Proving the No Retraction Theorem












3












$begingroup$


I try to understand a proof of the No Retraction Theorem which states that for any compact smooth manifold $M$ with boundary $partial M neq emptyset$, there is no smooth map
$$
f : M to partial M , f|_{partial M } = id.
$$



For the proof we suppose that such a map exists. Then by Sard's theorem we find a regular value $r in partial M$ and the primage of $r$ under $f$ is a 1-dimensional manifold. The author now continues to say that the boundary of this preimage has to be a subset of $partial M$. My questions is: Why do we know that?






general-topology functions differential-topology







share|cite|improve this question









$endgroup$

















    3












    $begingroup$


    I try to understand a proof of the No Retraction Theorem which states that for any compact smooth manifold $M$ with boundary $partial M neq emptyset$, there is no smooth map
    $$
    f : M to partial M , f|_{partial M } = id.
    $$



    For the proof we suppose that such a map exists. Then by Sard's theorem we find a regular value $r in partial M$ and the primage of $r$ under $f$ is a 1-dimensional manifold. The author now continues to say that the boundary of this preimage has to be a subset of $partial M$. My questions is: Why do we know that?






    general-topology functions differential-topology







    share|cite|improve this question









    $endgroup$















      3












      3








      3


      1



      $begingroup$


      I try to understand a proof of the No Retraction Theorem which states that for any compact smooth manifold $M$ with boundary $partial M neq emptyset$, there is no smooth map
      $$
      f : M to partial M , f|_{partial M } = id.
      $$



      For the proof we suppose that such a map exists. Then by Sard's theorem we find a regular value $r in partial M$ and the primage of $r$ under $f$ is a 1-dimensional manifold. The author now continues to say that the boundary of this preimage has to be a subset of $partial M$. My questions is: Why do we know that?






      general-topology functions differential-topology







      share|cite|improve this question









      $endgroup$




      I try to understand a proof of the No Retraction Theorem which states that for any compact smooth manifold $M$ with boundary $partial M neq emptyset$, there is no smooth map
      $$
      f : M to partial M , f|_{partial M } = id.
      $$



      For the proof we suppose that such a map exists. Then by Sard's theorem we find a regular value $r in partial M$ and the primage of $r$ under $f$ is a 1-dimensional manifold. The author now continues to say that the boundary of this preimage has to be a subset of $partial M$. My questions is: Why do we know that?






      general-topology functions differential-topology




      general-topology functions differential-topology






      share|cite|improve this question













      share|cite|improve this question











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      asked 1 hour ago









      MuziMuzi

      387218




      387218






















          2 Answers
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          $begingroup$

          This is due to the form that the regular value theorem takes for manifolds with boundary:




          Let $f:Mto N$ be a smooth map between manifolds with boundary and let $partial N=emptyset$. Let $cin N$ be a regular value for $f$. Then, $f^{-1}(c)$ is a submanifold of $M$ such that $dim f^{-1}(c)=dim M-dim N$ and $partial f^{-1}(c)= f^{-1}(c)cap partial M$.







          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            That helps a lot. I didn't know this version of the regular value theorem
            $endgroup$
            – Muzi
            55 mins ago



















          4












          $begingroup$

          The conclusion of the Sard's Theorem argument is a bit stronger than what you say.



          The real conclusion is that the map $f : M to partial M$ has a regular value $r in partial M$.



          It follows, first, that $f^{-1}(r)$ is a 1-manifold with boundary.



          Now suppose that you take $x in f^{-1}(r) cap text{interior}(M)$. The derivative map $D_x f : T_x M to T_r M$ has rank $1$, so its kernel is a 1-dimensional subspace of $M$. Applying the implicit function theorem, and using that $x$ is in the interior of $M$, it follows that $f^{-1}(r)$ intersected with a small neighborhood of $x$ is a 1-manifold without boundary containing $r$. Thus, $x$ cannot be a boundary point of the 1-manifold $f^{-1}(r)$, therefore all of the boundary points of $f^{-1}(r)$ are contained in $partial M$.



