Proof. Assume that f is onto and Ker f = {0}. Let A be a non-empty subset
of L. We have always f(A∗) ⊆ {f(A)}∗. Let x ∈ {f(A)}∗ ⊆ L′. Since f is
onto, there exists y ∈ L such that f(y) = x. By knowing that f(y) ∈ {f(A)}∗.
Then f(y) ∧ m = 0 for all m ∈ f(A). Let a ∈ A. Then f(y) ∧ f(a) = 0′. That is
f(y ∧ a) = 0′ which means y ∧ a ∈ Ker f = {0}. Then y ∧ a = 0. Hence y ∈ A∗.
Therefore {f(A)}∗ ⊆ f(A∗). Thus {f(A)}∗ = f(A∗). Therefore f is annihilator
preserving.