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Is a manifold-with-boundary with given interior and non-empty boundary essentially unique?



Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?Contractible manifold with boundary - is it a disc?Manifolds with two coordinate chartsDoes a *topological* manifold have an exhaustion by compact submanifolds with boundary?If 2-manifolds are homeomorphic and smooth, are they diffeomorphic?Exotic line arrangementsVolume form on a hyperbolic manifold with geodesic boundaryOn compact, orientable 3-manifolds with non-empty boundaryExtension of a group action beyond the boundaryFinding a specific Global Smooth FunctionRemove a disc from a manifold. When is the resulting sphere nullhomotopic?










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$begingroup$


Let $M$ be a compact connected manifold-with-boundary such that $circ M neq emptyset$, where $circ M$ is the boundary of $M$. Let $N$ be a compact connected manifold-with-boundary such that $circ N neq emptyset$ and $bullet M approx bullet N$, where $bullet M$ denotes the interior of $M$ and $approx$ denotes homeomorphic. Does it necessarily hold that $N approx M$?



(I have asked this question before here, but there were no replies.)










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    4












    $begingroup$


    Let $M$ be a compact connected manifold-with-boundary such that $circ M neq emptyset$, where $circ M$ is the boundary of $M$. Let $N$ be a compact connected manifold-with-boundary such that $circ N neq emptyset$ and $bullet M approx bullet N$, where $bullet M$ denotes the interior of $M$ and $approx$ denotes homeomorphic. Does it necessarily hold that $N approx M$?



    (I have asked this question before here, but there were no replies.)










    share|cite|improve this question









    New contributor




    kaba is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$














      4












      4








      4





      $begingroup$


      Let $M$ be a compact connected manifold-with-boundary such that $circ M neq emptyset$, where $circ M$ is the boundary of $M$. Let $N$ be a compact connected manifold-with-boundary such that $circ N neq emptyset$ and $bullet M approx bullet N$, where $bullet M$ denotes the interior of $M$ and $approx$ denotes homeomorphic. Does it necessarily hold that $N approx M$?



      (I have asked this question before here, but there were no replies.)










      share|cite|improve this question









      New contributor




      kaba is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      Let $M$ be a compact connected manifold-with-boundary such that $circ M neq emptyset$, where $circ M$ is the boundary of $M$. Let $N$ be a compact connected manifold-with-boundary such that $circ N neq emptyset$ and $bullet M approx bullet N$, where $bullet M$ denotes the interior of $M$ and $approx$ denotes homeomorphic. Does it necessarily hold that $N approx M$?



      (I have asked this question before here, but there were no replies.)







      differential-topology manifolds






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      kaba is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      share|cite|improve this question








      edited Apr 15 at 22:33









      YCor

      29.1k486141




      29.1k486141






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      asked Apr 15 at 20:21









      kabakaba

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          1 Answer
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          11












          $begingroup$

          No, there are examples detected by Whitehead torsion. If $P$ is a compact connected $(n-1)$-manifold with empty boundary, then (assuming $nge 6$) for every element $tau$ of the Whitehead group of $pi_1(P)$ there is an $h$-cobordism $M$ on $P$ such that $tau$ is the Whitehead torsion of the pair $(M,P)$. The interior of $M$ will be isomorphic to $Ptimesmathbb R$, but if $tau$ is nontrivial then $M$ will not be isomorphic to $Ptimes I$.






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            Thank you. I do not yet understand the answer; I will have to read about Whitehead torsion tomorrow. Would the same answer apply if $M$ and $N$ were orientable?
            $endgroup$
            – kaba
            Apr 15 at 23:45










          • $begingroup$
            Yes, it has nothing to do with orientability.
            $endgroup$
            – Tom Goodwillie
            Apr 16 at 0:30











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          1 Answer
          1






          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          11












          $begingroup$

          No, there are examples detected by Whitehead torsion. If $P$ is a compact connected $(n-1)$-manifold with empty boundary, then (assuming $nge 6$) for every element $tau$ of the Whitehead group of $pi_1(P)$ there is an $h$-cobordism $M$ on $P$ such that $tau$ is the Whitehead torsion of the pair $(M,P)$. The interior of $M$ will be isomorphic to $Ptimesmathbb R$, but if $tau$ is nontrivial then $M$ will not be isomorphic to $Ptimes I$.






