representing covering space by permutations

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When I read the section of representing covering space by permutations of Hatcher's Algebraic Topology,he points that:




n-sheeted covering spaces of $X$(path-connected,locallypath-connected,semilocally simply connected) are classified by equivalence classes(cojugate class) of homomorphism $pi_1(X,x_0) to Sigma_n$,where $Sigma_n$ is n-symmetric group.




Then does it imply that given a homomorphism $pi_1(X,x_0) to Sigma_n$,there is corresponding covering space?If so,how to deal with it?I can't find it in the book.










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    3












    $begingroup$


    When I read the section of representing covering space by permutations of Hatcher's Algebraic Topology,he points that:




    n-sheeted covering spaces of $X$(path-connected,locallypath-connected,semilocally simply connected) are classified by equivalence classes(cojugate class) of homomorphism $pi_1(X,x_0) to Sigma_n$,where $Sigma_n$ is n-symmetric group.




    Then does it imply that given a homomorphism $pi_1(X,x_0) to Sigma_n$,there is corresponding covering space?If so,how to deal with it?I can't find it in the book.










    share|cite|improve this question









    $endgroup$














      3












      3








      3





      $begingroup$


      When I read the section of representing covering space by permutations of Hatcher's Algebraic Topology,he points that:




      n-sheeted covering spaces of $X$(path-connected,locallypath-connected,semilocally simply connected) are classified by equivalence classes(cojugate class) of homomorphism $pi_1(X,x_0) to Sigma_n$,where $Sigma_n$ is n-symmetric group.




      Then does it imply that given a homomorphism $pi_1(X,x_0) to Sigma_n$,there is corresponding covering space?If so,how to deal with it?I can't find it in the book.










      share|cite|improve this question









      $endgroup$




      When I read the section of representing covering space by permutations of Hatcher's Algebraic Topology,he points that:




      n-sheeted covering spaces of $X$(path-connected,locallypath-connected,semilocally simply connected) are classified by equivalence classes(cojugate class) of homomorphism $pi_1(X,x_0) to Sigma_n$,where $Sigma_n$ is n-symmetric group.




      Then does it imply that given a homomorphism $pi_1(X,x_0) to Sigma_n$,there is corresponding covering space?If so,how to deal with it?I can't find it in the book.







      algebraic-topology






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      asked Jan 27 at 15:57









      Daniel XuDaniel Xu

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          2 Answers
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          $begingroup$

          Consider a representation $h:pi_1(M)rightarrow Sigma_n$ and $U_n=1,...,n$ the following $n$-cover can be associated: the quotient of $hat Xtimes U_n$ by the diagonal action of $pi_1(M)$ where $hat X$ is the universal cover of $X$.



          Conversely given a $n$-cover $p:Nrightarrow M$ you can associate to it its holonomy obtained by the action of $pi_1(M)$ on the fibre of any element of $M$, these actions are conjugate.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            It's worth pointing out that this is an instance of the "Borel construction": given a group $G$ and two $G$-spaces $X$, $Y$, their Borel construction is $Xtimes Y/sim$ where $(x,y)sim(xg, gy)$ (often people require $X$ to be a right $G$-space and $Y$ a left $G$-space for notation's sake). A typical situation is when $X$ is a principal $G$-bundle and $Y$ is an arbitrary $G$-space, and the effect of the construction is "changing the fibres" of $X$ from $G$ to $Y$. In our case the principal $pi_1M$ bundle is $hatX$ and the new fibres are $Y = U_n$.
            $endgroup$
            – William
            Jan 27 at 17:23



















          1












          $begingroup$

          One (maybe roundabout) way to see this is with classifying spaces and homotopy theory, at least when $X$ is a CW-space. (You will encounter all of these ideas in Hatcher at some point if you have not already.) Tsemo Aristide's answer is certainly cleaner and works for a broader class of spaces, but the concepts here might help in other situations.



