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In the above pictures, we can see two definitions of that a map $F:A\to N$ is smooth on $A$, are they equivalent?

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  • $\begingroup$ @JackLee Please take a look at my question. $\endgroup$ – Born to be proud Dec 11 '18 at 13:35
  • $\begingroup$ They're not always equivalent in case $N$ has nonempty boundary. Here's a counterexample: Let $M=\mathbb R$, $N = [0,\infty)$, $A = [0,\infty)\subseteq M$, and let $F\colon A\to N$ be the identity map. Then $F$ is not smooth by the definition in the text, because there is no neighborhood of $0$ in $\mathbb R$ on which $F$ has a smooth extension as a map into $N$. But the coordinate representation of $F$ using standard coordinates in domain and codomain (which is essentially the same map) has an extension to an open subset of $\mathbb R$ as a map into $\mathbb R$. $\endgroup$ – Jack Lee Dec 11 '18 at 23:15
  • $\begingroup$ See also this question. $\endgroup$ – Jack Lee Dec 11 '18 at 23:17
  • $\begingroup$ @JackLee Thank you very much! $\endgroup$ – Born to be proud Dec 11 '18 at 23:54

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