partial converse of existence of covering spaces












5












$begingroup$


suppose $X$ and $Y$ are two spaces and let $Z$ be a covering spaces for both $X$ and $Y$, then is it true that there is some space $W$ such that $X$ and $Y$ is a covering space of $W$???
(sorry but I didn't get any answer... my intuition says that it might be correct or rather I need to put some more little conditions)
and sorry if somebody thinks that the title is not appropriate.






general-topology algebraic-topology







share|cite|improve this question











$endgroup$

















    5












    $begingroup$


    suppose $X$ and $Y$ are two spaces and let $Z$ be a covering spaces for both $X$ and $Y$, then is it true that there is some space $W$ such that $X$ and $Y$ is a covering space of $W$???
    (sorry but I didn't get any answer... my intuition says that it might be correct or rather I need to put some more little conditions)
    and sorry if somebody thinks that the title is not appropriate.






    general-topology algebraic-topology







    share|cite|improve this question











    $endgroup$















      5












      5








      5


      3



      $begingroup$


      suppose $X$ and $Y$ are two spaces and let $Z$ be a covering spaces for both $X$ and $Y$, then is it true that there is some space $W$ such that $X$ and $Y$ is a covering space of $W$???
      (sorry but I didn't get any answer... my intuition says that it might be correct or rather I need to put some more little conditions)
      and sorry if somebody thinks that the title is not appropriate.






      general-topology algebraic-topology







      share|cite|improve this question











      $endgroup$




      suppose $X$ and $Y$ are two spaces and let $Z$ be a covering spaces for both $X$ and $Y$, then is it true that there is some space $W$ such that $X$ and $Y$ is a covering space of $W$???
      (sorry but I didn't get any answer... my intuition says that it might be correct or rather I need to put some more little conditions)
      and sorry if somebody thinks that the title is not appropriate.






      general-topology algebraic-topology




      general-topology algebraic-topology






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Feb 1 '15 at 16:33









      Stefan Hamcke

      21.7k42879




      21.7k42879










      asked Feb 1 '15 at 14:57









      Anubhav MukherjeeAnubhav Mukherjee

      4,9861826




      4,9861826






















          2 Answers
          2






          active

          oldest

          votes


















          11












          $begingroup$

          Let $X$ and $Y$ be the two graphs having two vertices and three edges. There is a common two-sheeted covering space $Z$ of $X$ and $Y$ which is a graph with four vertices and six edges. Exercise: Find $Z$ and show that $X$ and $Y$ are not covering spaces of any other spaces (besides themselves, of course).



          Interesting side note: The covering spaces $Zto X$ and $Zto Y$ are defined by free actions of ${mathbb Z}_2$ on $Z$. These generate a ${mathbb Z}_2times{mathbb Z}_2$ action, but it is not a free action since the product of the two generators has a fixed point.






          share|cite|improve this answer









          $endgroup$









          • 2




            $begingroup$
            wonderful!! it is just too beautiful... Z is tow circles connecting their noth poles by an edge and south poles by an edge...and if W exists then it has to be wedge of 3 circles then both X and Y be two sheeted covering of W but any two sheeted covering of W contain 6 edges...so contradiction
            $endgroup$
            – Anubhav Mukherjee
            Feb 11 '15 at 23:20





















          6












          $begingroup$

          No, it is not true.



          The open 2-dimensional disc $Z=D^2$ is a covering space of the genus 1 closed surface $X=F_1 = S^1 times S^1$ and of the genus 2 closed surface $Y=F_2 = F_1 # F_1$ (that symbol $#$ means connected sum). One sees this best using geometry. The torus $F_1$ has a Euclidean structure and so there is a locally isometric universal covering map from the Euclidean plane $mathbb{R}^2 to F_1$. And the surface $F_2$ has a hyperbolic structure and so there is a locally isometric universal covering map from the hyperbolic plane $mathbb{H}^2 to F_2$. And, of course, each of $mathbb{R}^2$, $mathbb{H}^2$ is homeomorphic to $D^2$.