          One can make a still stronger conclusion, namely that $f^{-1}(r)$ is a "properly embedded" submanifold of $M$, which means that each point on $partial M cap f^{-1}(r)$ has a neighborhood in which the intersection of that neighborhood with $f^{-1}(r)$ looks like the intersection of the half-disc ${(x,y) mid x^2 + y^2 < 0, |y| ge 1}$ with the line $x=0$.






          share|cite|improve this answer









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            2 Answers
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            active

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            2 Answers
            2






            active

            oldest

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            active

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            active

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            3












            $begingroup$

            This is due to the form that the regular value theorem takes for manifolds with boundary:




            Let $f:Mto N$ be a smooth map between manifolds with boundary and let $partial N=emptyset$. Let $cin N$ be a regular value for $f$. Then, $f^{-1}(c)$ is a submanifold of $M$ such that $dim f^{-1}(c)=dim M-dim N$ and $partial f^{-1}(c)= f^{-1}(c)cap partial M$.







            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              That helps a lot. I didn't know this version of the regular value theorem
              $endgroup$
              – Muzi
              55 mins ago
















            3












            $begingroup$

            This is due to the form that the regular value theorem takes for manifolds with boundary:




            Let $f:Mto N$ be a smooth map between manifolds with boundary and let $partial N=emptyset$. Let $cin N$ be a regular value for $f$. Then, $f^{-1}(c)$ is a submanifold of $M$ such that $dim f^{-1}(c)=dim M-dim N$ and $partial f^{-1}(c)= f^{-1}(c)cap partial M$.







            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              That helps a lot. I didn't know this version of the regular value theorem
              $endgroup$
              – Muzi
              55 mins ago














            3












            3








            3





            $begingroup$

            This is due to the form that the regular value theorem takes for manifolds with boundary:




            Let $f:Mto N$ be a smooth map between manifolds with boundary and let $partial N=emptyset$. Let $cin N$ be a regular value for $f$. Then, $f^{-1}(c)$ is a submanifold of $M$ such that $dim f^{-1}(c)=dim M-dim N$ and $partial f^{-1}(c)= f^{-1}(c)cap partial M$.







            share|cite|improve this answer









            $endgroup$



            This is due to the form that the regular value theorem takes for manifolds with boundary:




            Let $f:Mto N$ be a smooth map between manifolds with boundary and let $partial N=emptyset$. Let $cin N$ be a regular value for $f$. Then, $f^{-1}(c)$ is a submanifold of $M$ such that $dim f^{-1}(c)=dim M-dim N$ and $partial f^{-1}(c)= f^{-1}(c)cap partial M$.








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            share|cite|improve this answer



            share|cite|improve this answer










            answered 1 hour ago









            Saucy O'PathSaucy O'Path

            5,9091626




            5,9091626












            • $begingroup$
              That helps a lot. I didn't know this version of the regular value theorem
              $endgroup$
              – Muzi
              55 mins ago


















            • $begingroup$
              That helps a lot. I didn't know this version of the regular value theorem
              $endgroup$
              – Muzi
              55 mins ago
















            $begingroup$
            That helps a lot. I didn't know this version of the regular value theorem
            $endgroup$
            – Muzi
            55 mins ago




            $begingroup$
            That helps a lot. I didn't know this version of the regular value theorem
            $endgroup$
            – Muzi
            55 mins ago











            4












            $begingroup$

            The conclusion of the Sard's Theorem argument is a bit stronger than what you say.



            The real conclusion is that the map $f : M to partial M$ has a regular value $r in partial M$.



            It follows, first, that $f^{-1}(r)$ is a 1-manifold with boundary.



            Now suppose that you take $x in f^{-1}(r) cap text{interior}(M)$. The derivative map $D_x f : T_x M to T_r M$ has rank $1$, so its kernel is a 1-dimensional subspace of $M$. Applying the implicit function theorem, and using that $x$ is in the interior of $M$, it follows that $f^{-1}(r)$ intersected with a small neighborhood of $x$ is a 1-manifold without boundary containing $r$. Thus, $x$ cannot be a boundary point of the 1-manifold $f^{-1}(r)$, therefore all of the boundary points of $f^{-1}(r)$ are contained in $partial M$.



