Adjunction spaces and gluing along subsets

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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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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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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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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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Consider the quotient space XX formed from two copies of D1×D1D^1 \times D^1 using f:{−1,1}×D1→{−1,1}×D1f : \left\{ -1, 1\right\} \times D^1 \to \left\{ -1, 1\right\} \times D^1 by f(−1,x)=(1,−x)f(-1, x) = (1, -x), f(1,y)=(−1,y)f(1, y) = (-1, y), X=(D1×D1)∪f(D1×D1)X = (D^1 \times D^1) \cup_f (D^1 \times D^1). Decide whether XX is homeomorphic to the annulus or the Möbius band, and prove your assertion.

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Show that D2∪fD2D^2 \cup_f D^2, where f:K→Kf : K \to K is f(x)=xf(x) = x, and K={(x1,x2)∈S1:x1≥0}K = \left\{ (x_1, x_2) \in S^1 : x_1 \geq 0\right\} is homeomorphic to D2D^2.

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Remark.
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First choose a homeomorphism hh from D2D^2 to D1×D1D^1 \times D^1 where h(K)={(x1,x2)∈D1×D1:x1=1}h(K) = \left\{ (x_1, x_2) \in D^1 \times D^1 : x_1 = 1\right\}. Use this to get a homeomorphism D2∪fD2≃(D1×D1)∪g(D1×D1)D^2 \cup_f D^2 \simeq (D^1 \times D^1) \cup_g (D^1 \times D^1), where g:h(K)→h(K)g : h(K) \to h(K) is g(x)=xg(x) = x.

Construct a homeomorphism between the quotient space D11∪idD21D_{1}^{1} \cup_{\mathrm{id}} D_{2}^{1} and the circle S1S^{1}, where Di1D_{i}^{1} denotes a copy of the interval [−1,1][-1,1] and the identification glues corresponding endpoints of the two copies together via the identity.

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References

[Law06] Lawson T. Topology: A Geometric Approach. Oxford University Press, 2006.