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I'm new to this notation: $$ F=(F_1,F_2, \dots, F_m) $$ where $F:\mathbb{R}^n \rightarrow \mathbb{R}^m$.

No boldface or variables. Is this a abbreviation for a vector field?

I mean, is the explicit form actually a vector $\mathbf{x}=(x_1,x_2,\dots, x_n)$ in $\mathbb{R}^n$ and $$ F(\mathbf{x})=(F_1(\mathbf{x}), F_2(\mathbf{x}), \dots, F_m(\mathbf{x})) \iff $$ $$ F(x_1,x_2,\dots, x_n)=(F_1(x_1,x_2,\dots, x_n), F_2(x_1,x_2,\dots, x_n), \dots, F_m(x_1,x_2,\dots, x_n)) $$ in $\mathbb{R}^m$? Is this a correct interpretation of the notation?

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    $\begingroup$ Yep, and each $\;F_i\;$ is a scalar function: $\;F_i:\Bbb R^n\to\Bbb R\;$ . $\endgroup$ – DonAntonio May 26 '17 at 22:50
  • $\begingroup$ Yes, but I wouldn't describe $F$ as a vector field. It's just a function. $\endgroup$ – Qiaochu Yuan May 26 '17 at 22:54
  • $\begingroup$ @QiaochuYuan I think that is precisely the definition of vector field in general. $\endgroup$ – DonAntonio May 26 '17 at 22:58
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    $\begingroup$ @DonAntonio The problem is that $n$ may not be $m$, so that the function does not really assign a vector from $\Bbb R^n$. $\endgroup$ – edm May 26 '17 at 23:22
  • $\begingroup$ @edm Well, yes: that seems to be a condition among several authors. Perhaps it is matter of definition, that the map must be from $\;\Bbb R^n\;$ to itself. Thanks. $\endgroup$ – DonAntonio May 26 '17 at 23:38

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