Whether the induced map in de Rham cohomology is injective












2












$begingroup$


Let $M, N$ be smooth manifolds, and let $f: M rightarrow N$ be a surjective submersion, i.e. a surjective smooth map such that the differential $f_{*}$ is also surjective.



I have shown that for all $k geq 0$, the pullback map of $k$-forms $$ f^{*} : Omega^{k}(N) rightarrow Omega^{k} (M)$$ is injective.



However, the problem now asks me whether the induced map on de Rham cohomology $$ H_{dR}^k (N) rightarrow H_{dR}^k (M) : [omega] mapsto [f^{*} omega] $$ is also injective?



I was trying to prove this. I took $[omega] in H_{dR}^{k} (N)$ and assumed $[f^{*} omega] = [0]$. This means $f^{*} omega sim 0$ or $$ f^{*} omega = d tau $$ for some $(k-1)$ form $tau$ on $M$.



I want to conclude from this somehow that $[omega ] = [0]$ or $omega = d sigma$ for some $(k-1)$ form $sigma$ on $N$. But I'm not sure if the statement is even true. I tried to find a counter example, but couldn't.



Any help is appreciated!






differential-geometry homology-cohomology differential-forms







share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    Let $M, N$ be smooth manifolds, and let $f: M rightarrow N$ be a surjective submersion, i.e. a surjective smooth map such that the differential $f_{*}$ is also surjective.



    I have shown that for all $k geq 0$, the pullback map of $k$-forms $$ f^{*} : Omega^{k}(N) rightarrow Omega^{k} (M)$$ is injective.



    However, the problem now asks me whether the induced map on de Rham cohomology $$ H_{dR}^k (N) rightarrow H_{dR}^k (M) : [omega] mapsto [f^{*} omega] $$ is also injective?



    I was trying to prove this. I took $[omega] in H_{dR}^{k} (N)$ and assumed $[f^{*} omega] = [0]$. This means $f^{*} omega sim 0$ or $$ f^{*} omega = d tau $$ for some $(k-1)$ form $tau$ on $M$.



    I want to conclude from this somehow that $[omega ] = [0]$ or $omega = d sigma$ for some $(k-1)$ form $sigma$ on $N$. But I'm not sure if the statement is even true. I tried to find a counter example, but couldn't.



    Any help is appreciated!






    differential-geometry homology-cohomology differential-forms







    share|cite|improve this question









    $endgroup$















      2












      2








      2


      1



      $begingroup$


      Let $M, N$ be smooth manifolds, and let $f: M rightarrow N$ be a surjective submersion, i.e. a surjective smooth map such that the differential $f_{*}$ is also surjective.



      I have shown that for all $k geq 0$, the pullback map of $k$-forms $$ f^{*} : Omega^{k}(N) rightarrow Omega^{k} (M)$$ is injective.



      However, the problem now asks me whether the induced map on de Rham cohomology $$ H_{dR}^k (N) rightarrow H_{dR}^k (M) : [omega] mapsto [f^{*} omega] $$ is also injective?



      I was trying to prove this. I took $[omega] in H_{dR}^{k} (N)$ and assumed $[f^{*} omega] = [0]$. This means $f^{*} omega sim 0$ or $$ f^{*} omega = d tau $$ for some $(k-1)$ form $tau$ on $M$.



      I want to conclude from this somehow that $[omega ] = [0]$ or $omega = d sigma$ for some $(k-1)$ form $sigma$ on $N$. But I'm not sure if the statement is even true. I tried to find a counter example, but couldn't.



      Any help is appreciated!






      differential-geometry homology-cohomology differential-forms







      share|cite|improve this question









      $endgroup$




      Let $M, N$ be smooth manifolds, and let $f: M rightarrow N$ be a surjective submersion, i.e. a surjective smooth map such that the differential $f_{*}$ is also surjective.



      I have shown that for all $k geq 0$, the pullback map of $k$-forms $$ f^{*} : Omega^{k}(N) rightarrow Omega^{k} (M)$$ is injective.



      However, the problem now asks me whether the induced map on de Rham cohomology $$ H_{dR}^k (N) rightarrow H_{dR}^k (M) : [omega] mapsto [f^{*} omega] $$ is also injective?



      I was trying to prove this. I took $[omega] in H_{dR}^{k} (N)$ and assumed $[f^{*} omega] = [0]$. This means $f^{*} omega sim 0$ or $$ f^{*} omega = d tau $$ for some $(k-1)$ form $tau$ on $M$.



      I want to conclude from this somehow that $[omega ] = [0]$ or $omega = d sigma$ for some $(k-1)$ form $sigma$ on $N$. But I'm not sure if the statement is even true. I tried to find a counter example, but couldn't.



      Any help is appreciated!






      differential-geometry homology-cohomology differential-forms




      differential-geometry homology-cohomology differential-forms






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 2 at 14:51









      KamilKamil

      2,05221649




      2,05221649






















          1 Answer
          1






          active

          oldest

          votes


















          4












          $begingroup$

          Consider the convering of the torus $T^2$ by $mathbb{R}^2$, $H^2_{DR}(T^2)neq 0$ and $H_{DR}^2(mathbb{R}^2)=0$ since it is contractible so the induced map on cohomology is not injective.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks. Do you take $N = T^{2}$ in this case? How do you see the map $f: mathbb{R}^2 rightarrow T^2$ is a submersion?
            $endgroup$
            – Kamil
            Jan 3 at 10:15






          • 1




            $begingroup$
            @Kamil: (Smooth) covering maps are always local diffeomorphisms, hence, in particular, submersions.
            $endgroup$
            – Ted Shifrin
            Jan 3 at 19:31












