Proposition 1.5.21.  Let X be a set, A ⊆ X, B be a set, Y ⊆ B, f : X → Y be a function such that f∣AB is surjective. Then:
f is surjective
Proof.  Let y ∈ Y. Then ∃ x ∈ X : f(x) = y:
f∣AB is surjective ⇒def ∀ b ∈ B : ∃ a ∈ A : f∣AB(a) = b ⇒y ∈ B ∃ a ∈ A : f∣AB(a) = y ⇒def f(a) = y
Take x := a.