\to V,$ $(x,y)\longmapsto m(x,y):=x\odot y,~$ satisfy the following conditions :

1.  $x\odot(y+z-\beta(x))=x\odot y+x\odot z-x$, for all $x,y,z\in V$, such that $\alpha(y)=\beta(x)=\alpha(z)$.

2.  $x\odot(ky+(1-k)\beta(x))=k(x\odot y)+(1-k)x$, for all $x,y\in V$, such that $\alpha(y)=\beta(x)$.

3.  $(y+z-\alpha(x))\odot x=y\odot x+z\odot x-x$, for all $x,y,z\in V$, such that $\alpha(x)=\beta(y)=\beta(z)$.

4.  $(ky+(1-k)\alpha(x))\odot x=k(y\odot x)+(1-k)x$ for all $x,y\in V$, such that $\alpha(x)=\beta(y)$.

When there can be no confusion we put $xy$ or $x\cdot y$ instead of $x\odot y$.

From Definition 3.1 follows the following corollary.

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

$(i)~~~$ The source and target $ \alpha, \beta : V \to V_{0} $ are linear epimorphisms.

$(ii)~~$ The inversion $ \iota : V \to V $ is a linear automorphism.

$(iii)~$ The fibres $\alpha^{-1}(0)$ and $ \beta^{-1}(0) $ and the isotropy group\
$ V(0):= \alpha^{-1}(0)\cap \beta^{-1}(0) $ are vector subspaces of the vector space $V$.

Let $V$ be a vector space over a field $K$. If we define the maps $ \alpha_{0}, \beta_{0}, \iota_{0}: V\longrightarrow V,\ \
\alpha_{0}(x)=\beta_{0}(x)=0,\ \iota_{0}(x)=-x,$ and the multiplication law $m_{0}(x,y)=x+y$, then $(V, \alpha_{0},
\beta_{0}, m_{0}, \iota_{0}, V_{0}=\{0\})$ is a vector groupoid called [*vector groupoid with a single unit*]{}. We will denote this vector groupoid by $(V, + 