Exercises:Saul - Algebraic Topology - 8/Exercise 8.5

Exercises

Exercise 8.5

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

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

Precursors

We make extensive use of the following theorem:

[<collapsible-expand>]

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

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

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

Also:

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

• There exists a unique $n\in\mathbb{N}_0$ 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 $\mathbb{R}^n$
    2. that the neighbourhood is homeomorphic to an open ball (of some radius, with some centre) in $\mathbb{R}^n$
       • 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 $\mathbb{R}^n$.
• 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 $W$ sets. Purple is $U_1$, $V_1$ and $f(U_1)$, cyan is $U_2$ and $V_2$. We take the intersection of $f(U_1)$ and $U_2$ to get an open set, that's homeomorphic to its image under $\varphi_2$ and we can also pull it back via the inverse homeomorphism to $f$ and then through $\varphi_1$

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

This completes the proof

Short outline

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