Solution: X = Z/4, sigma(x,y) = (y, x+1)   [rack x <| y = x+1]; bijective, YBE: True
  involutive: False  idempotent: False  fixed pairs of sigma: []
A1  sigma(u_a (x) u_b) = i^a u_b (x) u_a for all a,b in Z/4: True
A2  dim R = 14; u_a u_b in R for (a,b) in {(0,1),(1,0),(1,1)} (so these products vanish in A): [((0, 1), True), ((1, 0), True), ((1, 1), True)]
A3  QS_3 on W (columns = images of u1u1u0, u1u0u1, u0u1u1):
        [(1-1i), (-1+1i), (1-1i)]
        [(-1-1i), (2+0i), (-1+1i)]
        [(-1+1i), (-1-1i), (1-1i)]
    rank = 2
    QS_3(kappa) for kappa = u1u1u0 + (1+i) u1u0u1 + i u0u1u1 : [(0+0i), (0+0i), (0+0i)]  (zero: True )
    u0u1u1 in Im QS_3 ? False
A4  multidegree 2a1+a0: dims (C_3, C_2, C_1, C_0, A_3) = (2, 6, 3, 0, 0)  Euler char -C3+C2-C1+C0-A = 1 (nonzero => Im f not exact)
B   Re z has 64 nonzero integer coordinates, Im z has 64
B1  rank QS_2 = 14; Re z in A(x)B_2(x)A: True; Im z in A(x)B_2(x)A: True
B2  b'(Re z) = 0: True;  b'(Im z) = 0: True
B3  rank_Q of boundaries b'(1(x)B_3(x)1) = 47 (= dim B_3, the map is injective in weight 3)
    rank_Q of boundaries + {Re z, Im z} = 49  => Re z, Im z independent mod boundaries: True
CONCLUSION: H_2(A (x) B (x) A) != 0 in weight 3 over Q; FGG Question 44 has a negative answer.
