us choose a point $p \in \mathbb{R}$, a chart $\varphi_1 \colon U_1 \rightarrow \mathbb{R}$ around $p \in \mathbb{R}$ and a chart $\varphi_2 \colon U_2 \rightarrow \mathbb{R}$ around $f(p) \in \mathbb{R}$. By definition, $f$ is smooth if and only if $\varphi_2 \circ f \circ \varphi_1^{-1} \colon V_1 \rightarrow V_2$ is smooth in the regular sense where $\varphi_i(U_i) = V_i$. In particular, $\varphi_2 \circ f \circ \varphi_1^{-1}$ is continuous. As charts are required to be homeomorphisms, this means that $f$ is also continuous.
If you want an example of a non-smooth (but continuous) function $f \colon \mathbb{R} \rightarrow \mathbb{R}$ that is smooth as a function between manifolds $f \colon (\mathbb{R}, \mathcal{A}_1) \rightarrow (\mathbb{R}, \mathcal{A}_2)$, see here.