Exercises:Saul - Algebraic Topology - 8

Exercises

Exercise 8.5

[ilmath]\newcommand\J{\mathcal{J} }[/ilmath]Suppose [ilmath](M,\J_M)[/ilmath] is a topological [ilmath]m[/ilmath]-manifold, and [ilmath](\mathbb{R}^m,\J_m)[/ilmath] is a topological space of the standard [ilmath]m[/ilmath]-dimensional Euclidean space, then suppose [ilmath](N,\J_N)[/ilmath] is a topological [ilmath]n[/ilmath]-manifold, and [ilmath](\mathbb{R}^n,\J_n)[/ilmath] is [ilmath]n[/ilmath]-dimensional Euclidean space with its usual topology.

Suppose that [ilmath]f:M\rightarrow N[/ilmath] is a homeomorphism, so [ilmath]M\cong_f N[/ilmath], show that if this is so then we must have [ilmath]m\eq n[/ilmath] - a usual logical implication question.

Precursors

We make extensive use of the following theorem:

Given a homeomorphism all subspaces of the domain are homeomorphic to their image under the homeomorphism itself

Let [ilmath](X,\mathcal{ J })[/ilmath] and [ilmath](Y,\mathcal{ K })[/ilmath] be topological spaces and suppose that [ilmath]f:X\rightarrow Y[/ilmath] is a homeomorphism between them, so [ilmath]X\cong_f Y[/ilmath], then:

• [ilmath]\forall A\in\mathcal{P}(X)[A\cong_{f\vert_{A}^{\text{Im} } } f(A)][/ilmath]
  • In words: For all subspaces of [ilmath]X[/ilmath], suppose in particular [ilmath]A[/ilmath] is a subspace, then [ilmath]f\vert_{A}^\text{Im}:A\rightarrow f(A)[/ilmath] - the restriction onto its image of [ilmath]f[/ilmath] to [ilmath]A[/ilmath] - is a homeomorphism between [ilmath]A[/ilmath] and [ilmath]f(A)\subseteq Y[/ilmath]
    • So [ilmath]A\cong_{f\vert_A^\text{Im} }f(A)[/ilmath] explicitly

Also:

Note: there are 3 common and equivalent definitions of locally euclidean (of fixed dimension), they vary as follows:

• There exists a unique [ilmath]n\in\mathbb{N}_0[/ilmath] such that:
  • For all points of the manifold there is an open neighbourhood to the point such that
    1. that the neighbourhood is homeomorphic to an open set of [ilmath]\mathbb{R}^n[/ilmath]
    2. that the neighbourhood is homeomorphic to an open ball (of some radius, with some centre) in [ilmath]\mathbb{R}^n[/ilmath]
       • that the neighbourhood is homeomorphic to the open unit ball centred at the origin - this is easy as any open ball centred anywhere is homeomorphic to this open ball
    3. that the neighbourhood is homeomorphic to [ilmath]\mathbb{R}^n[/ilmath].
• We take it as known that these are equivalent, thus we may choose any. I use the first one (homeomorphic to any open set) as the others are trivially instances of this

Proof

The red sets are our [ilmath]W[/ilmath] sets. Purple is [ilmath]U_1[/ilmath], [ilmath]V_1[/ilmath] and [ilmath]f(U_1)[/ilmath], cyan is [ilmath]U_2[/ilmath] and [ilmath]V_2[/ilmath]. We take the intersection of [ilmath]f(U_1)[/ilmath] and [ilmath]U_2[/ilmath] to get an open set, that's homeomorphic to its image under [ilmath]\varphi_2[/ilmath] and we can also pull it back via the inverse homeomorphism to [ilmath]f[/ilmath] and then through [ilmath]\varphi_1[/ilmath]