          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "69"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3059555%2fwhether-the-induced-map-in-de-rham-cohomology-is-injective%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          4












          $begingroup$

          Consider the convering of the torus $T^2$ by $mathbb{R}^2$, $H^2_{DR}(T^2)neq 0$ and $H_{DR}^2(mathbb{R}^2)=0$ since it is contractible so the induced map on cohomology is not injective.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks. Do you take $N = T^{2}$ in this case? How do you see the map $f: mathbb{R}^2 rightarrow T^2$ is a submersion?
            $endgroup$
            – Kamil
            Jan 3 at 10:15






          • 1




            $begingroup$
            @Kamil: (Smooth) covering maps are always local diffeomorphisms, hence, in particular, submersions.
            $endgroup$
            – Ted Shifrin
            Jan 3 at 19:31
















          4












          $begingroup$

          Consider the convering of the torus $T^2$ by $mathbb{R}^2$, $H^2_{DR}(T^2)neq 0$ and $H_{DR}^2(mathbb{R}^2)=0$ since it is contractible so the induced map on cohomology is not injective.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks. Do you take $N = T^{2}$ in this case? How do you see the map $f: mathbb{R}^2 rightarrow T^2$ is a submersion?
            $endgroup$
            – Kamil
            Jan 3 at 10:15






          • 1




            $begingroup$
            @Kamil: (Smooth) covering maps are always local diffeomorphisms, hence, in particular, submersions.
            $endgroup$
            – Ted Shifrin
            Jan 3 at 19:31














          4












          4








          4





          $begingroup$

          Consider the convering of the torus $T^2$ by $mathbb{R}^2$, $H^2_{DR}(T^2)neq 0$ and $H_{DR}^2(mathbb{R}^2)=0$ since it is contractible so the induced map on cohomology is not injective.






          share|cite|improve this answer









          $endgroup$



          Consider the convering of the torus $T^2$ by $mathbb{R}^2$, $H^2_{DR}(T^2)neq 0$ and $H_{DR}^2(mathbb{R}^2)=0$ since it is contractible so the induced map on cohomology is not injective.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 2 at 15:05









          Tsemo AristideTsemo Aristide

          60k11446




          60k11446












          • $begingroup$
            Thanks. Do you take $N = T^{2}$ in this case? How do you see the map $f: mathbb{R}^2 rightarrow T^2$ is a submersion?
            $endgroup$
            – Kamil
            Jan 3 at 10:15






          • 1




            $begingroup$
            @Kamil: (Smooth) covering maps are always local diffeomorphisms, hence, in particular, submersions.
            $endgroup$
            – Ted Shifrin
            Jan 3 at 19:31


















          • $begingroup$
            Thanks. Do you take $N = T^{2}$ in this case? How do you see the map $f: mathbb{R}^2 rightarrow T^2$ is a submersion?
            $endgroup$
            – Kamil
            Jan 3 at 10:15






          • 1




            $begingroup$
            @Kamil: (Smooth) covering maps are always local diffeomorphisms, hence, in particular, submersions.
            $endgroup$
            – Ted Shifrin
            Jan 3 at 19:31
















          $begingroup$
          Thanks. Do you take $N = T^{2}$ in this case? How do you see the map $f: mathbb{R}^2 rightarrow T^2$ is a submersion?
          $endgroup$
          – Kamil
          Jan 3 at 10:15




          $begingroup$
          Thanks. Do you take $N = T^{2}$ in this case? How do you see the map $f: mathbb{R}^2 rightarrow T^2$ is a submersion?
          $endgroup$
          – Kamil
          Jan 3 at 10:15




          1




          1




          $begingroup$
          @Kamil: (Smooth) covering maps are always local diffeomorphisms, hence, in particular, submersions.
          $endgroup$
          – Ted Shifrin
          Jan 3 at 19:31




          $begingroup$
          @Kamil: (Smooth) covering maps are always local diffeomorphisms, hence, in particular, submersions.
          $endgroup$
          – Ted Shifrin
          Jan 3 at 19:31


















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Mathematics Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid …



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3059555%2fwhether-the-induced-map-in-de-rham-cohomology-is-injective%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          -

          Popular posts from this blog

          Wiesbaden

          Image
          Wappen Deutschlandkarte 50.0825 8.24 117 Koordinaten: 50° 5′  N , 8° 14′  O Basisdaten Bundesland: Hessen Regierungsbezirk: Darmstadt Höhe: 117 m ü. NHN Fläche: 203,93 km 2 Einwohner: 278.654 (31. Dez. 2017) [1] Bevölkerungsdichte: 1366 Einwohner je km 2 Postleitzahlen: 65183–65207, 55246, 55252 Vorlage:Infobox Gemeinde in Deutschland/Wartung/PLZ enthält Text Vorwahlen: 0611, 06122, 06127, 06134 Kfz-Kennzeichen: WI Gemeindeschlüssel: 06 4 14 000 LOCODE: DE WIB NUTS: DE714 Stadtgliederung: 26 Ortsbezirke Adresse der Stadtverwaltung: Schlossplatz 6 65183 Wiesbaden Webpräsenz: www.wiesbaden.de Oberbürgermeister: Sven Gerich (SPD) Lage der Landeshauptstadt Wiesbaden in Hessen und im Regierungsbezirk Darmstadt Offizielles Logo der Landeshauptstadt Wiesbaden Flagge der Landeshauptstadt Wiesbaden Wiesbaden is
          Read more

          Marschland

          Image
          Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
          Read more

          Dieringhausen

          Image
          Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
          Read more