Proposition 1.5.19.  Let X be a set, A ⊆ X, B be a set, Y ⊆ B, f:X → Y be an injectivefunction. Then:
Proof.  Let a, b ∈ A such that fAB(a) = fAB(b). Then a = b:
fAB(a) = fAB(b)deff(a) = f(b)
f is injectivedef∀ x, y ∈ X s.t. f(x) = f(y) : x = ya,b ∈ X, f(a) = f(b)a = b