• Let [ilmath]p\in M[/ilmath] be an arbitrary point
  • Let [ilmath](U_1,\varphi_1:U_1\rightarrow V_1)[/ilmath] be a chart about the point [ilmath]p[/ilmath] - so [ilmath]p\in U_1[/ilmath] and [ilmath]U_1\cong_{\varphi_1} V_1[/ilmath] and [ilmath]V_1\in\J_m [/ilmath]
    • By "Given a homeomorphism all subspaces of the domain are homeomorphic to their image under the homeomorphism itself" we see [ilmath]U_1\cong_{f\vert^\text{Im}_{U_1} } f(U_1)[/ilmath]
      • As [ilmath]p\in U_1[/ilmath] we see [ilmath]f(p)\in f(U_1)[/ilmath]
      • As [ilmath]f[/ilmath] is a homeomorphism it is an open map so [ilmath]f(U_1)\in\mathcal{J}_N[/ilmath] that is [ilmath]f(U_1)[/ilmath] is open in [ilmath]N[/ilmath]
        • Let [ilmath](U_2,\varphi_2:U_2\rightarrow V_2)[/ilmath] be a chart about the point [ilmath]f(p)[/ilmath], notice [ilmath]U_2\cong_{\varphi_2} V_2\in\mathcal{J}_n[/ilmath]
          • Note that [ilmath]\varphi_2(f(p))\in V_2[/ilmath] (as [ilmath]f(p)\in U_2[/ilmath], the domain)
          • Define [ilmath]W_3:\eq U_2\cap f(U_1)[/ilmath] - note that this is open in [ilmath]N[/ilmath] as the intersection of a finite number (2) of sets is open in a topology
            • Note that [ilmath]f(p)\in U_2[/ilmath] and [ilmath]f(p)\in f(U_1)[/ilmath], so [ilmath]f(p)\in W_3[/ilmath]
            • Again using "Given a homeomorphism all subspaces of the domain are homeomorphic to their image under the homeomorphism itself" we see [ilmath]W_3\cong_{\varphi_2\vert^\text{Im}_{W_3} } \varphi_2(W_3)[/ilmath]
            • Define [ilmath]W_4:\eq \varphi_2(W_3)[/ilmath], so we have [ilmath]W_3\cong W_4[/ilmath]
              • Notice that as [ilmath]f(p)\in W_3[/ilmath] we have [ilmath]\varphi_2(f(p))\in W_4[/ilmath]
              • As "the intersection of sets is a subset of each set" we notice that [ilmath]W_3\subseteq f(U_1)[/ilmath]
              • Using "Given a homeomorphism all subspaces of the domain are homeomorphic to their image under the homeomorphism itself" we see [ilmath]W_3\cong f^{-1}(W_3)[/ilmath] (as [ilmath]f[/ilmath] is a homeomorphism [ilmath]f^{-1} [/ilmath] is a function and itself a homeomorphism, also as [ilmath]W_3[/ilmath] is open [ilmath]f^{-1}(W_3)[/ilmath] is open (by continuity of [ilmath]f[/ilmath])
                • As [ilmath]W_3\subseteq f(U_1)[/ilmath] notice [ilmath]f^{-1}(W_3)\subseteq U_1[/ilmath] - we can only do this because [ilmath]f[/ilmath] is a bijection
                  • Define [ilmath]W_2:\eq f^{-1}(W_3)[/ilmath], so [ilmath]W_2\cong W_3[/ilmath] (and by transitivity: [ilmath]W_2\cong W_4[/ilmath])
                    • As [ilmath]f(p)\in W_3[/ilmath] we see [ilmath]f^{-1}(f(p))\in W_2[/ilmath], so [ilmath]p\in W_2[/ilmath] (as for a bijection we actually have an inverse function)
                      • We also have [ilmath]\varphi_1(p)\in V_1[/ilmath] which we will use shortly.
                    • As mentioned: [ilmath]W_2\subseteq U_1[/ilmath] - this will be very important
                    • Using "Given a homeomorphism all subspaces of the domain are homeomorphic to their image under the homeomorphism itself" we see that [ilmath]W_2\cong_{\varphi_1\vert^\text{Im}_{W_2} } \varphi_1(W_2)\subseteq V_1[/ilmath], and by continuity of the inverse (as [ilmath]\varphi[/ilmath] is a homeomorphism remember) we see [ilmath]\varphi_1(W_2)[/ilmath] is open.
                      • Define [ilmath]W_1:\eq\varphi_1(W_2)[/ilmath], so we have [ilmath]W_1\cong W_2[/ilmath] (and by transitivity: [ilmath]W_1\cong W_4[/ilmath])
                        • As [ilmath]p\in W_2[/ilmath] we see [ilmath]\varphi_1(p)\in W_1[/ilmath]
                        • Note that [ilmath]W_1\in\J_m[/ilmath] and [ilmath]W_4\in\J_n[/ilmath]
                        • We invoke Hatcher - p126.6 - Theorem 2.26: If [ilmath]U\subseteq\mathbb{R}^m[/ilmath] and [ilmath]V\subseteq\mathbb{R}^n[/ilmath] are non-empty open sets then if [ilmath]U\cong V[/ilmath] we have [ilmath]m\eq n[/ilmath]
                          • Note that [ilmath]W_1\cong W_4[/ilmath] - [ilmath]W_1[/ilmath] and [ilmath]W_4[/ilmath] are homeomorphic, note also that they're non-empty, as [ilmath]\varphi_1(p)\in W_4[/ilmath] and [ilmath]\varphi_2(f(p))\in W_4[/ilmath]
                            • We apply the theorem directly:
                              • [ilmath]W_1\cong W_4\implies m\eq n[/ilmath]

This completes the proof

Short outline

• Let [ilmath]p\in M[/ilmath] be given
  • Let [ilmath](U_1,\varphi_1:U_1\rightarrow V_1\in\J_m)[/ilmath] be a chart about [ilmath]p[/ilmath]
    • Then [ilmath]V_1\cong U_1[/ilmath] and [ilmath]U_1\cong f(U_1)[/ilmath]
    • Let [ilmath](U_2,\varphi_2:U_2\rightarrow V_2\in\J_n)[/ilmath] be a chart about [ilmath]f(p)[/ilmath]
      • Then [ilmath]W_3:\eq U_2\cap f(U_1)[/ilmath] is open
        • By the subspace homeomorphism theorem we use a lot: [ilmath]W_3\cong W_4:\eq\varphi_2(W_3)[/ilmath]
        • Same theorem, using [ilmath]f^{-1} [/ilmath] [ilmath]W_3\cong W_2:\eq f^{-1}(W_3) [/ilmath]
        • Using it again: [ilmath]W_2\cong W_1:\eq\varphi_1(W_2)[/ilmath]
          • Thus [ilmath]W_1\cong W_2\cong W_3\cong W_4[/ilmath] or just [ilmath]W_1\cong W_4[/ilmath]
            • We notice this is non-empty by showing that some (pre)image of some chain of functions of the point [ilmath]p[/ilmath] will be in each of these
              • Ie:
                • [ilmath]f(p)\in W_3[/ilmath] as [ilmath]f(p)\in f(U_1)[/ilmath] and [ilmath]f(p)\in U_2[/ilmath]
                • and so forth (these are in the main proof)
              • We then apply Hatcher - Theorem 2.26 directly to show [ilmath]m\eq n[/ilmath]