Injection

Question

Let f : [0, 1] → R be an injection. Prove that g : [0, π/2] → R defined by g(x) = f(sin x) is also an injection.

Answer 1

sin(0)=0 and sin(π/2)=1 so since sine is a continuous function, the range is [0,1]. For every angle within the interval [0,π/2] there is the unique sine of the angle in the range [0,1], that is, f(sin(x)) ∀x∈[0,π/2] is f:[0,1]→ℝ, implying g:[0,π/2]→ℝ injectively.