Proving that hopf map from $S^3 to S^2 $ is not null homotopic












1














I want to prove that hopf map from $S^3 to S^2 $ is not null homotopic. Is there some elementary proof of this fact?






algebraic-topology homotopy-theory







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  • What facts do you know? It's hard to know what tools to use without a context. This result is quite overdetermined.
    – Justin Young
    Oct 9 '16 at 15:57










  • @Justin Young I am looking for a proof which only requires basic algebraic topology (by this I mean content of Hatcher's Algebraic topology book)
    – happymath
    Oct 11 '16 at 4:50










  • In that case, the answer below is best. The key point is that the cup product in $mathbb CP^2$ is non-trivial, as opposed to $S^2 vee S^4$.
    – Justin Young
    Oct 11 '16 at 14:05


















1














I want to prove that hopf map from $S^3 to S^2 $ is not null homotopic. Is there some elementary proof of this fact?






algebraic-topology homotopy-theory







share|cite|improve this question






















  • What facts do you know? It's hard to know what tools to use without a context. This result is quite overdetermined.
    – Justin Young
    Oct 9 '16 at 15:57










  • @Justin Young I am looking for a proof which only requires basic algebraic topology (by this I mean content of Hatcher's Algebraic topology book)
    – happymath
    Oct 11 '16 at 4:50










  • In that case, the answer below is best. The key point is that the cup product in $mathbb CP^2$ is non-trivial, as opposed to $S^2 vee S^4$.
    – Justin Young
    Oct 11 '16 at 14:05
















1












1








1


1





I want to prove that hopf map from $S^3 to S^2 $ is not null homotopic. Is there some elementary proof of this fact?






algebraic-topology homotopy-theory







share|cite|improve this question













I want to prove that hopf map from $S^3 to S^2 $ is not null homotopic. Is there some elementary proof of this fact?






algebraic-topology homotopy-theory




algebraic-topology homotopy-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Oct 1 '16 at 9:36









happymath

3,81011439




3,81011439












  • What facts do you know? It's hard to know what tools to use without a context. This result is quite overdetermined.
    – Justin Young
    Oct 9 '16 at 15:57










  • @Justin Young I am looking for a proof which only requires basic algebraic topology (by this I mean content of Hatcher's Algebraic topology book)
    – happymath
    Oct 11 '16 at 4:50










  • In that case, the answer below is best. The key point is that the cup product in $mathbb CP^2$ is non-trivial, as opposed to $S^2 vee S^4$.
    – Justin Young
    Oct 11 '16 at 14:05




















  • What facts do you know? It's hard to know what tools to use without a context. This result is quite overdetermined.
    – Justin Young
    Oct 9 '16 at 15:57










  • @Justin Young I am looking for a proof which only requires basic algebraic topology (by this I mean content of Hatcher's Algebraic topology book)
    – happymath
    Oct 11 '16 at 4:50










  • In that case, the answer below is best. The key point is that the cup product in $mathbb CP^2$ is non-trivial, as opposed to $S^2 vee S^4$.
    – Justin Young
    Oct 11 '16 at 14:05


















What facts do you know? It's hard to know what tools to use without a context. This result is quite overdetermined.
– Justin Young
Oct 9 '16 at 15:57




What facts do you know? It's hard to know what tools to use without a context. This result is quite overdetermined.
– Justin Young
Oct 9 '16 at 15:57












@Justin Young I am looking for a proof which only requires basic algebraic topology (by this I mean content of Hatcher's Algebraic topology book)
– happymath
Oct 11 '16 at 4:50




@Justin Young I am looking for a proof which only requires basic algebraic topology (by this I mean content of Hatcher's Algebraic topology book)
– happymath
Oct 11 '16 at 4:50












In that case, the answer below is best. The key point is that the cup product in $mathbb CP^2$ is non-trivial, as opposed to $S^2 vee S^4$.
– Justin Young
Oct 11 '16 at 14:05






In that case, the answer below is best. The key point is that the cup product in $mathbb CP^2$ is non-trivial, as opposed to $S^2 vee S^4$.
– Justin Young
Oct 11 '16 at 14:05












3 Answers
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4














If it were nullhomotopic, what do you know about the homotopy type of its mapping cone?
On the other hand, the Hopf map is the attaching map of the $4$-cell in $Bbb CP^2$, so its mapping cone is just $Bbb CP^2$.






share|cite|improve this answer





























    2














    Briefly: This follows since the Hopf map $pi:S^3to S^2$ is surjective and satisfies the homotopy lifting property: if $pi$ were nullhomotopic, we could use the homotopy lifting property to construct a homotopy of $text{id}_{S^3}$ to a non-surjective map, which is impossible.



    In more detail: The Hopf map $picolon S^3to S^2$ is a surjective submersion. Hence since $S^3$ is compact, Ehresmann's Lemma implies that $pi$ is a fiber bundle, and in particular a Hurewicz fibration. Therefore, $pi$ satisfies the homotopy lifting property.



