# Smoothness is local

Let us consider a map $$f:M\to R$$ where $$M$$ is a smooth manifold. If every point $$p\in M$$ has a neighborhood $$U$$ such that $$f|_U$$ is smooth, prove that $$f$$ is a smooth function.

My idea is to prove that any two coordinate charts from any two atlases are smoothly compatible (if $$f|_U$$ is smoothly than $$f\circ \varphi^{-1}$$ is smoothly for any $$\varphi$$ from the atlas that defines the smooth structure on $$U$$). Is that ok? If yes, how can I prove that?

Thank you!

I think that is better to prove the statement by definition. Given $$p\in M$$ and $$U$$ neighborhood of $$p$$ such that $$f|_U$$ is smooth, there exists coordinate charts $$(U\cap U_\alpha, \varphi_\alpha|_{U\cap U_\alpha})$$ (where $$(U_\alpha, \varphi_\alpha)$$ is a chart of $$M$$) and $$(V,\psi)$$ of $$p$$ and $$f|_U(p)=f(p)$$ respectively such that $$\psi\circ f\circ \varphi_\alpha^{-1}|_{U\cap U_\alpha}:\varphi_\alpha(U\cap U_\alpha)\to \psi(V)$$ is smooth, but this is the definition of smoothness of $$f:M\to N$$ using the charts $$(U\cap U_\alpha, \varphi_\alpha|_{U\cap U_\alpha})$$ and $$(V,\psi)$$ where $$f(U\cap U_\alpha)=f|_U (U\cap U_\alpha)\subset V$$.
• A comment about question math.stackexchange.com/q/3493048 : I have appreciated you behavior (withdrawal was a solution ; improvement was another one). I have given an answer using ill-known (even for me) Grassmann-Pluecker relationships, but it is a long proof that works when $P,Q,R$ are independent... Dec 31, 2019 at 23:29
• Since the question is about real valued smooth functions, there is no need to use the definition for smooth functions between smooth manifolds, i.e., there is only need for expressions of the form $f\circ \varphi^{-1}$. Moreover, the proof is incomplete since you have only found a pair of charts that satisfy the definition (where as the definition requires for them all), although it can be easily solved by using an arbitrary chart $(W, \varphi)$ and the smooth compatibility so that you get a composition of smooth functions $f\circ\varphi_{\alpha}^{-1}\circ(\varphi_{\alpha}\circ\varphi)$ Nov 18, 2023 at 11:28