Homology of a simplicial complex union a simplex












1














Let $L=Kcupsigma$, where $L, K$ are abstract simplicial complexes, and $sigma$ is a simplex whose faces are all in $K$. (This condition is necessary for $L$ to be a simplicial complex.)



Furthermore, we have that $sigmanotin K$, which also implies that $sigma$ is a maximal face of $L$. (If $sigmasubsetneqtauin L$, then $tauin K$ implies $sigmain K$ which is a contradiction.)



Is there any relation between the homologies of $L$ and $K$? Is it possible to have a "formula" relating $H_*(L)$ and $H_*(K$)?



Thanks.






algebraic-topology homology-cohomology simplicial-complex







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    1














    Let $L=Kcupsigma$, where $L, K$ are abstract simplicial complexes, and $sigma$ is a simplex whose faces are all in $K$. (This condition is necessary for $L$ to be a simplicial complex.)



    Furthermore, we have that $sigmanotin K$, which also implies that $sigma$ is a maximal face of $L$. (If $sigmasubsetneqtauin L$, then $tauin K$ implies $sigmain K$ which is a contradiction.)



    Is there any relation between the homologies of $L$ and $K$? Is it possible to have a "formula" relating $H_*(L)$ and $H_*(K$)?



    Thanks.






    algebraic-topology homology-cohomology simplicial-complex







    share|cite|improve this question

























      1












      1








      1


      1





      Let $L=Kcupsigma$, where $L, K$ are abstract simplicial complexes, and $sigma$ is a simplex whose faces are all in $K$. (This condition is necessary for $L$ to be a simplicial complex.)



      Furthermore, we have that $sigmanotin K$, which also implies that $sigma$ is a maximal face of $L$. (If $sigmasubsetneqtauin L$, then $tauin K$ implies $sigmain K$ which is a contradiction.)



      Is there any relation between the homologies of $L$ and $K$? Is it possible to have a "formula" relating $H_*(L)$ and $H_*(K$)?



      Thanks.






      algebraic-topology homology-cohomology simplicial-complex







      share|cite|improve this question













      Let $L=Kcupsigma$, where $L, K$ are abstract simplicial complexes, and $sigma$ is a simplex whose faces are all in $K$. (This condition is necessary for $L$ to be a simplicial complex.)



      Furthermore, we have that $sigmanotin K$, which also implies that $sigma$ is a maximal face of $L$. (If $sigmasubsetneqtauin L$, then $tauin K$ implies $sigmain K$ which is a contradiction.)



      Is there any relation between the homologies of $L$ and $K$? Is it possible to have a "formula" relating $H_*(L)$ and $H_*(K$)?



      Thanks.






      algebraic-topology homology-cohomology simplicial-complex




      algebraic-topology homology-cohomology simplicial-complex






      share|cite|improve this question













      share|cite|improve this question











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      asked Nov 18 at 16:37









      yoyostein

      7,86483768




      7,86483768






















          2 Answers
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          active

          oldest

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          2














          Let me elaborate a bit more on Lord Shark the Unknown's answer. We can get a precise answer because the maps in the sequence $$0to H_n(K)to H_n(L)toBbb Zto H_{n-1}(K)to H_{n-1}(L)to0$$ can be identified explicitly. Note that the $mathbb{Z}$ here is generated by the homology class of a generator of $Y_n$, which is just the image of $sigmain C_n(L)$ under the quotient map $C_n(L)to Y$. So, by definition of the long exact sequence, the map $Bbb Zto H_{n-1}(K)$ is given by lifting this generator to $sigmain C_n(L)$, taking the boundary, and then lifting that boundary to $C_{n-1}(K)$. In other words, it is exactly the class $[partialsigma]in H_{n-1}(K)$.