          "Intuitively" the idea is that since $BSigma_n$ only has non-vanishing homotopy in degree 1, homotopy classes of maps into it only depend on homotopical information up to degree 2, and in particular homotopy classes of maps to $BSigma_n$ are the same for spaces that have the same $2$-skeleton. Then you have to know that $X$ and $Bpi_1X$ have the same $2$-skeleton, and about how group homomorphisms correspond to maps between classifying spaces. I will elaborate.



          An $n$-sheeted covering is the same thing as a fibre bundle whose fibre is a set of cardinality $n$ and whose structure group is $Sigma_n$; therefore they are classified by homotopy classes of maps $[X, BSigma_n]$ where $Bcolon Grp to Top$ is a classifying space functor. Since $Sigma_n$ is a discrete group it follows that $BSigma_n sim K(Sigma_n, 1)$, the "Eilenberge-Maclane space" defined up to homotopy equivalence by the properties $pi_1K(Sigma_n, 1)cong Sigma_n$ and $pi_iK(Sigma_n, 1)=0$ for other values of $i$. Since the higher homotopy groups of $BSigma_n$ all vanish, a result from obstruction theory is that



          $[X,BSigma_n] cong [X^(2), BSigma_n]$



          where $X^(2)$ is the $2$-skeleton of $X$. That is, the (isomorphism class of the) covering space space over $X$ is determined by its restriction to the $2$-skeleton.



          Here's where things get a bit funny: the $2$-skeleton of $X$ is also the $2$-skeleton of a model of $Bpi_1 X$. This is because since $pi_1X$ is discrete its classifying space is again an Eilenberg-Maclane space in degree 1, so we can construct a CW model from a group presentation of $pi_1X$ by taking a 1-cell for every generator and attaching 2-cell along every relation, and then adding higher-dimensional cells to kill off any higher homotopy we may have introduced. But the $2$-skeleton of $X$ determines a presentation of $pi_1X$ so it is also the $2$-skeleton of the Eilenberg-Maclane construction. Therefore we get



          $[X,BSigma_n] cong [X^(2), BSigma_n] = [(Bpi_1X)^(2), BSigma_n]cong [Bpi_1 X, BSigma_n]$



          Now the last step is to establish the correspondence between $[BG, BH]$ and conjugacy classes of homomorphisms $Gto H$. I will see if I can remember a clean way of showing this and make an edit later... Again, I believe it is also in Hatcher.






          share|cite|improve this answer









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            $begingroup$

            Consider a representation $h:pi_1(M)rightarrow Sigma_n$ and $U_n=1,...,n$ the following $n$-cover can be associated: the quotient of $hat Xtimes U_n$ by the diagonal action of $pi_1(M)$ where $hat X$ is the universal cover of $X$.



            Conversely given a $n$-cover $p:Nrightarrow M$ you can associate to it its holonomy obtained by the action of $pi_1(M)$ on the fibre of any element of $M$, these actions are conjugate.






            share|cite|improve this answer









            $endgroup$












            • $begingroup$
              It's worth pointing out that this is an instance of the "Borel construction": given a group $G$ and two $G$-spaces $X$, $Y$, their Borel construction is $Xtimes Y/sim$ where $(x,y)sim(xg, gy)$ (often people require $X$ to be a right $G$-space and $Y$ a left $G$-space for notation's sake). A typical situation is when $X$ is a principal $G$-bundle and $Y$ is an arbitrary $G$-space, and the effect of the construction is "changing the fibres" of $X$ from $G$ to $Y$. In our case the principal $pi_1M$ bundle is $hatX$ and the new fibres are $Y = U_n$.
              $endgroup$
              – William
              Jan 27 at 17:23
















            4












            $begingroup$

            Consider a representation $h:pi_1(M)rightarrow Sigma_n$ and $U_n=1,...,n$ the following $n$-cover can be associated: the quotient of $hat Xtimes U_n$ by the diagonal action of $pi_1(M)$ where $hat X$ is the universal cover of $X$.



