View source for Exercises:Saul - Algebraic Topology - 8/Exercise 8.5

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==Exercises==
===Exercise 8.5===
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<m>\newcommand\J{\mathcal{J} }</m>Suppose {{M|(M,\J_M)}} is a [[topological n-manifold|topological {{M|m}}-manifold]], and {{M|(\mathbb{R}^m,\J_m)}} is a [[topological space]] of the standard {{M|m}}-dimensional [[Euclidean space]]<ref group="Note">We mention the topology as {{M|V_1\in\J_m}} makes it obvious {{M|V_1\subseteq\mathbb{R}^m}} and {{M|V_1}} is an [[open set]] of {{M|\mathbb{R}^n}}</ref>, then suppose {{M|(N,\J_N)}} is a topological {{N|manifold}}, and {{M|(\mathbb{R}^n,\J_n)}} is {{n|dimensional}} Euclidean space with its usual topology. 

Suppose that {{M|f:M\rightarrow N}} is a [[homeomorphism]], so {{M|M\cong_f N}}, show that if this is so then we must have {{M|m\eq n}} - a usual [[logical implication]] question.
====Precursors====
We make extensive use of the following theorem:
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''[[Given a homeomorphism all subspaces of the domain are homeomorphic to their image under the homeomorphism itself]]''
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{{:Given a homeomorphism all subspaces of the domain are homeomorphic to their image under the homeomorphism itself/Statement}}
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Also:
: '''Note: ''' there are 3 common and equivalent definitions of [[locally euclidean of dimension n|locally euclidean (of fixed dimension)]], they vary as follows:
:* There exists a unique {{M|n\in\mathbb{N}_0}} such that:
:** For all points of the manifold there is an [[open neighbourhood]] to the point such that
:**# that the neighbourhood is homeomorphic to an [[open set]] of {{M|\mathbb{R}^n}}
:**# that the neighbourhood is homeomorphic to an [[open ball]] (of some radius, with some centre) in {{M|\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
:**# that the neighbourhood is homeomorphic to {{M|\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====
[[File:ManifoldsAndOpenSets.JPG|thumb|The red sets are our {{M|W}} sets. Purple is {{M|U_1}}, {{M|V_1}} and {{M|f(U_1)}}, cyan is {{M|U_2}} and {{M|V_2}}. We take the intersection of {{M|f(U_1)}} and {{M|U_2}} to get an open set, that's homeomorphic to its image under {{M|\varphi_2}} and we can also pull it back via the inverse homeomorphism to {{M|f}} and then through {{M|\varphi_1}}]]
* Let {{M|p\in M}} be an arbitrary point
** Let {{M|(U_1,\varphi_1:U_1\rightarrow V_1)}} be a [[chart]] about the point {{M|p}} - so {{M|p\in U_1}} and {{M|U_1\cong_{\varphi_1} V_1}} and {{M|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 {{M|U_1\cong_{f\vert^\text{Im}_{U_1} } f(U_1)}}
**** As {{M|p\in U_1}} we see {{M|f(p)\in f(U_1)}}
**** As {{M|f}} is a [[homeomorphism]] it is an [[open map]] so {{M|f(U_1)\in\mathcal{J}_N}} that is {{M|f(U_1)}} is open in {{M|N}}
***** Let {{M|(U_2,\varphi_2:U_2\rightarrow V_2)}} be a chart about the point {{M|f(p)}}, notice {{M|U_2\cong_{\varphi_2} V_2\in\mathcal{J}_n}}
****** Note that {{M|\varphi_2(f(p))\in V_2}} (as {{M|f(p)\in U_2}}, the domain)
****** Define {{M|W_3:\eq U_2\cap f(U_1)}} - note that this is [[open set|open]] in {{M|N}} as the intersection of a finite number (2) of sets is open in a [[topology]]
******* Note that {{M|f(p)\in U_2}} and {{M|f(p)\in f(U_1)}}, so {{M|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 {{M|W_3\cong_{\varphi_2\vert^\text{Im}_{W_3} } \varphi_2(W_3)}}
******* Define {{M|W_4:\eq \varphi_2(W_3)}}, so we have {{M|W_3\cong W_4}}
******** Notice that as {{M|f(p)\in W_3}} we have {{M|\varphi_2(f(p))\in W_4}}
******** As "''[[the intersection of sets is a subset of each set]]''" we notice that {{M|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 {{M|W_3\cong f^{-1}(W_3)}} (as {{M|f}} is a [[homeomorphism]] {{M|f^{-1} }} is a [[function]] and itself a homeomorphism, also as {{M|W_3}} is open {{M|f^{-1}(W_3)}} is open (by [[continuity]] of {{M|f}})
********* As {{M|W_3\subseteq f(U_1)}} notice {{M|f^{-1}(W_3)\subseteq U_1}}<ref group="Note">this is easy to show because {{M|f}} is a [[bijection]] but {{Caveat|May not be true in general!}} consider a non-injective function with {{M|f(a)\eq f(b)}} for {{M|a\neq b}} and {{M|b\notin W_3}} for example, it is also conceivable that {{M|U_1\neq f^{-1}(f(U_1))}} - however as we have a bijection it is okay</ref> - we can only do this because {{M|f}} is a [[bijection]]
********** Define {{M|W_2:\eq f^{-1}(W_3)}}, so {{M|W_2\cong W_3}} (and by transitivity: {{M|W_2\cong W_4}})
*********** As {{M|f(p)\in W_3}} we see {{M|f^{-1}(f(p))\in W_2}}, so {{M|p\in W_2}} (as for a [[bijection]] we actually have an [[inverse function]])
************ We also have {{M|\varphi_1(p)\in V_1}} which we will use shortly.
*********** As mentioned: {{M|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 {{M|W_2\cong_{\varphi_1\vert^\text{Im}_{W_2} } \varphi_1(W_2)\subseteq V_1}}, and by [[continuity]] of the inverse (as {{M|\varphi}} is a [[homeomorphism]] remember) we see {{M|\varphi_1(W_2)}} is open.
************ Define {{M|W_1:\eq\varphi_1(W_2)}}, so we have {{M|W_1\cong W_2}} (and by transitivity: {{M|W_1\cong W_4}})
************* As {{M|p\in W_2}} we see {{M|\varphi_1(p)\in W_1}}
************* Note that {{M|W_1\in\J_m}} and {{M|W_4\in\J_n}}
************* We invoke '''''Hatcher - p126.6 - Theorem 2.26: ''' If {{M|U\subseteq\mathbb{R}^m}} and {{M|V\subseteq\mathbb{R}^n}} are non-empty [[open sets]] then if {{M|U\cong V}} we have {{M|m\eq n}} ''
************** Note that {{M|W_1\cong W_4}} - {{M|W_1}} and {{M|W_4}} are [[homeomorphic]], note also that they're non-empty, as {{M|\varphi_1(p)\in W_4}} and {{M|\varphi_2(f(p))\in W_4}}
*************** We apply the theorem directly:
**************** {{M|W_1\cong W_4\implies m\eq n}}
This completes the proof
=====Short outline=====
* Let {{M|p\in M}} be given
** Let {{M|(U_1,\varphi_1:U_1\rightarrow V_1\in\J_m)}} be a chart about {{M|p}}
*** Then {{M|V_1\cong U_1}} and {{M|U_1\cong f(U_1)}}
*** Let {{M|(U_2,\varphi_2:U_2\rightarrow V_2\in\J_n)}} be a chart about {{M|f(p)}}
**** Then {{M|W_3:\eq U_2\cap f(U_1)}} is open
***** By the subspace homeomorphism theorem we use a lot: {{M|W_3\cong W_4:\eq\varphi_2(W_3)}}
***** Same theorem, using {{M|f^{-1} }} {{M|W_3\cong W_2:\eq f^{-1}(W_3) }}
***** Using it again: {{M|W_2\cong W_1:\eq\varphi_1(W_2)}}
****** Thus {{M|W_1\cong W_2\cong W_3\cong W_4}} or just {{M|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 {{M|p}} will be in each of these
******** Ie:
********* {{M|f(p)\in W_3}} as {{M|f(p)\in f(U_1)}} and {{M|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|m\eq n}}

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==Notes==
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==References==
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