Euclidean Group Maps: Proving Injectivity, Surjectivity, and Inverses

[Thales Castro]

Homework Statement

The n-dimentional Euclidean group $E^{n}$ is made of an n-dimentional translation $a: x \mapsto x+a$ ($x,a \in \mathbb{R}^{n}$ ) and a $O(n)$ rotation $R: x \mapsto Rx$, $R \in O(n)$. A general element $(R,a)$ of $E^{n}$ acts on $x$ by $(R,a): x \mapsto Rx + a$. The product is defined by $(R_{2},a_{2}) \times (R_{1},a_{1}): x \mapsto R_{2}(R_{1}x + a_{1}) + a_{2}$, that is, $(R_{2},a_{2}) \circ (R_{1},a_{1}) = (R_{2}R_{1},R_{2}a_{1}+a_{2})$. Show that the maps $a$, $R$, and $(R,a)$ are bijections. Find their inverse maps.

The Attempt at a Solution

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It is not very hard to prove that they're all injective. For instance, if $x,x' \in \mathbb{R}^{n}$ with $x \neq x'$, we can do a proof by contradiction. If the map is not injective, then it is possible that $x + a = x' + a$, which implies that $x = x'$. So, we can conclude that the translation map is injective. The proofs for $R$ and $(R,a)$ are analogous.
I don't know how to prove that they're surjective. It makes intuitive sense to me (if we think of a plane, for example, it's easy to see that every point on the plane after a rotation/translation must come from a point on the "standard" plane), but I'm not sure how to formally write this.
Another question is: do I really need to prove that they are injective and then prove that they are surjective? Would it be enough to just find an inverse map and prove that the composition is the identity?

Thanks in advance!

 

[fresh_42]

The inverses of the translation and the rotation are more or less easy to find. This is shorter than to bother about surjectivity. In a similar way you could try to solve $(R_2,a_2)\cdot (R_1,a_1) = 1 = (1,0)$.

 

[Thales Castro]

The inverses of the translation and the rotation are more or less easy to find. This is shorter than to bother about surjectivity. In a similar way you could try to solve $(R_2,a_2)\cdot (R_1,a_1) = 1 = (1,0)$.

Thanks for the reply!

I understand that it is much easier in this case to just find the inverse map. But, for the sake of the exercise, I still want to find a proof that the maps are surjective, I just don't know how to formally write it.

 

[fresh_42]

Well, in case of $x \mapsto x+a$, you only have to show, that for every $y$ there is an $x$ that maps on $y$. It already is there: $x=y-a$ does it. Similar for rotations. So to find a preimage point that maps to a given image point, is basically the same as to apply the inverse function.

The combined functions $(R,a)$ may be a bit more difficult, but in principle it's the same. A function $f : X \longrightarrow Y$ is surjective, if every $y \in Y$ is hit, i.e. has an $x \in X$ with $f(x)=y$. To find such an $x$ given the $y$, we have to invert $f$. In case $f$ is only surjective and not injective, there can be found more than one candidate for $x$, e.g. $x \mapsto x^2$, but here it is only a single $x$, so $x = f^{-1}(y)$ and we have inverted $f$ anyways.