~$ and

$ {\widetilde}{m}(f(x),f(y))=
{\widetilde}{m}( (\alpha(x),\beta(x)), (\alpha(y),\beta(y))) =
(\alpha(x),\beta(y)).$

Hence the equality (ii) from Definition 2.2 is verified.

Let now two elements $ x,y \in V$ such that $(f(x),f(y))\in
(V_{0}\times V_{0})_{(2)}.$ Then $ {\widetilde}{\beta}(f(x)) =
{\widetilde}{\alpha}(f(y))$. Since $ f(x) = (\alpha(x), \beta(x)) $ and $
f(y) = (\alpha(y), \beta(y)) $ we deduce that $ (\beta(x),
\beta(x))=(\alpha(y), \alpha(y)) $. Therefore $ \beta(x) = \alpha
(y) $ and $(x,y)\in G_{(V)}.$ Therefore the condition (2.2) from Definition 2.3 is satisfied.

Hence $ f : V \to V_{0}\times V_{0} $ is a groupoid homomorphism.

Let $ x, y \in V $ and $ a,b\in K$. Since $ \alpha, \beta $ are linear maps, we have\
$ f(ax+by) = (\alpha(ax+by), \beta(ax+by)) = ( a \alpha(x) +
b\alpha(y), a \beta(x) + b\beta(y))= $\
$= a(\alpha(x) , \beta(x)) + b (\alpha(y) , \beta(y))= a f(x) + b
f(y) $, i.e. $f$ is a linear map.

Therefore, the conditions from Definition 3.2 are verified. Hence $f$ is a vector groupoid homomorphism.$\b$

ALGEBRAIC CONSTRUCTIONS OF VECTOR GROUPOIDS
===========================================

In this section we shall give some important ways of building up new vector groupoids.

[**1. Direct product of two vector groupoids**]{}. Let given the vector groupoids $ ( V, \alpha_{V}, 