            Conversely given a $n$-cover $p:Nrightarrow M$ you can associate to it its holonomy obtained by the action of $pi_1(M)$ on the fibre of any element of $M$, these actions are conjugate.






            share|cite|improve this answer









            $endgroup$












            • $begingroup$
              It's worth pointing out that this is an instance of the "Borel construction": given a group $G$ and two $G$-spaces $X$, $Y$, their Borel construction is $Xtimes Y/sim$ where $(x,y)sim(xg, gy)$ (often people require $X$ to be a right $G$-space and $Y$ a left $G$-space for notation's sake). A typical situation is when $X$ is a principal $G$-bundle and $Y$ is an arbitrary $G$-space, and the effect of the construction is "changing the fibres" of $X$ from $G$ to $Y$. In our case the principal $pi_1M$ bundle is $hatX$ and the new fibres are $Y = U_n$.
              $endgroup$
              – William
              Jan 27 at 17:23














            4












            4








            4





            $begingroup$

            Consider a representation $h:pi_1(M)rightarrow Sigma_n$ and $U_n=1,...,n$ the following $n$-cover can be associated: the quotient of $hat Xtimes U_n$ by the diagonal action of $pi_1(M)$ where $hat X$ is the universal cover of $X$.



            Conversely given a $n$-cover $p:Nrightarrow M$ you can associate to it its holonomy obtained by the action of $pi_1(M)$ on the fibre of any element of $M$, these actions are conjugate.






            share|cite|improve this answer









            $endgroup$



            Consider a representation $h:pi_1(M)rightarrow Sigma_n$ and $U_n=1,...,n$ the following $n$-cover can be associated: the quotient of $hat Xtimes U_n$ by the diagonal action of $pi_1(M)$ where $hat X$ is the universal cover of $X$.



            Conversely given a $n$-cover $p:Nrightarrow M$ you can associate to it its holonomy obtained by the action of $pi_1(M)$ on the fibre of any element of $M$, these actions are conjugate.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 27 at 16:34









            Tsemo AristideTsemo Aristide

            58.2k11445




            58.2k11445











            • $begingroup$
              It's worth pointing out that this is an instance of the "Borel construction": given a group $G$ and two $G$-spaces $X$, $Y$, their Borel construction is $Xtimes Y/sim$ where $(x,y)sim(xg, gy)$ (often people require $X$ to be a right $G$-space and $Y$ a left $G$-space for notation's sake). A typical situation is when $X$ is a principal $G$-bundle and $Y$ is an arbitrary $G$-space, and the effect of the construction is "changing the fibres" of $X$ from $G$ to $Y$. In our case the principal $pi_1M$ bundle is $hatX$ and the new fibres are $Y = U_n$.
              $endgroup$
              – William
              Jan 27 at 17:23

















            • $begingroup$
              It's worth pointing out that this is an instance of the "Borel construction": given a group $G$ and two $G$-spaces $X$, $Y$, their Borel construction is $Xtimes Y/sim$ where $(x,y)sim(xg, gy)$ (often people require $X$ to be a right $G$-space and $Y$ a left $G$-space for notation's sake). A typical situation is when $X$ is a principal $G$-bundle and $Y$ is an arbitrary $G$-space, and the effect of the construction is "changing the fibres" of $X$ from $G$ to $Y$. In our case the principal $pi_1M$ bundle is $hatX$ and the new fibres are $Y = U_n$.
              $endgroup$
              – William
              Jan 27 at 17:23
















            $begingroup$
            It's worth pointing out that this is an instance of the "Borel construction": given a group $G$ and two $G$-spaces $X$, $Y$, their Borel construction is $Xtimes Y/sim$ where $(x,y)sim(xg, gy)$ (often people require $X$ to be a right $G$-space and $Y$ a left $G$-space for notation's sake). A typical situation is when $X$ is a principal $G$-bundle and $Y$ is an arbitrary $G$-space, and the effect of the construction is "changing the fibres" of $X$ from $G$ to $Y$. In our case the principal $pi_1M$ bundle is $hatX$ and the new fibres are $Y = U_n$.
            $endgroup$
            – William
            Jan 27 at 17:23