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            Thank you. I do not yet understand the answer; I will have to read about Whitehead torsion tomorrow. Would the same answer apply if $M$ and $N$ were orientable?
            $endgroup$
            – kaba
            Apr 15 at 23:45










          • $begingroup$
            Yes, it has nothing to do with orientability.
            $endgroup$
            – Tom Goodwillie
            Apr 16 at 0:30















          11












          $begingroup$

          No, there are examples detected by Whitehead torsion. If $P$ is a compact connected $(n-1)$-manifold with empty boundary, then (assuming $nge 6$) for every element $tau$ of the Whitehead group of $pi_1(P)$ there is an $h$-cobordism $M$ on $P$ such that $tau$ is the Whitehead torsion of the pair $(M,P)$. The interior of $M$ will be isomorphic to $Ptimesmathbb R$, but if $tau$ is nontrivial then $M$ will not be isomorphic to $Ptimes I$.






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            Thank you. I do not yet understand the answer; I will have to read about Whitehead torsion tomorrow. Would the same answer apply if $M$ and $N$ were orientable?
            $endgroup$
            – kaba
            Apr 15 at 23:45










          • $begingroup$
            Yes, it has nothing to do with orientability.
            $endgroup$
            – Tom Goodwillie
            Apr 16 at 0:30













          11












          11








          11





          $begingroup$

          No, there are examples detected by Whitehead torsion. If $P$ is a compact connected $(n-1)$-manifold with empty boundary, then (assuming $nge 6$) for every element $tau$ of the Whitehead group of $pi_1(P)$ there is an $h$-cobordism $M$ on $P$ such that $tau$ is the Whitehead torsion of the pair $(M,P)$. The interior of $M$ will be isomorphic to $Ptimesmathbb R$, but if $tau$ is nontrivial then $M$ will not be isomorphic to $Ptimes I$.






          share|cite|improve this answer











          $endgroup$



          No, there are examples detected by Whitehead torsion. If $P$ is a compact connected $(n-1)$-manifold with empty boundary, then (assuming $nge 6$) for every element $tau$ of the Whitehead group of $pi_1(P)$ there is an $h$-cobordism $M$ on $P$ such that $tau$ is the Whitehead torsion of the pair $(M,P)$. The interior of $M$ will be isomorphic to $Ptimesmathbb R$, but if $tau$ is nontrivial then $M$ will not be isomorphic to $Ptimes I$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Apr 15 at 22:14

























          answered Apr 15 at 21:15









          Tom GoodwillieTom Goodwillie

          40.4k3111201




          40.4k3111201











          • $begingroup$
            Thank you. I do not yet understand the answer; I will have to read about Whitehead torsion tomorrow. Would the same answer apply if $M$ and $N$ were orientable?
            $endgroup$
            – kaba
            Apr 15 at 23:45










          • $begingroup$
            Yes, it has nothing to do with orientability.
            $endgroup$
            – Tom Goodwillie
            Apr 16 at 0:30
















          • $begingroup$
            Thank you. I do not yet understand the answer; I will have to read about Whitehead torsion tomorrow. Would the same answer apply if $M$ and $N$ were orientable?
            $endgroup$
            – kaba
            Apr 15 at 23:45










          • $begingroup$
            Yes, it has nothing to do with orientability.
            $endgroup$
            – Tom Goodwillie
            Apr 16 at 0:30















          $begingroup$
          Thank you. I do not yet understand the answer; I will have to read about Whitehead torsion tomorrow. Would the same answer apply if $M$ and $N$ were orientable?
          $endgroup$
          – kaba
          Apr 15 at 23:45




          $begingroup$
          Thank you. I do not yet understand the answer; I will have to read about Whitehead torsion tomorrow. Would the same answer apply if $M$ and $N$ were orientable?
          $endgroup$
          – kaba
          Apr 15 at 23:45












          $begingroup$
          Yes, it has nothing to do with orientability.
          $endgroup$
          – Tom Goodwillie
          Apr 16 at 0:30




          $begingroup$
          Yes, it has nothing to do with orientability.
          $endgroup$
          – Tom Goodwillie
          Apr 16 at 0:30










          kaba is a new contributor. Be nice, and check out our Code of Conduct.









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          kaba is a new contributor. Be nice, and check out our Code of Conduct.












          kaba is a new contributor. Be nice, and check out our Code of Conduct.











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