    Assume (to obtain a contradiction) that there is a nullhomotopy $pi_t:S^3to S^2$ with $pi_0 = pi$ and $pi_{1}$ a constant map.
    By the homotopy lifting property, there exists a homotopy $h_tcolon S^3 to S^3$ satisfying $h_0 = text{id}_{S^3}$ and $picirc h_t = pi_t$ for all $t$. Since $pi$ is surjective and $pi_1$ is not, $pi circ h_1 = pi_1$ implies that $h_1$ is not surjective. Hence the Brouwer degree $text{deg}(h_1) = 0$. But by homotopy invariance of the degree, $text{deg}(h_1) = text{deg}(h_0) = text{deg}(text{id}_{S^3}) = 1$, so we have obtained a contradiction.






    share|cite|improve this answer





























      1














      The exercises in the last chapter of Milnor's book, Topology from the Differentiable Viewpoint, may provide an elementary proof by the use of the idea of Linking number. You prove in a sequence of exercises that the linking number is a homotopy invariant. The only part, honestly, that I have not completely written down in detail is proving that for the hopf fibration the linking number is non zero - this would prove that the hopf map is not homotopic to the constant map.






      share|cite|improve this answer





















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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        4














        If it were nullhomotopic, what do you know about the homotopy type of its mapping cone?
        On the other hand, the Hopf map is the attaching map of the $4$-cell in $Bbb CP^2$, so its mapping cone is just $Bbb CP^2$.






        share|cite|improve this answer


























          4














          If it were nullhomotopic, what do you know about the homotopy type of its mapping cone?
          On the other hand, the Hopf map is the attaching map of the $4$-cell in $Bbb CP^2$, so its mapping cone is just $Bbb CP^2$.






          share|cite|improve this answer
























            4












            4








            4






            If it were nullhomotopic, what do you know about the homotopy type of its mapping cone?
            On the other hand, the Hopf map is the attaching map of the $4$-cell in $Bbb CP^2$, so its mapping cone is just $Bbb CP^2$.






            share|cite|improve this answer












            If it were nullhomotopic, what do you know about the homotopy type of its mapping cone?
            On the other hand, the Hopf map is the attaching map of the $4$-cell in $Bbb CP^2$, so its mapping cone is just $Bbb CP^2$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Oct 1 '16 at 9:45









            iwriteonbananas

            2,96711532




            2,96711532























                2














                Briefly: This follows since the Hopf map $pi:S^3to S^2$ is surjective and satisfies the homotopy lifting property: if $pi$ were nullhomotopic, we could use the homotopy lifting property to construct a homotopy of $text{id}_{S^3}$ to a non-surjective map, which is impossible.



                In more detail: The Hopf map $picolon S^3to S^2$ is a surjective submersion. Hence since $S^3$ is compact, Ehresmann's Lemma implies that $pi$ is a fiber bundle, and in particular a Hurewicz fibration. Therefore, $pi$ satisfies the homotopy lifting property.



                Assume (to obtain a contradiction) that there is a nullhomotopy $pi_t:S^3to S^2$ with $pi_0 = pi$ and $pi_{1}$ a constant map.
                By the homotopy lifting property, there exists a homotopy $h_tcolon S^3 to S^3$ satisfying $h_0 = text{id}_{S^3}$ and $picirc h_t = pi_t$ for all $t$. Since $pi$ is surjective and $pi_1$ is not, $pi circ h_1 = pi_1$ implies that $h_1$ is not surjective. Hence the Brouwer degree $text{deg}(h_1) = 0$. But by homotopy invariance of the degree, $text{deg}(h_1) = text{deg}(h_0) = text{deg}(text{id}_{S^3}) = 1$, so we have obtained a contradiction.






                share|cite|improve this answer


























                  2














                  Briefly: This follows since the Hopf map $pi:S^3to S^2$ is surjective and satisfies the homotopy lifting property: if $pi$ were nullhomotopic, we could use the homotopy lifting property to construct a homotopy of $text{id}_{S^3}$ to a non-surjective map, which is impossible.



                  In more detail: The Hopf map $picolon S^3to S^2$ is a surjective submersion. Hence since $S^3$ is compact, Ehresmann's Lemma implies that $pi$ is a fiber bundle, and in particular a Hurewicz fibration. Therefore, $pi$ satisfies the homotopy lifting property.