          But $F_1$ and $F_2$ cannot cover the same space $W$. For suppose they did, compactness of $F_1$ and $F_2$ would imply that the covering maps $F_1 mapsto W$ and $F_2 mapsto W$ are each of finite degree. The fundamental group $pi_1 W$ could therefore contain finite index subgroups $A_1,A_2$ isomorphic to $pi_1F_1$, $pi_1F_2$ respectively. Since $pi_1F_1 = mathbb{Z} oplus mathbb{Z}$ is abelian, $A_1 cap A_2$ is a finite index abelian subgroup of $A_2$. But $pi_1F_2$ has no finite index abelian subgroup, in fact its only abelian subgroups are trivial or infinite cyclic of infinite index.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            $pi_1(F_2) = <a,b,c,d/ aba^-1b^-1cdc^-1d^-1>$ but from here how do I conclude the last line that it has no finite index ableian subgroup ?? can you plese help me?
            $endgroup$
            – Anubhav Mukherjee
            Feb 1 '15 at 18:33










          • $begingroup$
            @AnubhaV: that is also best proved using geometry: every discrete abelian subgroup of the group of isometries of $mathbb{H}^2$ fixes some line in $mathbb{H}^2$, but the action of the deck transformation group on $mathbb{H}^2$ has no finite index subgroup fixing a line.
            $endgroup$
            – Lee Mosher
            Feb 1 '15 at 21:06










          • $begingroup$
            here if I restricted Z as a finite sheeted covering space...Now can you give me an counter examle...thanks for helping
            $endgroup$
            – Anubhav Mukherjee
            Feb 6 '15 at 14:07










          • $begingroup$
            @AnubhaV: That's a strong restriction which really makes it a different question. So you should post it as a different question (with a link to this one and an explanation of why you have added the restriction to finite sheeted covering spaces).
            $endgroup$
            – Lee Mosher
            Feb 6 '15 at 20:51










          • $begingroup$
            thanks...I am doing so now...and sir please help me to post properly the question...I dont know how to properly edit questions
            $endgroup$
            – Anubhav Mukherjee
            Feb 6 '15 at 21:56













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          2 Answers
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          2 Answers
          2






          active

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          active

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          active

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          11












          $begingroup$

          Let $X$ and $Y$ be the two graphs having two vertices and three edges. There is a common two-sheeted covering space $Z$ of $X$ and $Y$ which is a graph with four vertices and six edges. Exercise: Find $Z$ and show that $X$ and $Y$ are not covering spaces of any other spaces (besides themselves, of course).



          Interesting side note: The covering spaces $Zto X$ and $Zto Y$ are defined by free actions of ${mathbb Z}_2$ on $Z$. These generate a ${mathbb Z}_2times{mathbb Z}_2$ action, but it is not a free action since the product of the two generators has a fixed point.






          share|cite|improve this answer









          $endgroup$









          • 2




            $begingroup$
            wonderful!! it is just too beautiful... Z is tow circles connecting their noth poles by an edge and south poles by an edge...and if W exists then it has to be wedge of 3 circles then both X and Y be two sheeted covering of W but any two sheeted covering of W contain 6 edges...so contradiction
            $endgroup$
            – Anubhav Mukherjee
            Feb 11 '15 at 23:20


















          11












          $begingroup$

          Let $X$ and $Y$ be the two graphs having two vertices and three edges. There is a common two-sheeted covering space $Z$ of $X$ and $Y$ which is a graph with four vertices and six edges. Exercise: Find $Z$ and show that $X$ and $Y$ are not covering spaces of any other spaces (besides themselves, of course).



          Interesting side note: The covering spaces $Zto X$ and $Zto Y$ are defined by free actions of ${mathbb Z}_2$ on $Z$. These generate a ${mathbb Z}_2times{mathbb Z}_2$ action, but it is not a free action since the product of the two generators has a fixed point.






          share|cite|improve this answer









          $endgroup$









          • 2




            $begingroup$
            wonderful!! it is just too beautiful... Z is tow circles connecting their noth poles by an edge and south poles by an edge...and if W exists then it has to be wedge of 3 circles then both X and Y be two sheeted covering of W but any two sheeted covering of W contain 6 edges...so contradiction
            $endgroup$
            – Anubhav Mukherjee
            Feb 11 '15 at 23:20
















          11












          11








          11





          $begingroup$

          Let $X$ and $Y$ be the two graphs having two vertices and three edges. There is a common two-sheeted covering space $Z$ of $X$ and $Y$ which is a graph with four vertices and six edges. Exercise: Find $Z$ and show that $X$ and $Y$ are not covering spaces of any other spaces (besides themselves, of course).



