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I Why a bijective map may not preserve area?

  1. Jul 11, 2016 #1
    Consider the following map ##f## that maps an annulus to a larger annulus: ##f: (r, \theta)\to(r+1, \theta)##. ##f## maps the annulus in the region ##1\leq r\leq2## to the annulus in the region ##2\leq r\leq3##. Clearly, the area is not preserved.

    Next, is the converse true? That is, must an area-preserving map be a bijective map? How about a length-preserving map and a volume-preserving map? Must they be bijective?
     
  2. jcsd
  3. Jul 11, 2016 #2
    The two relation is not related originally. You cannot judge from intuition, but have to verify an idea from their definitions strictly.
     
  4. Jul 11, 2016 #3

    PeroK

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    No it's not true. You should be able to find a counterexample very easily.
     
  5. Jul 11, 2016 #4
    The true statements would be
    1. Not all injective map preserves area (or length or volume).
    2. All area-preserving maps are injective.

    Right?
     
  6. Jul 11, 2016 #5

    PeroK

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    Why would an area preserving map be an injection? You could map ##[0, 1]## onto itself without it being 1-1.

    PS: You are effectively saying that all mappings from a set to itself that are onto must be 1-1.
     
  7. Jul 11, 2016 #6

    mfb

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    There is no relation between area-conservation, being injective and being surjective. All 8 combinations are possible.
     
  8. Jul 11, 2016 #7
    https://en.m.wikipedia.org/wiki/Isometry

    Under formal definitions, it says an isometry (or a distance-preserving map) is automatically injective.
     
  9. Jul 11, 2016 #8

    PeroK

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    An isometry is a pointwise distance preserving map. For every two points ##x, y## the distance between ##f(x)## and ##f(y)## is preserved. That's a very specific additional condition on the map. In this case the 1-1 property is trivially true, since ##x = y \iff d(x, y) = 0 \iff d(f(x), f(y)) = 0 \iff f(x) = f(y)##

    Your initial example made us all think you were talking about the length/volume/area of the Domain and Range.
     
  10. Jul 11, 2016 #9
    Yes I was referring to pointwise distance-preserving maps. Would such a map (an isometry) also preserve area and volume (and higher-dimensional volume) point-wise or region-wise?

    Is there an easy proof?

    The questions in post #1 become "why a bijective map may not preserve distance point-wise?" and "must a region-wise area-preserving map be injective?".
     
    Last edited: Jul 11, 2016
  11. Jul 12, 2016 #10
    Hi Happiness,

    I will tackle your three questions separately.

    1) Your question was if an isometry would preserve ##n##-volume. The answer is yes, it would, but we must be a bit careful. If ##A \subset \mathbb{R}^n## then we define the volume of ##A## to be $$v(A) = \int_A 1$$. It can be shown (see Munkres' "Analysis on Manifolds", page 176) that a map ##h: \mathbb{R}^n \rightarrow \mathbb{R}^n## is an isometry if and only if ##h(x) = Ax + p## where ##A## is an orthogonal matrix. This is a pretty intuitive fact. We can now prove that an isometry preserves volume: We know that ##Dh(x) = A##. It is not hard to see that ##h## is a diffeomorphism. Thus by change of variables theorem, $$\begin{align*} v(h(A)) &= \int_{h(A)} 1 \\ &= \int_{A} \lvert \det(A) \rvert \\ &= v(A) \end{align*}$$ Thus ##h## preserves ##n##-volume.


    2) Your question was why doesn't a bijective map preserve point-wise distance. The answer to this is that it is just a feature of bijections. Think of the bijection that sends ##(0,1)## onto ##(-2,2)##. If you picture ##(0,1)## as a line segment that is made out of a stretchy material (think of a rubber band), then the bijection stretches it out until it is ##(-2,2)##. Even though this is a bijection, pointwise distance has not been preserved.


    3) Your question was must a region-wise area-preserving map be injective. If by this, you meant, "given a set ##A \subset \mathbb{R}^n##, if a map preserves its area, must that map be injective?" The answer to that question would be no. PeroK answered it above. Now if you meant, "must a map which preserves the volume of ALL regions (which have a well-defined volume) be injective?". The answer to that is, I don't know. I think that the answer may be no, it does not need to be injective. I am thinking that you could possibly have a map which sends each region to itself, plus a fixed set of measure zero. Such a map would not be injective and would preserve volume since zero sets do not affect volume. I would have to think about this a bit more to be certain though.
     
  12. Jul 13, 2016 #11
    Hi JonnyG, thanks for your reply!

    The book I'm reading justifies the change-of-variables theorem by assuming ##h## preserves volume. So if we use the theorem to prove its assumption, that is, ##h## preserves volume, we have a circular argument.

    The change-of-variables theorem from the book:
    Screen Shot 2016-07-13 at 2.45.27 pm.png
     
  13. Jul 13, 2016 #12
    The theorem can be, and usually is, proved without even mentioning isometries.
     
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