            $begingroup$
            It's worth pointing out that this is an instance of the "Borel construction": given a group $G$ and two $G$-spaces $X$, $Y$, their Borel construction is $Xtimes Y/sim$ where $(x,y)sim(xg, gy)$ (often people require $X$ to be a right $G$-space and $Y$ a left $G$-space for notation's sake). A typical situation is when $X$ is a principal $G$-bundle and $Y$ is an arbitrary $G$-space, and the effect of the construction is "changing the fibres" of $X$ from $G$ to $Y$. In our case the principal $pi_1M$ bundle is $hatX$ and the new fibres are $Y = U_n$.
            $endgroup$
            – William
            Jan 27 at 17:23












            1












            $begingroup$

            One (maybe roundabout) way to see this is with classifying spaces and homotopy theory, at least when $X$ is a CW-space. (You will encounter all of these ideas in Hatcher at some point if you have not already.) Tsemo Aristide's answer is certainly cleaner and works for a broader class of spaces, but the concepts here might help in other situations.



            "Intuitively" the idea is that since $BSigma_n$ only has non-vanishing homotopy in degree 1, homotopy classes of maps into it only depend on homotopical information up to degree 2, and in particular homotopy classes of maps to $BSigma_n$ are the same for spaces that have the same $2$-skeleton. Then you have to know that $X$ and $Bpi_1X$ have the same $2$-skeleton, and about how group homomorphisms correspond to maps between classifying spaces. I will elaborate.



            An $n$-sheeted covering is the same thing as a fibre bundle whose fibre is a set of cardinality $n$ and whose structure group is $Sigma_n$; therefore they are classified by homotopy classes of maps $[X, BSigma_n]$ where $Bcolon Grp to Top$ is a classifying space functor. Since $Sigma_n$ is a discrete group it follows that $BSigma_n sim K(Sigma_n, 1)$, the "Eilenberge-Maclane space" defined up to homotopy equivalence by the properties $pi_1K(Sigma_n, 1)cong Sigma_n$ and $pi_iK(Sigma_n, 1)=0$ for other values of $i$. Since the higher homotopy groups of $BSigma_n$ all vanish, a result from obstruction theory is that



            $[X,BSigma_n] cong [X^(2), BSigma_n]$



            where $X^(2)$ is the $2$-skeleton of $X$. That is, the (isomorphism class of the) covering space space over $X$ is determined by its restriction to the $2$-skeleton.



            Here's where things get a bit funny: the $2$-skeleton of $X$ is also the $2$-skeleton of a model of $Bpi_1 X$. This is because since $pi_1X$ is discrete its classifying space is again an Eilenberg-Maclane space in degree 1, so we can construct a CW model from a group presentation of $pi_1X$ by taking a 1-cell for every generator and attaching 2-cell along every relation, and then adding higher-dimensional cells to kill off any higher homotopy we may have introduced. But the $2$-skeleton of $X$ determines a presentation of $pi_1X$ so it is also the $2$-skeleton of the Eilenberg-Maclane construction. Therefore we get



            $[X,BSigma_n] cong [X^(2), BSigma_n] = [(Bpi_1X)^(2), BSigma_n]cong [Bpi_1 X, BSigma_n]$



            Now the last step is to establish the correspondence between $[BG, BH]$ and conjugacy classes of homomorphisms $Gto H$. I will see if I can remember a clean way of showing this and make an edit later... Again, I believe it is also in Hatcher.






            share|cite|improve this answer









            $endgroup$

















              1












              $begingroup$

              One (maybe roundabout) way to see this is with classifying spaces and homotopy theory, at least when $X$ is a CW-space. (You will encounter all of these ideas in Hatcher at some point if you have not already.) Tsemo Aristide's answer is certainly cleaner and works for a broader class of spaces, but the concepts here might help in other situations.