          Interesting side note: The covering spaces $Zto X$ and $Zto Y$ are defined by free actions of ${mathbb Z}_2$ on $Z$. These generate a ${mathbb Z}_2times{mathbb Z}_2$ action, but it is not a free action since the product of the two generators has a fixed point.






          share|cite|improve this answer









          $endgroup$



          Let $X$ and $Y$ be the two graphs having two vertices and three edges. There is a common two-sheeted covering space $Z$ of $X$ and $Y$ which is a graph with four vertices and six edges. Exercise: Find $Z$ and show that $X$ and $Y$ are not covering spaces of any other spaces (besides themselves, of course).



          Interesting side note: The covering spaces $Zto X$ and $Zto Y$ are defined by free actions of ${mathbb Z}_2$ on $Z$. These generate a ${mathbb Z}_2times{mathbb Z}_2$ action, but it is not a free action since the product of the two generators has a fixed point.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Feb 11 '15 at 22:36









          Allen HatcherAllen Hatcher

          85666




          85666








          • 2




            $begingroup$
            wonderful!! it is just too beautiful... Z is tow circles connecting their noth poles by an edge and south poles by an edge...and if W exists then it has to be wedge of 3 circles then both X and Y be two sheeted covering of W but any two sheeted covering of W contain 6 edges...so contradiction
            $endgroup$
            – Anubhav Mukherjee
            Feb 11 '15 at 23:20
















          • 2




            $begingroup$
            wonderful!! it is just too beautiful... Z is tow circles connecting their noth poles by an edge and south poles by an edge...and if W exists then it has to be wedge of 3 circles then both X and Y be two sheeted covering of W but any two sheeted covering of W contain 6 edges...so contradiction
            $endgroup$
            – Anubhav Mukherjee
            Feb 11 '15 at 23:20










          2




          2




          $begingroup$
          wonderful!! it is just too beautiful... Z is tow circles connecting their noth poles by an edge and south poles by an edge...and if W exists then it has to be wedge of 3 circles then both X and Y be two sheeted covering of W but any two sheeted covering of W contain 6 edges...so contradiction
          $endgroup$
          – Anubhav Mukherjee
          Feb 11 '15 at 23:20






          $begingroup$
          wonderful!! it is just too beautiful... Z is tow circles connecting their noth poles by an edge and south poles by an edge...and if W exists then it has to be wedge of 3 circles then both X and Y be two sheeted covering of W but any two sheeted covering of W contain 6 edges...so contradiction
          $endgroup$
          – Anubhav Mukherjee
          Feb 11 '15 at 23:20













          6












          $begingroup$

          No, it is not true.



          The open 2-dimensional disc $Z=D^2$ is a covering space of the genus 1 closed surface $X=F_1 = S^1 times S^1$ and of the genus 2 closed surface $Y=F_2 = F_1 # F_1$ (that symbol $#$ means connected sum). One sees this best using geometry. The torus $F_1$ has a Euclidean structure and so there is a locally isometric universal covering map from the Euclidean plane $mathbb{R}^2 to F_1$. And the surface $F_2$ has a hyperbolic structure and so there is a locally isometric universal covering map from the hyperbolic plane $mathbb{H}^2 to F_2$. And, of course, each of $mathbb{R}^2$, $mathbb{H}^2$ is homeomorphic to $D^2$.



          But $F_1$ and $F_2$ cannot cover the same space $W$. For suppose they did, compactness of $F_1$ and $F_2$ would imply that the covering maps $F_1 mapsto W$ and $F_2 mapsto W$ are each of finite degree. The fundamental group $pi_1 W$ could therefore contain finite index subgroups $A_1,A_2$ isomorphic to $pi_1F_1$, $pi_1F_2$ respectively. Since $pi_1F_1 = mathbb{Z} oplus mathbb{Z}$ is abelian, $A_1 cap A_2$ is a finite index abelian subgroup of $A_2$. But $pi_1F_2$ has no finite index abelian subgroup, in fact its only abelian subgroups are trivial or infinite cyclic of infinite index.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            $pi_1(F_2) = <a,b,c,d/ aba^-1b^-1cdc^-1d^-1>$ but from here how do I conclude the last line that it has no finite index ableian subgroup ?? can you plese help me?
            $endgroup$
            – Anubhav Mukherjee
            Feb 1 '15 at 18:33