          We conclude that $H_{n-1}(L)$ is exactly the quotient of $H_{n-1}(K)$ by the subgroup generated by $[partialsigma]$. We also conclude that the cokernel of $H_n(K)to H_n(L)$ is the group of $ninmathbb{Z}$ such that $n[partialsigma]$ is $0$ in $H_{n-1}(K)$. If $[partialsigma]$ is torsion this group is isomorphic to $mathbb{Z}$, and otherwise it is trivial. So we concldue that $H_n(L)cong H_n(K)oplus mathbb{Z}$ if $[partialsigma]$ is a torsion class in $H_{n-1}(K)$, and $H_n(L)cong H_n(K)$ if $[partialsigma]$ is not torsion.






          share|cite|improve this answer





















          • Thanks. I am confused on this example: Let $K={[a],[b],[c],[a,b],[a,c]}$, i.e. a triangle with 2 sides. Let $sigma={[b,c]}$. Then $L$ is a triangle homotopy equivalent to a circle. We see that $H_1(L)congmathbb{Z}$ and $H_1(K)cong 0$. However, I don't see how $[partialsigma]$ is a torsion class in $H_0(K)congmathbb{Z}$?
            – yoyostein
            Nov 19 at 2:26








          • 1




            $partialsigma=c-b=(c-a)-(b-a)$ is a difference of two boundaries in $K$.
            – Eric Wofsey
            Nov 19 at 2:29










          • Thanks! I get it now. So in fact $n[partialsigma]$ is 0 in $H_{0}(K)$ for all $ninmathbb{Z}$, since $[partialsigma]$ itself is 0 in $H_0(K)$.
            – yoyostein
            Nov 19 at 2:38



















          3














          Suppose that $sigma$ has dimension $n$. Consider the complexes of simplicial chains $C(K)$ and $C(L)$. Then there is an exact sequence of complexes
          $$0to C(K)to C(L)to Yto 0tag{*}$$
          where the complex $Y$ consists only of a $Bbb Z$ in dimension $n$. Then (*)
          has an associated long exact sequence of homology. But the homology of $Y$
          consists of just a $Bbb Z$ is dimension $n$. We conclude that $H_m(K)cong H_m(L)$
          for $mnotin{n-1,n}$ and that there is an exact sequence
          $$0to H_n(K)to H_n(L)toBbb Zto H_{n-1}(K)to H_{n-1}(L)to0.$$
          We can conclude several things from this, if we know something about the
          homology groups of $K$. So adding $sigma$ kills the homology of the cycle
          $partialsigma$ in $H_{n-1}(K)$. Moreover $H_n(L)$ will be either $H_n(K)$
          or the sum of $H_n(K)$ with a copy of $Bbb Z$ etc.






          share|cite|improve this answer





















          • Thanks. For the last sentence, can we know (possibly with some added assumptions or extra conditions) whether $H_n(L)=H_n(K)$ or sum of $H_n(K)$ with a copy of Z? Is there any condition that can tell us which case it falls into?
            – yoyostein
            Nov 18 at 17:06






          • 2




            It depends on whether the homology class of $partialsigma$ in $H_{n-1}(K)$ is torsion.
            – Lord Shark the Unknown
            Nov 18 at 17:13











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          2 Answers
          2






          active

          oldest

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          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2














          Let me elaborate a bit more on Lord Shark the Unknown's answer. We can get a precise answer because the maps in the sequence $$0to H_n(K)to H_n(L)toBbb Zto H_{n-1}(K)to H_{n-1}(L)to0$$ can be identified explicitly. Note that the $mathbb{Z}$ here is generated by the homology class of a generator of $Y_n$, which is just the image of $sigmain C_n(L)$ under the quotient map $C_n(L)to Y$. So, by definition of the long exact sequence, the map $Bbb Zto H_{n-1}(K)$ is given by lifting this generator to $sigmain C_n(L)$, taking the boundary, and then lifting that boundary to $C_{n-1}(K)$. In other words, it is exactly the class $[partialsigma]in H_{n-1}(K)$.



          We conclude that $H_{n-1}(L)$ is exactly the quotient of $H_{n-1}(K)$ by the subgroup generated by $[partialsigma]$. We also conclude that the cokernel of $H_n(K)to H_n(L)$ is the group of $ninmathbb{Z}$ such that $n[partialsigma]$ is $0$ in $H_{n-1}(K)$. If $[partialsigma]$ is torsion this group is isomorphic to $mathbb{Z}$, and otherwise it is trivial. So we concldue that $H_n(L)cong H_n(K)oplus mathbb{Z}$ if $[partialsigma]$ is a torsion class in $H_{n-1}(K)$, and $H_n(L)cong H_n(K)$ if $[partialsigma]$ is not torsion.






          share|cite|improve this answer





















          • Thanks. I am confused on this example: Let $K={[a],[b],[c],[a,b],[a,c]}$, i.e. a triangle with 2 sides. Let $sigma={[b,c]}$. Then $L$ is a triangle homotopy equivalent to a circle. We see that $H_1(L)congmathbb{Z}$ and $H_1(K)cong 0$. However, I don't see how $[partialsigma]$ is a torsion class in $H_0(K)congmathbb{Z}$?
            – yoyostein
            Nov 19 at 2:26








