Let ∼ be an equivalence relation on X. Let π : X → X/ ∼, π(x) = [x] = {y | y ∼ x}

[A] = ∪_{a∈A}[a] Defn: ∼ is open if U^{open} ⊂ X implies [U] open in X/ ∼. Lemma: ∼ is open iff π is an open mapping. Lemma: If ∼ is open and if X has a countable basis, then X/ ∼ has a countable basis. Lemma: Let ∼ be open. Then {(x, y) | x ∼ y} is closed in X ×X iff X/ ∼ is Hausdorff. Ex: P^{n}(R) = RP^{n} = n-dimensional real projective space is a smooth manifold. RP^{n} = S^{n}/(x ∼ −x) = (R^{n} − {0})/(x ∼ tx)