                  Assume (to obtain a contradiction) that there is a nullhomotopy $pi_t:S^3to S^2$ with $pi_0 = pi$ and $pi_{1}$ a constant map.
                  By the homotopy lifting property, there exists a homotopy $h_tcolon S^3 to S^3$ satisfying $h_0 = text{id}_{S^3}$ and $picirc h_t = pi_t$ for all $t$. Since $pi$ is surjective and $pi_1$ is not, $pi circ h_1 = pi_1$ implies that $h_1$ is not surjective. Hence the Brouwer degree $text{deg}(h_1) = 0$. But by homotopy invariance of the degree, $text{deg}(h_1) = text{deg}(h_0) = text{deg}(text{id}_{S^3}) = 1$, so we have obtained a contradiction.






                  share|cite|improve this answer
























                    2












                    2








                    2






                    Briefly: This follows since the Hopf map $pi:S^3to S^2$ is surjective and satisfies the homotopy lifting property: if $pi$ were nullhomotopic, we could use the homotopy lifting property to construct a homotopy of $text{id}_{S^3}$ to a non-surjective map, which is impossible.



                    In more detail: The Hopf map $picolon S^3to S^2$ is a surjective submersion. Hence since $S^3$ is compact, Ehresmann's Lemma implies that $pi$ is a fiber bundle, and in particular a Hurewicz fibration. Therefore, $pi$ satisfies the homotopy lifting property.



                    Assume (to obtain a contradiction) that there is a nullhomotopy $pi_t:S^3to S^2$ with $pi_0 = pi$ and $pi_{1}$ a constant map.
                    By the homotopy lifting property, there exists a homotopy $h_tcolon S^3 to S^3$ satisfying $h_0 = text{id}_{S^3}$ and $picirc h_t = pi_t$ for all $t$. Since $pi$ is surjective and $pi_1$ is not, $pi circ h_1 = pi_1$ implies that $h_1$ is not surjective. Hence the Brouwer degree $text{deg}(h_1) = 0$. But by homotopy invariance of the degree, $text{deg}(h_1) = text{deg}(h_0) = text{deg}(text{id}_{S^3}) = 1$, so we have obtained a contradiction.






                    share|cite|improve this answer












                    Briefly: This follows since the Hopf map $pi:S^3to S^2$ is surjective and satisfies the homotopy lifting property: if $pi$ were nullhomotopic, we could use the homotopy lifting property to construct a homotopy of $text{id}_{S^3}$ to a non-surjective map, which is impossible.



                    In more detail: The Hopf map $picolon S^3to S^2$ is a surjective submersion. Hence since $S^3$ is compact, Ehresmann's Lemma implies that $pi$ is a fiber bundle, and in particular a Hurewicz fibration. Therefore, $pi$ satisfies the homotopy lifting property.



                    Assume (to obtain a contradiction) that there is a nullhomotopy $pi_t:S^3to S^2$ with $pi_0 = pi$ and $pi_{1}$ a constant map.
                    By the homotopy lifting property, there exists a homotopy $h_tcolon S^3 to S^3$ satisfying $h_0 = text{id}_{S^3}$ and $picirc h_t = pi_t$ for all $t$. Since $pi$ is surjective and $pi_1$ is not, $pi circ h_1 = pi_1$ implies that $h_1$ is not surjective. Hence the Brouwer degree $text{deg}(h_1) = 0$. But by homotopy invariance of the degree, $text{deg}(h_1) = text{deg}(h_0) = text{deg}(text{id}_{S^3}) = 1$, so we have obtained a contradiction.







                    share|cite|improve this answer












                    share|cite|improve this answer



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                    answered Nov 26 '18 at 23:31









                    Matthew Kvalheim

                    661416




                    661416























                        1














                        The exercises in the last chapter of Milnor's book, Topology from the Differentiable Viewpoint, may provide an elementary proof by the use of the idea of Linking number. You prove in a sequence of exercises that the linking number is a homotopy invariant. The only part, honestly, that I have not completely written down in detail is proving that for the hopf fibration the linking number is non zero - this would prove that the hopf map is not homotopic to the constant map.






                        share|cite|improve this answer


























                          1














                          The exercises in the last chapter of Milnor's book, Topology from the Differentiable Viewpoint, may provide an elementary proof by the use of the idea of Linking number. You prove in a sequence of exercises that the linking number is a homotopy invariant. The only part, honestly, that I have not completely written down in detail is proving that for the hopf fibration the linking number is non zero - this would prove that the hopf map is not homotopic to the constant map.






                          share|cite|improve this answer
























                            1












                            1








                            1






                            The exercises in the last chapter of Milnor's book, Topology from the Differentiable Viewpoint, may provide an elementary proof by the use of the idea of Linking number. You prove in a sequence of exercises that the linking number is a homotopy invariant. The only part, honestly, that I have not completely written down in detail is proving that for the hopf fibration the linking number is non zero - this would prove that the hopf map is not homotopic to the constant map.






                            share|cite|improve this answer












                            The exercises in the last chapter of Milnor's book, Topology from the Differentiable Viewpoint, may provide an elementary proof by the use of the idea of Linking number. You prove in a sequence of exercises that the linking number is a homotopy invariant. The only part, honestly, that I have not completely written down in detail is proving that for the hopf fibration the linking number is non zero - this would prove that the hopf map is not homotopic to the constant map.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Feb 11 '17 at 22:54









                            Behnam Esmayli

                            1,946515




                            1,946515






























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