            One can make a still stronger conclusion, namely that $f^{-1}(r)$ is a "properly embedded" submanifold of $M$, which means that each point on $partial M cap f^{-1}(r)$ has a neighborhood in which the intersection of that neighborhood with $f^{-1}(r)$ looks like the intersection of the half-disc ${(x,y) mid x^2 + y^2 < 0, |y| ge 1}$ with the line $x=0$.






            share|cite|improve this answer









            $endgroup$


















              4












              $begingroup$

              The conclusion of the Sard's Theorem argument is a bit stronger than what you say.



              The real conclusion is that the map $f : M to partial M$ has a regular value $r in partial M$.



              It follows, first, that $f^{-1}(r)$ is a 1-manifold with boundary.



              Now suppose that you take $x in f^{-1}(r) cap text{interior}(M)$. The derivative map $D_x f : T_x M to T_r M$ has rank $1$, so its kernel is a 1-dimensional subspace of $M$. Applying the implicit function theorem, and using that $x$ is in the interior of $M$, it follows that $f^{-1}(r)$ intersected with a small neighborhood of $x$ is a 1-manifold without boundary containing $r$. Thus, $x$ cannot be a boundary point of the 1-manifold $f^{-1}(r)$, therefore all of the boundary points of $f^{-1}(r)$ are contained in $partial M$.



              One can make a still stronger conclusion, namely that $f^{-1}(r)$ is a "properly embedded" submanifold of $M$, which means that each point on $partial M cap f^{-1}(r)$ has a neighborhood in which the intersection of that neighborhood with $f^{-1}(r)$ looks like the intersection of the half-disc ${(x,y) mid x^2 + y^2 < 0, |y| ge 1}$ with the line $x=0$.






              share|cite|improve this answer









              $endgroup$
















                4












                4








                4





                $begingroup$

                The conclusion of the Sard's Theorem argument is a bit stronger than what you say.



                The real conclusion is that the map $f : M to partial M$ has a regular value $r in partial M$.



                It follows, first, that $f^{-1}(r)$ is a 1-manifold with boundary.



                Now suppose that you take $x in f^{-1}(r) cap text{interior}(M)$. The derivative map $D_x f : T_x M to T_r M$ has rank $1$, so its kernel is a 1-dimensional subspace of $M$. Applying the implicit function theorem, and using that $x$ is in the interior of $M$, it follows that $f^{-1}(r)$ intersected with a small neighborhood of $x$ is a 1-manifold without boundary containing $r$. Thus, $x$ cannot be a boundary point of the 1-manifold $f^{-1}(r)$, therefore all of the boundary points of $f^{-1}(r)$ are contained in $partial M$.



                One can make a still stronger conclusion, namely that $f^{-1}(r)$ is a "properly embedded" submanifold of $M$, which means that each point on $partial M cap f^{-1}(r)$ has a neighborhood in which the intersection of that neighborhood with $f^{-1}(r)$ looks like the intersection of the half-disc ${(x,y) mid x^2 + y^2 < 0, |y| ge 1}$ with the line $x=0$.






                share|cite|improve this answer









                $endgroup$



                The conclusion of the Sard's Theorem argument is a bit stronger than what you say.



                The real conclusion is that the map $f : M to partial M$ has a regular value $r in partial M$.



                It follows, first, that $f^{-1}(r)$ is a 1-manifold with boundary.



                Now suppose that you take $x in f^{-1}(r) cap text{interior}(M)$. The derivative map $D_x f : T_x M to T_r M$ has rank $1$, so its kernel is a 1-dimensional subspace of $M$. Applying the implicit function theorem, and using that $x$ is in the interior of $M$, it follows that $f^{-1}(r)$ intersected with a small neighborhood of $x$ is a 1-manifold without boundary containing $r$. Thus, $x$ cannot be a boundary point of the 1-manifold $f^{-1}(r)$, therefore all of the boundary points of $f^{-1}(r)$ are contained in $partial M$.



                One can make a still stronger conclusion, namely that $f^{-1}(r)$ is a "properly embedded" submanifold of $M$, which means that each point on $partial M cap f^{-1}(r)$ has a neighborhood in which the intersection of that neighborhood with $f^{-1}(r)$ looks like the intersection of the half-disc ${(x,y) mid x^2 + y^2 < 0, |y| ge 1}$ with the line $x=0$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 59 mins ago









                Lee MosherLee Mosher

                48.8k33682




                48.8k33682






























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