Why a bijective map may not preserve area?

[Happiness]

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?

 

[tommyxu3]

The two relation is not related originally. You cannot judge from intuition, but have to verify an idea from their definitions strictly.

 

[PeroK]

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?

No it's not true. You should be able to find a counterexample very easily.

 

[Happiness]

No it's not true. You should be able to find a counterexample very easily.

The true statements would be
1. Not all injective map preserves area (or length or volume).
2. All area-preserving maps are injective.

Right?

 

[PeroK]

The true statements would be
1. Not all injective map preserves area (or length or volume).
2. All area-preserving maps are injective.

Right?

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.

 

[mfb]

The true statements would be
1. Not all injective map preserves area (or length or volume).
2. All area-preserving maps are injective.

Right?

There is no relation between area-conservation, being injective and being surjective. All 8 combinations are possible.

 

[Happiness]

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.

https://en.m.wikipedia.org/wiki/Isometry

Under formal definitions, it says an isometry (or a distance-preserving map) is automatically injective.

 

[PeroK]

https://en.m.wikipedia.org/wiki/Isometry

Under formal definitions, it says an isometry (or a distance-preserving map) is automatically injective.

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.

 

[Happiness]

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.

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?".

 

[JonnyG]

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.

 

[Happiness]

Hi Happiness,

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.

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:

 

[JonnyG]

The theorem can be, and usually is, proved without even mentioning isometries.