groupoid.$\b$

In the following proposition, we give, in addition to those in Proposition 2.1, other rules of algebraic calculation in a vector groupoid.

In a vector groupoid $(V, \alpha, \beta, \odot, \iota, V_0)$ the following assertions hold :

1.  $0\cdot x=x,\ \forall\ x\in \alpha^{-1}(0)$;

2.  $x\cdot 0=x,\ \forall\ x\in \beta^{-1}(0)$;

3.  For all $x,\ y\in\beta^{-1}(0)$, we have $x-\alpha(x)=y-\alpha(y)\Longrightarrow x=y$;

4.  for all $x,y\in\alpha^{-1}(0)$, we have $x-\beta(x)=y-\beta(y)\Longrightarrow x=y$.

\(i) If $x\in\alpha^{-1}(0)$, then $\alpha(x)=0=\beta(0)$. So $(0,x)\in V_{(2)}$ and, using the condition [*(G2)*]{} from Definition 2.1, one obtains that $0\cdot x=\alpha(x)\cdot x=x$.\
(iv) Let $x,y\in\alpha^{-1}(0)$ such that $x-\beta(x)=y-\beta(y)$. Then $ \alpha(x)= \alpha(y) = 0 $ and $ x - y = \beta(x) -
\beta(y).$ Since $ \alpha $ is linear map and applying Proposition 2.1 (i), one obtains that $0=\alpha(x)- \alpha(y) =
\alpha(x-y)=\alpha (\beta(x)- \beta(y))=\beta(x)-\beta(y)= x-y $, and so $x=y$.

Similarly, we prove that the assertions (ii) and (iii) hold.

Let $(V, \alpha, \beta, \odot, \iota, V_0) $ be a vector groupoid. Then:

$(i)~~~t_{\beta}:\alpha^{-1}(0)\longrightarrow\beta^{-1}(0),~
t_{\beta}(x):=\beta(x)-x~$ is a linear isomorphism.

$(ii)~~t_{\alpha}:\beta^{-1}(0)\longrightarrow\alpha^{-1}(0),~
t_{\alpha}(x):=\alpha(x)-x~$ is a linear isomorphism.

$(i)$ Let be $ x_{1}, x_{2} \in V $ and $ k_{1},
k_{2} \in K$. Then $ t_{\beta}( k_{1} x_{1} + k_{2} x_{2})=$\
$=\beta( k_{1} x_{1} + k_{2} 