I'm now studying differential geometry, I wonder why the translation $T_a :\mathbb R^3 \to \mathbb R^3$ is a diffeomorphism. Specifically, I'm studying Isometries of $\mathbb R^3$ written by O'Nell.
I've been thinking about how to show $T_a$ is a diffeomorphism.
So.
First, since $T^{-1}_a = T_{-a},$ there exists the inverse. Hence, one to one and onto problem has been done.
However, How can I show $T_a$ is differentiable
$T_a$ is defined by $T_a = p+a$ for any $p$ in $\mathbb R^3.$
 A: To show that $T_a:\mathbb{R}^3\to\mathbb{R}^3$ is differentiable in $\mathbb{R}^3$, it suffices to show that all its components $(T_{a,1},T_{a,2},T_{a,3}):=T_a$ are differentiable. And they are, since they are all polynomials (in fact, they are of class $C^\infty$), since we can write:
$$T_a(p) = (x+a_1,y+a_2,z+a_3),$$
where $(x,y,z):=p$ and $(a_1,a_2,a_3):=a$.
Finally, just a quick reminder. In order to show that $T_a$ is indeed a diffeomorphism (of class $C^\infty$) you also must prove that its inverse is of class $C^\infty$. But, as you have mentioned, $T_a^{-1}=T_{-a}$, so it is already done.
A: A function $f: \mathbb{R}^m \to \mathbb{R}^n$ is said to be differentiable at a point $x_0 \in \mathbb{R}^m$ if there exists a linear map $J: \mathbb{R}^m \to \mathbb{R}^n$ such that
$$
\lim _{\mathbf{h} \rightarrow \mathbf{0}} \frac{\left\|\mathbf{f}\left(\mathbf{x}_{\mathbf{0}}+\mathbf{h}\right)-\mathbf{f}\left(\mathbf{x}_{\mathbf{0}}\right)-\mathbf{J}(\mathbf{h})\right\|_{\mathbf{R}^{n}}}{\|\mathbf{h}\|_{\mathbf{R}^{m}}}=0
$$
So let $f = T_a : \mathbb{R}^3 \to \mathbb{R}^3$ be given by $T_a(p) = p + a$. Then defining $J: \mathbb{R}^3 \to \mathbb{R}^3$ by $J(p) = p$, it's clear that this is satisfied for any $x_0 \in \mathbb{R}^3$, as desired.