          • 1




            $partialsigma=c-b=(c-a)-(b-a)$ is a difference of two boundaries in $K$.
            – Eric Wofsey
            Nov 19 at 2:29










          • Thanks! I get it now. So in fact $n[partialsigma]$ is 0 in $H_{0}(K)$ for all $ninmathbb{Z}$, since $[partialsigma]$ itself is 0 in $H_0(K)$.
            – yoyostein
            Nov 19 at 2:38
















          2














          Let me elaborate a bit more on Lord Shark the Unknown's answer. We can get a precise answer because the maps in the sequence $$0to H_n(K)to H_n(L)toBbb Zto H_{n-1}(K)to H_{n-1}(L)to0$$ can be identified explicitly. Note that the $mathbb{Z}$ here is generated by the homology class of a generator of $Y_n$, which is just the image of $sigmain C_n(L)$ under the quotient map $C_n(L)to Y$. So, by definition of the long exact sequence, the map $Bbb Zto H_{n-1}(K)$ is given by lifting this generator to $sigmain C_n(L)$, taking the boundary, and then lifting that boundary to $C_{n-1}(K)$. In other words, it is exactly the class $[partialsigma]in H_{n-1}(K)$.



          We conclude that $H_{n-1}(L)$ is exactly the quotient of $H_{n-1}(K)$ by the subgroup generated by $[partialsigma]$. We also conclude that the cokernel of $H_n(K)to H_n(L)$ is the group of $ninmathbb{Z}$ such that $n[partialsigma]$ is $0$ in $H_{n-1}(K)$. If $[partialsigma]$ is torsion this group is isomorphic to $mathbb{Z}$, and otherwise it is trivial. So we concldue that $H_n(L)cong H_n(K)oplus mathbb{Z}$ if $[partialsigma]$ is a torsion class in $H_{n-1}(K)$, and $H_n(L)cong H_n(K)$ if $[partialsigma]$ is not torsion.






          share|cite|improve this answer





















          • Thanks. I am confused on this example: Let $K={[a],[b],[c],[a,b],[a,c]}$, i.e. a triangle with 2 sides. Let $sigma={[b,c]}$. Then $L$ is a triangle homotopy equivalent to a circle. We see that $H_1(L)congmathbb{Z}$ and $H_1(K)cong 0$. However, I don't see how $[partialsigma]$ is a torsion class in $H_0(K)congmathbb{Z}$?
            – yoyostein
            Nov 19 at 2:26








          • 1




            $partialsigma=c-b=(c-a)-(b-a)$ is a difference of two boundaries in $K$.
            – Eric Wofsey
            Nov 19 at 2:29










          • Thanks! I get it now. So in fact $n[partialsigma]$ is 0 in $H_{0}(K)$ for all $ninmathbb{Z}$, since $[partialsigma]$ itself is 0 in $H_0(K)$.
            – yoyostein
            Nov 19 at 2:38














          2












          2








          2






          Let me elaborate a bit more on Lord Shark the Unknown's answer. We can get a precise answer because the maps in the sequence $$0to H_n(K)to H_n(L)toBbb Zto H_{n-1}(K)to H_{n-1}(L)to0$$ can be identified explicitly. Note that the $mathbb{Z}$ here is generated by the homology class of a generator of $Y_n$, which is just the image of $sigmain C_n(L)$ under the quotient map $C_n(L)to Y$. So, by definition of the long exact sequence, the map $Bbb Zto H_{n-1}(K)$ is given by lifting this generator to $sigmain C_n(L)$, taking the boundary, and then lifting that boundary to $C_{n-1}(K)$. In other words, it is exactly the class $[partialsigma]in H_{n-1}(K)$.



