“…Now we need to check that for the other four triples the map ϕ ε 1 , ε 2 , ε 3 is a homomorphism. It is clear for (ε 1 , ε 2 , ε 3 ) equal to (0, 0, 0) since we get the trivial map, and also for Page 6] and its kernel is the so-called virtual singular pure braid group, see [10,Definition 5]. Finally, that ϕ ε 1 , ε 2 , ε 3 is a homomorphism for (ε 1 , ε 2 , ε 3 ) equal to (1, 0, 1) or (0, 0, 1) follows from a simple computation using the canonical presentations of the groups involved.…”