          • $begingroup$
            @AnubhaV: that is also best proved using geometry: every discrete abelian subgroup of the group of isometries of $mathbb{H}^2$ fixes some line in $mathbb{H}^2$, but the action of the deck transformation group on $mathbb{H}^2$ has no finite index subgroup fixing a line.
            $endgroup$
            – Lee Mosher
            Feb 1 '15 at 21:06










          • $begingroup$
            here if I restricted Z as a finite sheeted covering space...Now can you give me an counter examle...thanks for helping
            $endgroup$
            – Anubhav Mukherjee
            Feb 6 '15 at 14:07










          • $begingroup$
            @AnubhaV: That's a strong restriction which really makes it a different question. So you should post it as a different question (with a link to this one and an explanation of why you have added the restriction to finite sheeted covering spaces).
            $endgroup$
            – Lee Mosher
            Feb 6 '15 at 20:51










          • $begingroup$
            thanks...I am doing so now...and sir please help me to post properly the question...I dont know how to properly edit questions
            $endgroup$
            – Anubhav Mukherjee
            Feb 6 '15 at 21:56


















          6












          $begingroup$

          No, it is not true.



          The open 2-dimensional disc $Z=D^2$ is a covering space of the genus 1 closed surface $X=F_1 = S^1 times S^1$ and of the genus 2 closed surface $Y=F_2 = F_1 # F_1$ (that symbol $#$ means connected sum). One sees this best using geometry. The torus $F_1$ has a Euclidean structure and so there is a locally isometric universal covering map from the Euclidean plane $mathbb{R}^2 to F_1$. And the surface $F_2$ has a hyperbolic structure and so there is a locally isometric universal covering map from the hyperbolic plane $mathbb{H}^2 to F_2$. And, of course, each of $mathbb{R}^2$, $mathbb{H}^2$ is homeomorphic to $D^2$.



          But $F_1$ and $F_2$ cannot cover the same space $W$. For suppose they did, compactness of $F_1$ and $F_2$ would imply that the covering maps $F_1 mapsto W$ and $F_2 mapsto W$ are each of finite degree. The fundamental group $pi_1 W$ could therefore contain finite index subgroups $A_1,A_2$ isomorphic to $pi_1F_1$, $pi_1F_2$ respectively. Since $pi_1F_1 = mathbb{Z} oplus mathbb{Z}$ is abelian, $A_1 cap A_2$ is a finite index abelian subgroup of $A_2$. But $pi_1F_2$ has no finite index abelian subgroup, in fact its only abelian subgroups are trivial or infinite cyclic of infinite index.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            $pi_1(F_2) = <a,b,c,d/ aba^-1b^-1cdc^-1d^-1>$ but from here how do I conclude the last line that it has no finite index ableian subgroup ?? can you plese help me?
            $endgroup$
            – Anubhav Mukherjee
            Feb 1 '15 at 18:33










          • $begingroup$
            @AnubhaV: that is also best proved using geometry: every discrete abelian subgroup of the group of isometries of $mathbb{H}^2$ fixes some line in $mathbb{H}^2$, but the action of the deck transformation group on $mathbb{H}^2$ has no finite index subgroup fixing a line.
            $endgroup$
            – Lee Mosher
            Feb 1 '15 at 21:06










          • $begingroup$
            here if I restricted Z as a finite sheeted covering space...Now can you give me an counter examle...thanks for helping
            $endgroup$
            – Anubhav Mukherjee
            Feb 6 '15 at 14:07










          • $begingroup$
            @AnubhaV: That's a strong restriction which really makes it a different question. So you should post it as a different question (with a link to this one and an explanation of why you have added the restriction to finite sheeted covering spaces).
            $endgroup$
            – Lee Mosher
            Feb 6 '15 at 20:51










          • $begingroup$
            thanks...I am doing so now...and sir please help me to post properly the question...I dont know how to properly edit questions
            $endgroup$
            – Anubhav Mukherjee
            Feb 6 '15 at 21:56
















          6












          6








          6





          $begingroup$

          No, it is not true.