          We conclude that $H_{n-1}(L)$ is exactly the quotient of $H_{n-1}(K)$ by the subgroup generated by $[partialsigma]$. We also conclude that the cokernel of $H_n(K)to H_n(L)$ is the group of $ninmathbb{Z}$ such that $n[partialsigma]$ is $0$ in $H_{n-1}(K)$. If $[partialsigma]$ is torsion this group is isomorphic to $mathbb{Z}$, and otherwise it is trivial. So we concldue that $H_n(L)cong H_n(K)oplus mathbb{Z}$ if $[partialsigma]$ is a torsion class in $H_{n-1}(K)$, and $H_n(L)cong H_n(K)$ if $[partialsigma]$ is not torsion.






          share|cite|improve this answer












          Let me elaborate a bit more on Lord Shark the Unknown's answer. We can get a precise answer because the maps in the sequence $$0to H_n(K)to H_n(L)toBbb Zto H_{n-1}(K)to H_{n-1}(L)to0$$ can be identified explicitly. Note that the $mathbb{Z}$ here is generated by the homology class of a generator of $Y_n$, which is just the image of $sigmain C_n(L)$ under the quotient map $C_n(L)to Y$. So, by definition of the long exact sequence, the map $Bbb Zto H_{n-1}(K)$ is given by lifting this generator to $sigmain C_n(L)$, taking the boundary, and then lifting that boundary to $C_{n-1}(K)$. In other words, it is exactly the class $[partialsigma]in H_{n-1}(K)$.



          We conclude that $H_{n-1}(L)$ is exactly the quotient of $H_{n-1}(K)$ by the subgroup generated by $[partialsigma]$. We also conclude that the cokernel of $H_n(K)to H_n(L)$ is the group of $ninmathbb{Z}$ such that $n[partialsigma]$ is $0$ in $H_{n-1}(K)$. If $[partialsigma]$ is torsion this group is isomorphic to $mathbb{Z}$, and otherwise it is trivial. So we concldue that $H_n(L)cong H_n(K)oplus mathbb{Z}$ if $[partialsigma]$ is a torsion class in $H_{n-1}(K)$, and $H_n(L)cong H_n(K)$ if $[partialsigma]$ is not torsion.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 18 at 17:14









          Eric Wofsey

          180k12204332




          180k12204332












          • Thanks. I am confused on this example: Let $K={[a],[b],[c],[a,b],[a,c]}$, i.e. a triangle with 2 sides. Let $sigma={[b,c]}$. Then $L$ is a triangle homotopy equivalent to a circle. We see that $H_1(L)congmathbb{Z}$ and $H_1(K)cong 0$. However, I don't see how $[partialsigma]$ is a torsion class in $H_0(K)congmathbb{Z}$?
            – yoyostein
            Nov 19 at 2:26








          • 1




            $partialsigma=c-b=(c-a)-(b-a)$ is a difference of two boundaries in $K$.
            – Eric Wofsey
            Nov 19 at 2:29










          • Thanks! I get it now. So in fact $n[partialsigma]$ is 0 in $H_{0}(K)$ for all $ninmathbb{Z}$, since $[partialsigma]$ itself is 0 in $H_0(K)$.
            – yoyostein
            Nov 19 at 2:38


















          • Thanks. I am confused on this example: Let $K={[a],[b],[c],[a,b],[a,c]}$, i.e. a triangle with 2 sides. Let $sigma={[b,c]}$. Then $L$ is a triangle homotopy equivalent to a circle. We see that $H_1(L)congmathbb{Z}$ and $H_1(K)cong 0$. However, I don't see how $[partialsigma]$ is a torsion class in $H_0(K)congmathbb{Z}$?
            – yoyostein
            Nov 19 at 2:26








          • 1




            $partialsigma=c-b=(c-a)-(b-a)$ is a difference of two boundaries in $K$.
            – Eric Wofsey
            Nov 19 at 2:29










          • Thanks! I get it now. So in fact $n[partialsigma]$ is 0 in $H_{0}(K)$ for all $ninmathbb{Z}$, since $[partialsigma]$ itself is 0 in $H_0(K)$.
            – yoyostein
            Nov 19 at 2:38
















          Thanks. I am confused on this example: Let $K={[a],[b],[c],[a,b],[a,c]}$, i.e. a triangle with 2 sides. Let $sigma={[b,c]}$. Then $L$ is a triangle homotopy equivalent to a circle. We see that $H_1(L)congmathbb{Z}$ and $H_1(K)cong 0$. However, I don't see how $[partialsigma]$ is a torsion class in $H_0(K)congmathbb{Z}$?
          – yoyostein
          Nov 19 at 2:26






