x_{2})- ( k_{1} x_{1} + k_{2} x_{2})=
k_{1}( \beta(x_{1}) - x_{1}) + k_{2}( \beta(x_{2}) - x_{2}) =$\
$=k_{1}t_{\beta}(x_{1}) + k_{2}t_{\beta}(x_{2}). $ Hence $
t_{\beta} $ is a linear map.

Let now $ x,y \in \alpha^{-1}(0) $ such that $ t_{\beta} (x) =
t_{\beta} (y)$. Applying Proposition 3.1(iv), one obtains $ x=y$, and so the map $ t_{\beta} $ is injective.

For any $ y\in \beta^{-1}(0)$ we take $ x = \alpha(y)-y $. Clearly $ x\in \alpha^{-1}(0).$ We have $ t_{\beta}(x) =
\beta(\alpha(y)-y) - (\alpha(y)-y ) = \alpha(y)- \beta(y) -
\alpha(y) + y = y $, since $ \beta(y) = 0$. Hence the map $
t_{\beta} $ is surjective. Therefore $ t_{\beta} $ is a linear isomorphism.

$(ii)~$ Similarly we prove that $ t_{\alpha}$ is a linear isomorphism.

Let $~(V, +, \cdot, \alpha, \beta, \odot, \iota, V_0)~$ be a vector groupoid over $ K $ and $~ u\in V_{0}~$ any unit of $~V$. The following assertions hold.

$(i)~~~$ The isotropy group $ V(u):=\{x\in V~|~
\alpha(x)=\beta(x)=u \}$ endowed with the laws $~\boxplus :
V\times V \to V~$ and $~\boxtimes : K\times V \to V~$ given by: $$x\boxplus y=x+y-u,~~~\forall~x,y\in V(u)\label{3.1}$$ $$k\boxtimes x = k x + (1-k)u,~~~\forall~k \in K,~x\in
V(u),\label{3.2}$$ has a structure 