          The open 2-dimensional disc $Z=D^2$ is a covering space of the genus 1 closed surface $X=F_1 = S^1 times S^1$ and of the genus 2 closed surface $Y=F_2 = F_1 # F_1$ (that symbol $#$ means connected sum). One sees this best using geometry. The torus $F_1$ has a Euclidean structure and so there is a locally isometric universal covering map from the Euclidean plane $mathbb{R}^2 to F_1$. And the surface $F_2$ has a hyperbolic structure and so there is a locally isometric universal covering map from the hyperbolic plane $mathbb{H}^2 to F_2$. And, of course, each of $mathbb{R}^2$, $mathbb{H}^2$ is homeomorphic to $D^2$.



          But $F_1$ and $F_2$ cannot cover the same space $W$. For suppose they did, compactness of $F_1$ and $F_2$ would imply that the covering maps $F_1 mapsto W$ and $F_2 mapsto W$ are each of finite degree. The fundamental group $pi_1 W$ could therefore contain finite index subgroups $A_1,A_2$ isomorphic to $pi_1F_1$, $pi_1F_2$ respectively. Since $pi_1F_1 = mathbb{Z} oplus mathbb{Z}$ is abelian, $A_1 cap A_2$ is a finite index abelian subgroup of $A_2$. But $pi_1F_2$ has no finite index abelian subgroup, in fact its only abelian subgroups are trivial or infinite cyclic of infinite index.






          share|cite|improve this answer











          $endgroup$



          No, it is not true.



          The open 2-dimensional disc $Z=D^2$ is a covering space of the genus 1 closed surface $X=F_1 = S^1 times S^1$ and of the genus 2 closed surface $Y=F_2 = F_1 # F_1$ (that symbol $#$ means connected sum). One sees this best using geometry. The torus $F_1$ has a Euclidean structure and so there is a locally isometric universal covering map from the Euclidean plane $mathbb{R}^2 to F_1$. And the surface $F_2$ has a hyperbolic structure and so there is a locally isometric universal covering map from the hyperbolic plane $mathbb{H}^2 to F_2$. And, of course, each of $mathbb{R}^2$, $mathbb{H}^2$ is homeomorphic to $D^2$.



          But $F_1$ and $F_2$ cannot cover the same space $W$. For suppose they did, compactness of $F_1$ and $F_2$ would imply that the covering maps $F_1 mapsto W$ and $F_2 mapsto W$ are each of finite degree. The fundamental group $pi_1 W$ could therefore contain finite index subgroups $A_1,A_2$ isomorphic to $pi_1F_1$, $pi_1F_2$ respectively. Since $pi_1F_1 = mathbb{Z} oplus mathbb{Z}$ is abelian, $A_1 cap A_2$ is a finite index abelian subgroup of $A_2$. But $pi_1F_2$ has no finite index abelian subgroup, in fact its only abelian subgroups are trivial or infinite cyclic of infinite index.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Feb 1 '15 at 16:35

























          answered Feb 1 '15 at 16:28









          Lee MosherLee Mosher

          49.1k33685




          49.1k33685












          • $begingroup$
            $pi_1(F_2) = <a,b,c,d/ aba^-1b^-1cdc^-1d^-1>$ but from here how do I conclude the last line that it has no finite index ableian subgroup ?? can you plese help me?
            $endgroup$
            – Anubhav Mukherjee
            Feb 1 '15 at 18:33










          • $begingroup$
            @AnubhaV: that is also best proved using geometry: every discrete abelian subgroup of the group of isometries of $mathbb{H}^2$ fixes some line in $mathbb{H}^2$, but the action of the deck transformation group on $mathbb{H}^2$ has no finite index subgroup fixing a line.
            $endgroup$
            – Lee Mosher
            Feb 1 '15 at 21:06










          • $begingroup$
            here if I restricted Z as a finite sheeted covering space...Now can you give me an counter examle...thanks for helping
            $endgroup$
            – Anubhav Mukherjee
            Feb 6 '15 at 14:07










          • $begingroup$
            @AnubhaV: That's a strong restriction which really makes it a different question. So you should post it as a different question (with a link to this one and an explanation of why you have added the restriction to finite sheeted covering spaces).
            $endgroup$
            – Lee Mosher
            Feb 6 '15 at 20:51










          • $begingroup$
            thanks...I am doing so now...and sir please help me to post properly the question...I dont know how to properly edit questions
            $endgroup$
            – Anubhav Mukherjee
            Feb 6 '15 at 21:56




















