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Existence of an open set such that the map is submersive



The 2019 Stack Overflow Developer Survey Results Are InGiven a smooth map which is open, is it a submersion?Regular value: intuition about surjectivity conditionExistence of a open set between a compact and an open setThe map $f:Mrightarrow N$ is differentiable, where $M,N$ are $m$-manifolds.Proper smooth map has its regular values an open setA level set of a smooth map, in which the rank is constant, but is not an embedded submanifoldIf the intersection number of two submanifolds is zero, we can deform them so that they are disjoint.Smooth map on a “non-open” subsetIs every diffeomorphism an open map?$F:Mto N$ is surjective if $int_M F^* eta ne 0$ for some $eta in Omega^n(N)$










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Let $f:Mto N$ be a smooth surjective map between two compact manifolds. Is there always an open set $Usubset M$, s.t. on $U$, $f$ is submersive?










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    $begingroup$


    Let $f:Mto N$ be a smooth surjective map between two compact manifolds. Is there always an open set $Usubset M$, s.t. on $U$, $f$ is submersive?










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      $begingroup$


      Let $f:Mto N$ be a smooth surjective map between two compact manifolds. Is there always an open set $Usubset M$, s.t. on $U$, $f$ is submersive?










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      Let $f:Mto N$ be a smooth surjective map between two compact manifolds. Is there always an open set $Usubset M$, s.t. on $U$, $f$ is submersive?







      differential-geometry manifolds compact-manifolds






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      asked Mar 23 at 18:52









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          By Sard's theorem there exists a regular value. Pick a point in its preimage and use the fact that having full rank is an open property to conclude that you find a open set where the differential has full rank.






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            $begingroup$

            By Sard's theorem there exists a regular value. Pick a point in its preimage and use the fact that having full rank is an open property to conclude that you find a open set where the differential has full rank.






            share|cite|improve this answer









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              3












              $begingroup$

              By Sard's theorem there exists a regular value. Pick a point in its preimage and use the fact that having full rank is an open property to conclude that you find a open set where the differential has full rank.






              share|cite|improve this answer









              $endgroup$















                3












                3








                3





                $begingroup$

                By Sard's theorem there exists a regular value. Pick a point in its preimage and use the fact that having full rank is an open property to conclude that you find a open set where the differential has full rank.






                share|cite|improve this answer









                $endgroup$



                By Sard's theorem there exists a regular value. Pick a point in its preimage and use the fact that having full rank is an open property to conclude that you find a open set where the differential has full rank.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Mar 23 at 19:06









                Severin SchravenSeverin Schraven

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