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







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    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











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      share|cite|improve this question










      asked Jan 2 at 14:51









      KamilKamil

      2,05221649




      2,05221649






















          1 Answer
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          $begingroup$

          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












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          1 Answer
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          1 Answer
          1






          active

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          active

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          active

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          $begingroup$

          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


















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