Let $A$ be an $m\times n$ Boolean matrix. Then the mapping $v\mapsto Av$ is 1:1 iff $n\leq m$ and some subset of $n$ rows of gives a permutation matrix. 

The reason is duality of modules over the Boolean semiring shows that $A$ is 1:1 iff the transpose is onto. Since the standard basis vectors of a free $\mathbb B$-module are join irreducible a Boolean matrix gives an onto map iff each standard basis vector appears as a column.