Proposition 1.5.24.  Let X, Y, Z be sets, f : X → Y be an injective function, g : Y → Z be an injective function. Then:
g ∘ f is injective
Proof.  Let a, b ∈ X such that g(f(a)) = g(f(b)). Then a = b:
g is injective ⇒def ∀ c, d ∈ Y s.t. g(c) = g(d) : c = d ⇒f(a),f(b) ∈ Y, g(f(a)) = g(f(b)) f(a) = f(b)
f is injective ⇒def ∀ x, y ∈ X s.t. f(x) = f(y) : x = y ⇒a,b ∈ X, f(a) = f(b) a = b