I) Let α: S → T. Suppose α is surjective and there exist maps β_{1}, β_{2} of T into a set U such that β_{1} ≠ β_{2} but β_{1}α = β_{2}α. Then there is a t ∈ T such that β_{1}(t) ≠ β_{2}(t). And there is an s ∈ S such that α(s) = t. So β_{1}(α(s)) ≠ β_{2}(α(s)). Contradiction. Conversely, by contraposition, if α is not surjective define β_{1}: T → U as the union of { (t, u_{0}) | t ∈ im α, u_{0} an arbitrary fixed element in U } and { (t, u_{1}) | t ∉ im α, u_{1} an arbitrary fixed element in U }. Define β_{2} as β_{1} with (t, u_{1}) replaced by (t, u_{2}), u_{1} ≠ u_{2} for every t ∉ im α. Thus β_{1}α = β_{2}α and β_{1} ≠ β_{2}.

II) Suppose α is injective and there exist maps γ_{1}, γ_{2} of a set U into S such that γ_{1} ≠ γ_{2} but αγ_{1} = αγ_{2}. Then there is a u ∈ U such that γ_{1}(u) ≠ γ_{2}(u). So α(γ_{1}(u)) ≠ α(γ_{2}(u)). Contradiction. Conversely, by contraposition, if α is not injective there exists s_{1}, s_{2} such that s_{1} ≠ s_{2} and α(s_{1}) = α(s_{2}). Define γ_{1}: U → S as { (u, s_{1}) | u ∈ U }. Define γ_{2}: U → S as { (u, s_{2}) | u ∈ U }. Thus γ_{1} ≠ γ_{2} but αγ_{1} = αγ_{2}.