              "Intuitively" the idea is that since $BSigma_n$ only has non-vanishing homotopy in degree 1, homotopy classes of maps into it only depend on homotopical information up to degree 2, and in particular homotopy classes of maps to $BSigma_n$ are the same for spaces that have the same $2$-skeleton. Then you have to know that $X$ and $Bpi_1X$ have the same $2$-skeleton, and about how group homomorphisms correspond to maps between classifying spaces. I will elaborate.



              An $n$-sheeted covering is the same thing as a fibre bundle whose fibre is a set of cardinality $n$ and whose structure group is $Sigma_n$; therefore they are classified by homotopy classes of maps $[X, BSigma_n]$ where $Bcolon Grp to Top$ is a classifying space functor. Since $Sigma_n$ is a discrete group it follows that $BSigma_n sim K(Sigma_n, 1)$, the "Eilenberge-Maclane space" defined up to homotopy equivalence by the properties $pi_1K(Sigma_n, 1)cong Sigma_n$ and $pi_iK(Sigma_n, 1)=0$ for other values of $i$. Since the higher homotopy groups of $BSigma_n$ all vanish, a result from obstruction theory is that



              $[X,BSigma_n] cong [X^(2), BSigma_n]$



              where $X^(2)$ is the $2$-skeleton of $X$. That is, the (isomorphism class of the) covering space space over $X$ is determined by its restriction to the $2$-skeleton.



              Here's where things get a bit funny: the $2$-skeleton of $X$ is also the $2$-skeleton of a model of $Bpi_1 X$. This is because since $pi_1X$ is discrete its classifying space is again an Eilenberg-Maclane space in degree 1, so we can construct a CW model from a group presentation of $pi_1X$ by taking a 1-cell for every generator and attaching 2-cell along every relation, and then adding higher-dimensional cells to kill off any higher homotopy we may have introduced. But the $2$-skeleton of $X$ determines a presentation of $pi_1X$ so it is also the $2$-skeleton of the Eilenberg-Maclane construction. Therefore we get



              $[X,BSigma_n] cong [X^(2), BSigma_n] = [(Bpi_1X)^(2), BSigma_n]cong [Bpi_1 X, BSigma_n]$



              Now the last step is to establish the correspondence between $[BG, BH]$ and conjugacy classes of homomorphisms $Gto H$. I will see if I can remember a clean way of showing this and make an edit later... Again, I believe it is also in Hatcher.






              share|cite|improve this answer









              $endgroup$















                1












                1








                1





                $begingroup$

                One (maybe roundabout) way to see this is with classifying spaces and homotopy theory, at least when $X$ is a CW-space. (You will encounter all of these ideas in Hatcher at some point if you have not already.) Tsemo Aristide's answer is certainly cleaner and works for a broader class of spaces, but the concepts here might help in other situations.



                "Intuitively" the idea is that since $BSigma_n$ only has non-vanishing homotopy in degree 1, homotopy classes of maps into it only depend on homotopical information up to degree 2, and in particular homotopy classes of maps to $BSigma_n$ are the same for spaces that have the same $2$-skeleton. Then you have to know that $X$ and $Bpi_1X$ have the same $2$-skeleton, and about how group homomorphisms correspond to maps between classifying spaces. I will elaborate.



                An $n$-sheeted covering is the same thing as a fibre bundle whose fibre is a set of cardinality $n$ and whose structure group is $Sigma_n$; therefore they are classified by homotopy classes of maps $[X, BSigma_n]$ where $Bcolon Grp to Top$ is a classifying space functor. Since $Sigma_n$ is a discrete group it follows that $BSigma_n sim K(Sigma_n, 1)$, the "Eilenberge-Maclane space" defined up to homotopy equivalence by the properties $pi_1K(Sigma_n, 1)cong Sigma_n$ and $pi_iK(Sigma_n, 1)=0$ for other values of $i$. Since the higher homotopy groups of $BSigma_n$ all vanish, a result from obstruction theory is that



                $[X,BSigma_n] cong [X^(2), BSigma_n]$



                where $X^(2)$ is the $2$-skeleton of $X$. That is, the (isomorphism class of the) covering space space over $X$ is determined by its restriction to the $2$-skeleton.



