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$F:M \rightarrow N$ is smooth if and only if $F: M \rightarrow N$ is continuous and for every chart $(U, \phi)$ on M, $(V, \psi)$ on N, then $ \psi \circ F \circ \phi^{-1}:\phi (U\cap F^{-1}(V)) \rightarrow \psi(V)$ is smooth. My question is: Let $a \in U \cap F^{-1}(V) \Rightarrow a \in U$ and $F(a) \in V$. Then $\phi (a) \in \phi(U)$, and $\psi (F (\phi^{-1} (\phi(a)))= \psi(F(a))$ since $\phi $ is homeomorphism. $F(a) \in V$, and $\psi: V \rightarrow \psi(V)$ is a chart, so it is smooth, so $\psi (F(a))$ is smooth. Then every map is smooth?

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Note that a chart being smooth doesn't make sense (a priori, unless you regard euclidean space as a smooth manifold itself). And points are also not smooth, so I guess you meant $\psi\circ F$ is smooth. But again this doesn't make sense a priori (unless regarding euclidean spaces as smooth manifolds). And even a in that case, this may not be true if $F$ is not smooth to begin with.

It is important to note that smooth is a notion that initially only makes sense for maps between open sets of euclidean spaces. The way we extend this notion to the notion of smooth map between manifolds is with the definition that you state at the beginning of the question. So always ask yourself what does every word mean when you say it. This is just a bit confusing, because smooth had an initial meaning and now we use the word smooth for something else a bit more abstract. But as I said before, the usual smooth maps between euclidean spaces are also naturally smooth in this sense, and this is why your (potentially confusing) statement "charts are smooth" is technically correct.

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  • $\begingroup$ But isnt it custom that the euclidean space is a smooth manifold with the identity chart? Also, when you want to prove a map $F: M \rightarrow N$ is smooth, don't you have to compute at a point in the underlying map between euclidean spaces? $\endgroup$ – zozo123 Sep 26 '18 at 0:59
  • $\begingroup$ Yes, that is what I tried to explain. So you should be careful with the distinction in the usage of the word smooth. What you said in your question, to check that $F$ is always smooth as you claim, is that $F$ is smooth and the chart is smooth so the composition is smooth, right? But this doesn't make sense, because to say that $F$ is smooth means precisely that this composition (also precomposed with the inverse of the other chart) is smooth. So you are being circular $\endgroup$ – Pedro Sep 26 '18 at 1:04
  • $\begingroup$ What I meant was since $F(a) \in V$, so it is a point in the domain $V$ of $\psi$, so that implies $\psi(F(a))$ is smooth? $\endgroup$ – zozo123 Sep 26 '18 at 1:09
  • $\begingroup$ A point cannot be smooth. Do you mean that the composition is smooth? $\endgroup$ – Pedro Sep 26 '18 at 1:09
  • $\begingroup$ I see. So what is the point of computing the function at a point? You will get a point in the end, then how to conclude that the map is smooth? $\endgroup$ – zozo123 Sep 26 '18 at 1:17

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