          • $begingroup$
            $pi_1(F_2) = <a,b,c,d/ aba^-1b^-1cdc^-1d^-1>$ but from here how do I conclude the last line that it has no finite index ableian subgroup ?? can you plese help me?
            $endgroup$
            – Anubhav Mukherjee
            Feb 1 '15 at 18:33










          • $begingroup$
            @AnubhaV: that is also best proved using geometry: every discrete abelian subgroup of the group of isometries of $mathbb{H}^2$ fixes some line in $mathbb{H}^2$, but the action of the deck transformation group on $mathbb{H}^2$ has no finite index subgroup fixing a line.
            $endgroup$
            – Lee Mosher
            Feb 1 '15 at 21:06










          • $begingroup$
            here if I restricted Z as a finite sheeted covering space...Now can you give me an counter examle...thanks for helping
            $endgroup$
            – Anubhav Mukherjee
            Feb 6 '15 at 14:07










          • $begingroup$
            @AnubhaV: That's a strong restriction which really makes it a different question. So you should post it as a different question (with a link to this one and an explanation of why you have added the restriction to finite sheeted covering spaces).
            $endgroup$
            – Lee Mosher
            Feb 6 '15 at 20:51










          • $begingroup$
            thanks...I am doing so now...and sir please help me to post properly the question...I dont know how to properly edit questions
            $endgroup$
            – Anubhav Mukherjee
            Feb 6 '15 at 21:56


















          $begingroup$
          $pi_1(F_2) = <a,b,c,d/ aba^-1b^-1cdc^-1d^-1>$ but from here how do I conclude the last line that it has no finite index ableian subgroup ?? can you plese help me?
          $endgroup$
          – Anubhav Mukherjee
          Feb 1 '15 at 18:33




          $begingroup$
          $pi_1(F_2) = <a,b,c,d/ aba^-1b^-1cdc^-1d^-1>$ but from here how do I conclude the last line that it has no finite index ableian subgroup ?? can you plese help me?
          $endgroup$
          – Anubhav Mukherjee
          Feb 1 '15 at 18:33












          $begingroup$
          @AnubhaV: that is also best proved using geometry: every discrete abelian subgroup of the group of isometries of $mathbb{H}^2$ fixes some line in $mathbb{H}^2$, but the action of the deck transformation group on $mathbb{H}^2$ has no finite index subgroup fixing a line.
          $endgroup$
          – Lee Mosher
          Feb 1 '15 at 21:06




          $begingroup$
          @AnubhaV: that is also best proved using geometry: every discrete abelian subgroup of the group of isometries of $mathbb{H}^2$ fixes some line in $mathbb{H}^2$, but the action of the deck transformation group on $mathbb{H}^2$ has no finite index subgroup fixing a line.
          $endgroup$
          – Lee Mosher
          Feb 1 '15 at 21:06












          $begingroup$
          here if I restricted Z as a finite sheeted covering space...Now can you give me an counter examle...thanks for helping
          $endgroup$
          – Anubhav Mukherjee
          Feb 6 '15 at 14:07




          $begingroup$
          here if I restricted Z as a finite sheeted covering space...Now can you give me an counter examle...thanks for helping
          $endgroup$
          – Anubhav Mukherjee
          Feb 6 '15 at 14:07












          $begingroup$
          @AnubhaV: That's a strong restriction which really makes it a different question. So you should post it as a different question (with a link to this one and an explanation of why you have added the restriction to finite sheeted covering spaces).
          $endgroup$
          – Lee Mosher
          Feb 6 '15 at 20:51




          $begingroup$
          @AnubhaV: That's a strong restriction which really makes it a different question. So you should post it as a different question (with a link to this one and an explanation of why you have added the restriction to finite sheeted covering spaces).
          $endgroup$
          – Lee Mosher
          Feb 6 '15 at 20:51












          $begingroup$
          thanks...I am doing so now...and sir please help me to post properly the question...I dont know how to properly edit questions
          $endgroup$
          – Anubhav Mukherjee
          Feb 6 '15 at 21:56






          $begingroup$
          thanks...I am doing so now...and sir please help me to post properly the question...I dont know how to properly edit questions
          $endgroup$
          – Anubhav Mukherjee
          Feb 6 '15 at 21:56




















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