Exercise 2.7 (Characterizations of smoothness)

Proposition 2.5

• Let $F$ be smooth. By definition, for every $p\in M$, we can find a chart $(U,{\phi }_p)$ containing $p$ and a chart $(V,{\psi }_p)$ containing $F(p)$ such that $F(U)\subseteq V$ and

  $$\psi \circ F\circ {\phi }^{-1}:\phi (U)\to \psi (V).$$

  is smooth in the calculus sense. By Proposition 2.4, $F$ is continuous; thus, $F^{-1}(V)$ is open. In particular, $U\cap f^{-1}(V)$ is open and the restriction

  $$\psi \circ F\circ {\phi }^{-1}:\phi (U\cap F^{-1}(V))\to \psi (V).$$

  is smooth.

• Suppose that (a) holds. By property (a), for each point $p\in M$, consider a chart $(U_p,{\phi }_p)$ containg $p$ and a chart $(V_p,{\psi }_p)$ containg $F(p)$ satisfying the desired properties. If we restrict our attention to $U_p^{\prime }:=U_p\cap F^{-1}(V_p)$ then
  (1) we obtain a new atlas ${\{(U_p^{\prime },{\phi }_p)\}}_{p\in M}$ since ${\{U_p^{\prime }\}}_{p\in M}$ is an open cover of $M$, and
  (2) $F$ is continuous when restricted to each $U_p^{\prime }$ since it is a composition of continuous functions

  $$F{\text{|}}_U={\psi }_p^{-1}\circ ({\psi }_p\circ F\circ {\phi }_p^{-1})\circ {\phi }_p:U_p\to V_p$$

  from which follows that $F$ is continuous globally.

  The same trick won’t work when taking ${\{(V_p,{\psi }_p)\}}_{p\in M}$ when constructing an atlas for $N$ since the coordinate domains do not necessarily cover $N$. But that’s not a problem since we can simply take the original atlas ${\{(V_{\beta },{\psi }_{\beta })\}}_{\beta \in B}$ of $N$ instead. Then for $p\in M$ and $\beta \in B$, the map

  $${\psi }_{\beta }\circ F\circ {\phi }_p^{-1}:{\phi }_p(U_p^{\prime }\cap F^{-1}(V_{\beta }))\to \psi (V_{\beta })$$

  has domain ${\phi }_p(U_p^{\prime }\cap F^{-1}(V_{\beta }))={\phi }_p(U_p\cap F^{-1}(V_p\cap V_{\beta }))$ which is exactly the composition

  $$\underset{\begin{array}{c}\text{smooth since}\\ \text{charts are compatible}\end{array}}{\underset{⏟}{{\psi }_{\beta }\circ {\psi }_p^{-1}}}\circ \underset{\begin{array}{c}\text{smooth by assumption}\end{array}}{\underset{⏟}{{\psi }_p\circ F\circ {\phi }_p^{-1}}}.$$

  Thus, ${\psi }_{\beta }\circ F\circ {\phi }_p^{-1}$ is smooth for the constructed atlases ${\{(U_p^{\prime },{\phi }_p)\}}_{p\in M}$ and ${\{(V_{\beta },{\psi }_{\beta })\}}_{\beta \in B}$.

• Suppose that (b) holds. Let $p\in M$ be arbitrary. In the given atlas, there is an $\alpha \in A$ such that $p\in U_{\alpha }$ and a $\beta \in B$ such that $F(p)\in V_{\beta }$. By assumption, $F^{-1}(V_{\beta })$ is open since $F$ is continuous and thus, $U\cap F^{-1}(V_{\beta })$ is open. Set $(U,\phi ):=(U_{\alpha }\cap F^{-1}(V_{\beta }),{\phi }_{\alpha }{\text{|}}_{U_{\alpha }\cap F^{-1}(V_{\beta })})$ and $(V,\psi )=(V_{\beta },{\psi }_{\beta })$. By assumption,

  $$\psi \circ F\circ {\phi }^{-1}:\phi (U)\to \psi (V)$$

  is smooth. Thus, $F$ is smooth.

Proposition 2.6

• Suppose that $F$ is locally smooth. Pick an arbitrary $p\in M$. By assumption, there exists a neighbourhood $\tilde{U}$ such that $F{\text{|}}_{\tilde{U}}$ is smooth on $\tilde{U}$. More specifically, for our fixed $p\in \tilde{U}$, there is a chart $(U,\phi )$ of $\tilde{U}$ containing $p$ and a chart $(V,\psi )$ of $N$ containing $F(p)$ such that $F(U)\subseteq V$ and $\psi \circ F\circ {\phi }^{-1}:\phi (U)\to \psi (V)$ is smooth. Since $U$ is an open subset of $\tilde{U}$ which is an open subset of $M$, $U$ is open in $M$ and $(U,\phi )$ is a chart of $M$. Since our choice of $p$ was arbitrary, $F$ is smooth.
• Suppose that $F$ is globally smooth. Let $\tilde{U}$ be an open subset of $M$. Let $p\in \tilde{U}$. By assumption, there is a chart $(U,\phi )$ of $M$ containing $p$ and a chart $(V,\psi )$ of $N$ containing $F(p)$ such that $F(U)\subseteq V$ and $\psi \circ F\circ {\phi }^{-1}:\phi (U)\to \psi (V)$ is smooth. Then the intersection $(\tilde{U}\cap U,\phi )$ is a chart of $M$ that satisfies all of the properties for local smoothness. Since our choice of $p\in \tilde{U}$ was arbitrary, $F{\text{|}}_{\tilde{U}}$ is smooth.