Let D be an oriented graph. The inversion of a set X of vertices in D consists in reversing the direction of all arcs with both ends in X. The inversion number of D, denoted by inv(D), is the minimum number of inversions needed to make D acyclic. Denoting by τ (D), τ (D), and ν(D) the cycle transversal number, the cycle arc-transversal number and the cycle packing number of D respectively, one shows that inv(D) ≤ τ (D), inv(D) ≤ 2τ (D) and there exists a function g such that inv(D) ≤ g(ν(D)). We conjecture that for any two oriented graphs L and R, inv(L → R) = inv(L) + inv(R) where L → R is the dijoin of L and R. This would imply that the first two inequalities are tight. We prove this conjecture when inv(L) ≤ 1 and inv(R) ≤ 2 and when inv(L) = inv(R) = 2 and L and R are strongly connected. We also show that the function g of the third inequality satisfies g(1) ≤ 4. We then consider the complexity of deciding whether inv(D) ≤ k for a given oriented graph D. We show that it is NP-complete for k = 1, which together with the above conjecture would imply that it is NP-complete for every k. This contrasts with a result of Belkhechine et al. [6] which states that deciding whether inv(T) ≤ k for a given tournament T is polynomial-time solvable.