Section 1.7

Quotient spaces

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Problem 1.7.1

Suppose q:X→Yq : X \to Y and YY has the quotient topology with respect to (X,q)(X, q) (i.e. U⊂YU \subset Y is open iff q−1(U)⊂Xq^{-1}(U) \subset X is open). Show C⊂YC \subset Y is closed iff q−1(C)⊂Xq^{-1}(C) \subset X is closed.

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Problem 1.7.2

Consider X=[0,1]X = [0, 1] with YY the quotient space formed by identifying 00 with 11, and X′=RX' = \mathbb {R} with Y′Y' the quotient space formed by identifying xx with x+nx + n for all n∈Zn \in \mathbb {Z}. Let q:X→Yq : X \to Y, q′:X′→Y′q' : X' \to Y' be the identification maps, and let i:X→X′i : X \to X' be the inclusion. Since [i(0)]=[0]=[1]=[i(1)][i(0)] = [0] = [1] = [i(1)], ii induces a map iˉ:Y→Y′\bar{i} : Y \to Y' defined by iˉ([x])=[i(x)]\bar{i}([x]) = [i(x)], which is a continuous bijection onto Y′Y'.

Construct an inverse for iˉ\bar{i} and show that it is continuous.

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Remark.
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Consider the discontinuous function from X′X' to XX defined by sending xx to x−[x]x - [x], where [x][x] denotes the greatest integer in xx; i.e. the unique integer satisfying [x]≤x<[x]+1[x] \leq x < [x] + 1.

Problem 1.7.3

Describe basic open sets in the quotient space [0,1]×[0,1]/(0,t)∼(1,t),(s,0)∼(s,1)[0, 1] \times [0, 1] / (0, t) \sim (1, t), (s, 0) \sim (s, 1) about each of the points [(0,0)][(0, 0)], [(12,0)][(\tfrac {1}{2}, 0)], [(0,12)][(0, \tfrac {1}{2})], and [(12,12)][(\tfrac {1}{2}, \tfrac {1}{2})]. Describe the inverse image q−1(U)q^{-1}(U) of each of these basic open sets.

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Problem 1.7.4

Show that the quotient space formed from a square by identifying all of the points in the bottom edge of the square to each other is homeomorphic to a triangle.

Triangle as a quotient space of the square.Triangle as a quotient space of the square.

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Remark.
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Start with the map from the rectangle to the triangle preserving yy-levels and sending the bottom edge of the rectangle to the bottom vertex of the triangle. See the figure above, where the bottom line that is to be collapsed to a point is thickened, as is the image point.

Problem 1.7.5

Let Ai∪fiBi=Ai⊔Bi/x∼fi(x),x∈Ki⊂BiA_i \cup_{f_i} B_i = A_i \sqcup B_i / x \sim f_i(x), x \in K_i \subset B_i be the quotient space of Ai∪BiA_i \cup B_i coming from identifying x∈Ki⊂Bix \in K_i \subset B_i with f(x)∈Aif(x) \in A_i via a homeomorphism fi:Ki→f(Ki)f_i : K_i \to f(K_i), i=1,2i = 1, 2.

Proposition 1.7.6: Suppose there are homeomorphisms FA:A1→A2F_A : A_1 \to A_2, FB:B1→B2F_B : B_1 \to B_2 with FB(K1)=K2F_B(K_1) = K_2 and FAf1=f2FBF_A f_1 = f_2 F_B. Then the map F:A1∪f1B1→A2∪f2B2F : A_1 \cup_{f_1} B_1 \to A_2 \cup_{f_2} B_2 given by F(x)=FA(x)F(x) = F_A(x) if x∈A1x \in A_1 and F(x)=FB(x)F(x) = F_B(x) if x∈B1x \in B_1 is a homeomorphism.

Prove Proposition 1.7.6.

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Problem 1.7.6

Consider the space XX formed from two copies of R=D1×D1R = D^1 \times D^1 by identifying {−1,1}×D1\left\{ -1, 1\right\} \times D^1 to itself via dd with d(1,y)=(1,−y)d(1, y) = (1, -y) and d(−1,y)=(−1,−y)d(-1, y) = (-1, -y); that is, X=R∪dRX = R \cup_d R. Construct a homeomorphism between XX and the annulus.

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Problem 1.7.7

Suppose X=A∪gBX = A \cup_g B, Y=A∪g′BY = A \cup_{g'} B, where g,g′:K⊂B→g(K),g′(K)⊂Ag, g' : K \subset B \to g(K), g'(K) \subset A are homeomorphisms. Suppose (g′)−1g:K→K=h∣K(g')^{-1} g : K \to K = h|_K, where h:B→Bh : B \to B is a homeomorphism. Show that the identity on AA and hh on BB piece together to give a homeomorphism from XX to YY.

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Problem 1.7.8

Show that the Möbius band can also be described as a quotient space D1×D1∪fD1×D1D^1 \times D^1 \cup_f D^1 \times D^1, with f(−1,t)=(−1,t)f(-1, t) = (-1, t), f(1,t)=(1,−t)f(1, t) = (1, -t).

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Problem 1.7.9

Identify all points in the lower half of the circle to each other. Show that the resulting quotient space S1/∼S^1 / \sim is homeomorphic to S1S^1.

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Remark.
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Find a continuous map from the circle to the circle which sends the lower half of the circle to a point and is 1--1 elsewhere.

Problem 1.7.10

Put an equivalence relation on the unit disk by making all points on the boundary circle equivalent to each other. Show that the resulting quotient space D2/∼D^2 / \sim is homeomorphic to S2S^2.

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Remark.
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Send diameters to great circles with the origin going to the south pole and the boundary circle going to the north pole.