Jets on affine spaces $\mathbb R^n$ I do not understand here in $0.1$ how
follows
$$J^r(\mathbb R^n,\mathbb R^m)=\mathbb R^m \times \mathbb R^n \times J_0^r(\mathbb R^m,\mathbb 
 R^n)_0$$
from
$$j_0^r t_{f(x)}\circ j_0^r (t_{-f(x)}\circ f\circ t_x)\circ j_x^r t_{-x}$$
EDIT: (translation of 0.1)
0.1 The variety $J^r(\mathbb R^n,\mathbb R^m)$. On affine spaces
$\mathbb R^n,\mathbb R^m$ we have distinguished global
coordinate maps (i.e. ordinary coordinates). Every jet $j^r_x f$
of a mapping $f:\mathbb R^m\to \mathbb R^n$
we can express as
$$j_0^r t_{f(x)}\circ j_0^r (t_{-f(x)}\circ f\circ t_x)\circ j_x^r t_{-x}$$
where $t_a$ means shift at $a$. SO we get
$$J^r(\mathbb R^n,\mathbb R^m)=\mathbb R^m \times \mathbb R^n \times J_0^r(\mathbb R^m,\mathbb 
 R^n)_0.$$
 A: A mapping $f\colon \mathbb{R}^m \to \mathbb{R}^n$ takes the value $f(x) \in \mathbb{R}^n$ at the point $x \in \mathbb{R}^m$.
For fixed $x\in \mathbb{R}^m$ we can shift in the domain and codomain to make zero map to zero: indeed, $$g(y) = f(x+y)-f(x)$$ has the property $g(0) = 0$, and (by replacing $y$ by $y-x$ above) we have $f(y) = g(y-x) + f(x)$.
In terms of compositions with translation functions, this is $$g = t_{-f(x)} \circ f \circ t_x \quad\text{and}\quad f = t_{f(x)} \circ g \circ t_{-x},$$
hence $$f = t_{f(x)} \circ (t_{-f(x)} \circ f \circ t_x) \circ t_{-x}$$
and the equality of jets $$j^r_xf = j_0^r t_{f(x)}\circ j_0^r (t_{-f(x)}\circ f\circ t_x)\circ j_x^r t_{-x}$$ follows.
Hence the $r$-jet $j^r_x f \in J^r_x(\mathbb{R}^m,\mathbb{R}^n)$ is completely determined by $j^r_0 g \in J^r_0(\mathbb{R}^m,\mathbb{R}^n)_0$ and the pair $(x,f(x)) \in \mathbb{R}^m \times \mathbb{R}^n$. Conversely, every $j^r_0 g \in J^r_0(\mathbb{R}^m,\mathbb{R}^n)_0$ and $(x,z) \in \mathbb{R}^m \times \mathbb{R}^n$ determine the $r$-jet of a function at $x$.
