Claim (a): Consider the function f: R→ [0, ∞) given by f(x) = x² for x = R. This f is surjective but not injective. It is not bijective. Proof of (a). To prove surjectivity of f, we must show that for every y € [0, ∞) there exists such that f(x) = y. Let y [0, ∞). The non-negative number y has a non-negative square root x = √20. Then we have f(x) x² (√y)² Since x ER, this shows that f is surjective. To prove non-injectivity of f, we must prove that there exists x₁, x2 E R such that x₁ + x2 and = = Consider for example x₁ = -3 and x₂ f(x₁) = f(-3) = (-3)² Since x₁x2, this shows that f is not injective. Finally, f is not bijective because it = 9 = = 2 = We have ƒ( - f(x₂).

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