                Here's where things get a bit funny: the $2$-skeleton of $X$ is also the $2$-skeleton of a model of $Bpi_1 X$. This is because since $pi_1X$ is discrete its classifying space is again an Eilenberg-Maclane space in degree 1, so we can construct a CW model from a group presentation of $pi_1X$ by taking a 1-cell for every generator and attaching 2-cell along every relation, and then adding higher-dimensional cells to kill off any higher homotopy we may have introduced. But the $2$-skeleton of $X$ determines a presentation of $pi_1X$ so it is also the $2$-skeleton of the Eilenberg-Maclane construction. Therefore we get



                $[X,BSigma_n] cong [X^(2), BSigma_n] = [(Bpi_1X)^(2), BSigma_n]cong [Bpi_1 X, BSigma_n]$



                Now the last step is to establish the correspondence between $[BG, BH]$ and conjugacy classes of homomorphisms $Gto H$. I will see if I can remember a clean way of showing this and make an edit later... Again, I believe it is also in Hatcher.






                share|cite|improve this answer









                $endgroup$



                One (maybe roundabout) way to see this is with classifying spaces and homotopy theory, at least when $X$ is a CW-space. (You will encounter all of these ideas in Hatcher at some point if you have not already.) Tsemo Aristide's answer is certainly cleaner and works for a broader class of spaces, but the concepts here might help in other situations.



                "Intuitively" the idea is that since $BSigma_n$ only has non-vanishing homotopy in degree 1, homotopy classes of maps into it only depend on homotopical information up to degree 2, and in particular homotopy classes of maps to $BSigma_n$ are the same for spaces that have the same $2$-skeleton. Then you have to know that $X$ and $Bpi_1X$ have the same $2$-skeleton, and about how group homomorphisms correspond to maps between classifying spaces. I will elaborate.



                An $n$-sheeted covering is the same thing as a fibre bundle whose fibre is a set of cardinality $n$ and whose structure group is $Sigma_n$; therefore they are classified by homotopy classes of maps $[X, BSigma_n]$ where $Bcolon Grp to Top$ is a classifying space functor. Since $Sigma_n$ is a discrete group it follows that $BSigma_n sim K(Sigma_n, 1)$, the "Eilenberge-Maclane space" defined up to homotopy equivalence by the properties $pi_1K(Sigma_n, 1)cong Sigma_n$ and $pi_iK(Sigma_n, 1)=0$ for other values of $i$. Since the higher homotopy groups of $BSigma_n$ all vanish, a result from obstruction theory is that



                $[X,BSigma_n] cong [X^(2), BSigma_n]$



                where $X^(2)$ is the $2$-skeleton of $X$. That is, the (isomorphism class of the) covering space space over $X$ is determined by its restriction to the $2$-skeleton.



                Here's where things get a bit funny: the $2$-skeleton of $X$ is also the $2$-skeleton of a model of $Bpi_1 X$. This is because since $pi_1X$ is discrete its classifying space is again an Eilenberg-Maclane space in degree 1, so we can construct a CW model from a group presentation of $pi_1X$ by taking a 1-cell for every generator and attaching 2-cell along every relation, and then adding higher-dimensional cells to kill off any higher homotopy we may have introduced. But the $2$-skeleton of $X$ determines a presentation of $pi_1X$ so it is also the $2$-skeleton of the Eilenberg-Maclane construction. Therefore we get



                $[X,BSigma_n] cong [X^(2), BSigma_n] = [(Bpi_1X)^(2), BSigma_n]cong [Bpi_1 X, BSigma_n]$



                Now the last step is to establish the correspondence between $[BG, BH]$ and conjugacy classes of homomorphisms $Gto H$. I will see if I can remember a clean way of showing this and make an edit later... Again, I believe it is also in Hatcher.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 27 at 17:14









                WilliamWilliam

                2,0851121




                2,0851121



























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