Let $g(x,y)$ be a scalar function and $G(x,y)=<A(x,y),B(x,y)>$ be a vector field in the $(x,y)$ plane.
My understanding is that the function composition
$$g\circ G=g(A(x,y),B(x,y))$$
but
$$G\circ g$$ is undefined/does not make sense.
Am I right?
1
- 1$\begingroup$ Yes. You are right. $\endgroup$Bobbie D– Bobbie D2016-08-30 13:29:56 +00:00Commented Aug 30, 2016 at 13:29
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1 Answer
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You are correct. Perhaps specifying the domain and target of each function will help clear up any doubts.
$$g: \Bbb R^2 \to \Bbb R \\ G: \Bbb R^2 \to \Bbb R^2$$
Therefore $$\require{AMScd} \begin{CD} \Bbb R^2 @>G>> \Bbb R^2 @>g>> \Bbb R\\ @. \large\searrow @. \large\nearrow \\ @. \stackrel{\large\longrightarrow}{g\circ G} \end{CD}$$ makes sense while trying to go the other way wouldn't. The problem with $G\circ g$ is that $g$ "outputs" a scalar where $G$ expects a vector "input".
