# Question about the definition of smooth map between manifolds

$$F:M \rightarrow N$$ 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?

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.
• 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? – zozo123 Sep 26 '18 at 0:59
• 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 – Pedro Sep 26 '18 at 1:04
• 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? – zozo123 Sep 26 '18 at 1:09