# Equivalent Characterizations of Smoothness.

Definition. Let $$M$$ and $$N$$ be smooth manifolds, and let $$F\colon M\to N$$ be any map. We say that $$F$$ is a smooth map if for every $$p\in M$$, there exists smooth chart $$\big(U,\varphi\big)$$ containing $$p$$ and $$\big(V,\psi\big)$$ containing $$F(p)$$ such that $$F(U)\subseteq V$$ and the composite map $$\psi\circ F\circ \varphi^{-1}\colon \varphi(U)\to\psi(V)$$ is smooth.

I have to show the following proposition, but many points are not clear to me.

Proposition. Suppose $$M$$ and $$N$$ are smooth manifolds with or without boundary, and $$F\colon M\to N$$ is a map. Then $$F$$ is smooth if and only if either of the following condition is satisfied:

$$(a)$$ For every $$p\in M$$, there exist 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 $$\psi\big(V\big)$$.

$$(b)$$ $$F$$ is continuous and there exists amooth atlases $$\{(U_\alpha,\varphi_\alpha)\}$$ and $$\{(V_\beta,\psi_\beta)\}$$ for $$M$$ and $$N$$ respectevely 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 $$\psi(V_\beta)$$.

Proof. I would like to proceed, hoping that it is the simplest way as follows: first prove that $$F$$ is smooth iff $$(a)$$ holds, and after showing that $$(a)$$ holds iff $$(b)$$ holds.

## F is smooth iff $$(a)$$ holds

($$\Rightarrow$$) If $$F$$ is smooth we have that $$F(U)\subseteq V$$, then $$U\subseteq F^{-1}(V)$$. Therefore $$U\cap F^{-1}(V)=U$$ that is open.

$$(\Rightarrow)$$ If $$(a)$$ holds for every $$p\in M$$, there exist 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 $$\psi\big(V\big)$$. It only remains to show that $$F(U)\subseteq V$$, now if $$\tilde{p}\in U$$ then there exists a smooth chart $$(\tilde{V},\tilde{\psi})$$ containing $$F(\tilde{p})$$

Question 1. How can I prove that $$F\big(\tilde{q}\big)\in V$$?

## (a)$$\iff$$(b)

$$(b)\Rightarrow (a)$$ it seems obvious.

Question 2. I have no idea how to show that $$(a)\Rightarrow (b)$$. Could you give me a hints?

Thanks!

• For Q1, you should first replace $U$ by $U\cap F^{-1}V$ and restricting $\varphi$, then everything follows. For Q2, prove smooth implies (b), which amounts to just patching the charts. May 12, 2019 at 10:25

Question 1: $$F(U)\subset V$$ is not always true. But you can take a new chart of the form $$(U^\prime,\varphi)$$ which has this property (take $$U^\prime=U\cap F^{-1}(V)$$).

Question 2: If you want to do it without "$$a\Leftrightarrow F \text{ smooth}$$", which I guess you do:

For $$p\in M$$ take $$(U_p,\varphi_p)$$ and $$(V_p,\psi_p)$$ some charts around $$p$$ and $$F(p)$$ respectively. If $$U_p^\prime=U_p\cap F^{-1}(V_p)$$ then $$\{ U_p^\prime\}_p$$ is an open cover for $$M$$ such that $$F_{\vert U_p^\prime}$$ is continuous, so $$F$$ is continuous.

Take the atlas $$\{(U_p^\prime,\varphi_p)\}_p$$ for $$M$$. Unfortunately $$\{(V_p,\psi_p)\}$$ may not be an atlas for $$N$$ because the domain of the charts may not cover $$N$$ is $$F$$ is not surjective. You can take the original atlas of $$N$$ instead, which we write $$\{(V_\alpha,\psi_\alpha)\}$$. Then for $$p$$ and $$\alpha$$ the map

$$\psi_\alpha\circ F\circ \varphi_p^{-1}:\varphi_p(U_p^\prime\cap F^{-1}( V_\alpha))\longrightarrow\psi(V_\alpha)$$

has domain $$\varphi_p(U_p^\prime\cap F^{-1}( V_\alpha))=\varphi_p(U_p\cap F^{-1}(V_p\cap V_\alpha))$$ so it is just the composition

$$\underbrace{\psi_\alpha\circ\psi_p^{-1}}_{\substack{\text{smooth because of the} \\ \text{coherence of the charts}}} \circ\underbrace{\psi_p\circ F\circ \varphi_p^{-1}}_{\substack{\text{smooth by assumption} }}.$$

• @AdamThanks for your ansewer. Perhaps it is more direct to prove that if $F$ is smooth then $(b)$ holds and viceversaa. What do you say? May 12, 2019 at 14:02
• @JackJ. Yes the best way might be to prove that "$F$ smooth" implies $(b)$ then $(b)\Rightarrow (a)$ and finally $(a)$ implies "$F$ smooth". But if you want to learn how to manipulate coordinate charts it's a good exercise to prove every directions! May 12, 2019 at 14:23
• @AdamI have another problem! If I start from $F$ smooth, then $(a)$ holds, because we have shown equivalence. From this point how do I get to $(b)$? May 12, 2019 at 14:33
• @JackJ. you mean you want to prove "$F$ smooth" implies $(b)$ ? May 12, 2019 at 14:37
• @JackJ. this is because $F_{\vert U_p^\prime}$ composed at the beginning and at the end by homeomorphisms (namely $\varphi_p$ and $\psi_p$) is continuous (by assumption it is even differentiable in the usual sense). (Ps: don't forget to accept the answer if it satisfies you) May 12, 2019 at 15:54