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This is homework. I just finished a question regarding double integration over the unit sphere involving pullbacks of differential forms to provide context (course is advanced Calculus).

The question is given as:

$$f:D->D, D:= {(x,y): x^2+y+^2<_1}$$

$$g(A)=A+a(A)(A-f(A))$$

My task is to find a(A) algebraically. I just am not quite sure where to even start off. At first I thought there might be a theorem in my book involving a a version of stoke's or green's theorem to apply but the rather specific D has me thinking otherwise. I mean sure I could algebraically rearrange g to get a(A) but that seems rather.. superficial and lacking of meaning...

Edit: i should note that $g(a):D->S1$(Unit circle)

Edit: Important details I left out, probably just wasn't thinking it thoroughly. f is C1 and f(A)!= A for each A element D.

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  • $\begingroup$ (Stokes, not Stoke.) $\endgroup$
    – Hans Lundmark
    Commented May 19, 2014 at 6:37
  • $\begingroup$ Barely comprehensible. Do you want construct a retraction $g:D\longrightarrow S^1$ form $f$? Then the formula should be $g(A)=A+a(A)(f(A)-A)$ (NOT $g(a)$) and the idea is: if $f$ has no fixed points, the vector $f(A)-A$ determines a unique point $g(A)$ over $S^1$ (draw it) with $g(A)-A$ some scalar multiple of $f(A)-A$. $\endgroup$
    – Martín-Blas Pérez Pinilla
    Commented May 19, 2014 at 7:11
  • $\begingroup$ You were right, it should be g(A). I have a good visualization of the drawing, though I am not sure this helps me with this problems unless I can simply use simple algebraic manipulation which definitely seems to be an incorrect way of thinking. I mean I could try to parametrize that vector to try to go somewhere with this... $\endgroup$
    – CTfunction
    Commented May 19, 2014 at 7:25

1 Answer 1

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Hint: as $$g(A)-A=a(A)(f(A)−A),$$ taking norms: $$\|g(A)-A\|=|a(A)|\|(f(A)−A)\|$$ and $$|a(A)|=\cdots$$ And the sign of $|a(A)|$ will be...

Also remember that $\|g(A)\|=1$.

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  • $\begingroup$ Thanks for the hint, I see where I need to go now. The last question I need is if a(A) is an element of C1. I am thinking that I can argue that it is since f(A) is C1, and the norm of g(A) is 1. Is this faulty logic? $\endgroup$
    – CTfunction
    Commented May 19, 2014 at 13:03
  • $\begingroup$ The euclidean norm is defined by $\|x\|^2=\langle x,x\rangle=\cdots$. $\endgroup$
    – Martín-Blas Pérez Pinilla
    Commented May 19, 2014 at 14:23

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