4a.) Claim: g : R^{2} → R, g(x, y) = x + y^{2} is a quotient map. If x ∈ R, g(x, 0) = x. Hence g is onto. Since g(x, y) = π_{1}(x, y) + π_{2}(x, y)^{2} and π_{i} are continuous, products and sums of continuous functions are continuous, g is continuous. Suppose g^{−1}(U) is open in R^{2}. We need to show U is open in R. Take z ∈ U. Then (z, 0) ∈ g^{−1}(z) and there exists (a, b) × (c, d) such that (z, 0) ∈ (a, b) × (c, d) ⊂ g^{−1}(U). Claim: z ∈ (a, b) ⊂ U. (z, 0) ∈ (a, b) × (c, d) implies z ∈ (a, b) Suppose w ∈ (a, b). Then a < w < b Thus, (w, 0) ∈ (a, b) × (c, d) ⊂ g^{−1}(U) Thus w = g((w, 0)) ∈ g(g^{−1}(U)) = U. Hence (a, b) ⊂ U. Hence U is open in R. Thus g is a quotient map. R^{2}/ ∼ = {g^{−1}{z}) | z ∈ R} where R^{2}/ ∼ has the quotient topology from p : R^{2} → R^{2}/ ∼. Thus g induces a bijective map f : R^{2}/ ∼ → R which is a homeomorphism since g is a quotient map (cor 22.3). Thus R^{2}/ ∼ is homeomorphic to R with standard topology. 4b.) Claim: g : R^{2} → [0, ∞), g(x, y) = x^{2} + y^{2} is a quotient map. If x ∈ [0, ∞), g(\sqrt{x}, 0) = x. Hence g is onto. Since g(x, y) = π_{1}(x, y)^{2} + π_{2}(x, y)^{2} and π_{i} are continuous, products and sums of continuous functions are continuous, g is continuous. Suppose g^{−1}(U) is open in R^{2}. We need to show U is open in [0, ∞). Take z ∈ U. Then (\sqrt{z}, 0) ∈ g^{−1}(z) and there exists (a, b) × (c, d) such that (\sqrt{z}, 0) ∈ (a, b) × (c, d) ⊂ g^{−1}(U). Let e = max\{0, a\} Claim: (e^{2}, b^{2}) ⊂ U. Suppose w ∈ (e^{2}, b^{2}). Then since e ≥ 0, e < \sqrt{w} < b. Thus, (\sqrt{w}, 0) ∈ (e, b) × (c, d) ⊂ (a, b) × (c, d) ⊂ g^{−1}(U) Thus w = g((\sqrt{w}, 0)) ∈ g(g^{−1}(U)) = U. Hence (e^{2}, b^{2}) ⊂ U. If z = 0, then a < \sqrt{z} = 0 and hence e = max\{0, a\} = 0 . Since z ∈ U, z ∈ [0, b^{2}) = [e^{2}, b^{2}) ⊂ U where [0, b^{2}) is open in [0, ∞). If z > 0, \sqrt{z} ∈ (e, b) ⊂ (a, b). e < \sqrt{z} < b implies e^{2} < z < b^{2} since e ≥ 0. Thus z ∈ (e^{2}, b^{2}). Hence z ∈ (e^{2}, b^{2}) ⊂ U where (e^{2}, b^{2}) is open in [0, ∞). Hence U is open in [0, ∞). Thus g is a quotient map. R^{2}/ ∼ = {g^{−1}{z}) | z ∈ [0∞)} where R^{2}/ ∼ has the quotient topology from p : R^{2} → R^{2}/ ∼. Thus g induces a bijective map f : R^{2}/ ∼ → [0, ∞) which is a homeomorphism since g is a quotient map (cor 22.3). Thus R^{2}/ ∼ is homeomorphic to [0, ∞).