          Thanks. I am confused on this example: Let $K={[a],[b],[c],[a,b],[a,c]}$, i.e. a triangle with 2 sides. Let $sigma={[b,c]}$. Then $L$ is a triangle homotopy equivalent to a circle. We see that $H_1(L)congmathbb{Z}$ and $H_1(K)cong 0$. However, I don't see how $[partialsigma]$ is a torsion class in $H_0(K)congmathbb{Z}$?
          – yoyostein
          Nov 19 at 2:26






          1




          1




          $partialsigma=c-b=(c-a)-(b-a)$ is a difference of two boundaries in $K$.
          – Eric Wofsey
          Nov 19 at 2:29




          $partialsigma=c-b=(c-a)-(b-a)$ is a difference of two boundaries in $K$.
          – Eric Wofsey
          Nov 19 at 2:29












          Thanks! I get it now. So in fact $n[partialsigma]$ is 0 in $H_{0}(K)$ for all $ninmathbb{Z}$, since $[partialsigma]$ itself is 0 in $H_0(K)$.
          – yoyostein
          Nov 19 at 2:38




          Thanks! I get it now. So in fact $n[partialsigma]$ is 0 in $H_{0}(K)$ for all $ninmathbb{Z}$, since $[partialsigma]$ itself is 0 in $H_0(K)$.
          – yoyostein
          Nov 19 at 2:38











          3














          Suppose that $sigma$ has dimension $n$. Consider the complexes of simplicial chains $C(K)$ and $C(L)$. Then there is an exact sequence of complexes
          $$0to C(K)to C(L)to Yto 0tag{*}$$
          where the complex $Y$ consists only of a $Bbb Z$ in dimension $n$. Then (*)
          has an associated long exact sequence of homology. But the homology of $Y$
          consists of just a $Bbb Z$ is dimension $n$. We conclude that $H_m(K)cong H_m(L)$
          for $mnotin{n-1,n}$ and that there is an exact sequence
          $$0to H_n(K)to H_n(L)toBbb Zto H_{n-1}(K)to H_{n-1}(L)to0.$$
          We can conclude several things from this, if we know something about the
          homology groups of $K$. So adding $sigma$ kills the homology of the cycle
          $partialsigma$ in $H_{n-1}(K)$. Moreover $H_n(L)$ will be either $H_n(K)$
          or the sum of $H_n(K)$ with a copy of $Bbb Z$ etc.






          share|cite|improve this answer





















          • Thanks. For the last sentence, can we know (possibly with some added assumptions or extra conditions) whether $H_n(L)=H_n(K)$ or sum of $H_n(K)$ with a copy of Z? Is there any condition that can tell us which case it falls into?
            – yoyostein
            Nov 18 at 17:06






          • 2




            It depends on whether the homology class of $partialsigma$ in $H_{n-1}(K)$ is torsion.
            – Lord Shark the Unknown
            Nov 18 at 17:13
















          3














          Suppose that $sigma$ has dimension $n$. Consider the complexes of simplicial chains $C(K)$ and $C(L)$. Then there is an exact sequence of complexes
          $$0to C(K)to C(L)to Yto 0tag{*}$$
          where the complex $Y$ consists only of a $Bbb Z$ in dimension $n$. Then (*)
          has an associated long exact sequence of homology. But the homology of $Y$
          consists of just a $Bbb Z$ is dimension $n$. We conclude that $H_m(K)cong H_m(L)$
          for $mnotin{n-1,n}$ and that there is an exact sequence
          $$0to H_n(K)to H_n(L)toBbb Zto H_{n-1}(K)to H_{n-1}(L)to0.$$
          We can conclude several things from this, if we know something about the
          homology groups of $K$. So adding $sigma$ kills the homology of the cycle
          $partialsigma$ in $H_{n-1}(K)$. Moreover $H_n(L)$ will be either $H_n(K)$
          or the sum of $H_n(K)$ with a copy of $Bbb Z$ etc.






          share|cite|improve this answer





















          • Thanks. For the last sentence, can we know (possibly with some added assumptions or extra conditions) whether $H_n(L)=H_n(K)$ or sum of $H_n(K)$ with a copy of Z? Is there any condition that can tell us which case it falls into?
            – yoyostein
            Nov 18 at 17:06






          • 2




            It depends on whether the homology class of $partialsigma$ in $H_{n-1}(K)$ is torsion.
            – Lord Shark the Unknown
            Nov 18 at 17:13














          3












          3








          3






          Suppose that $sigma$ has dimension $n$. Consider the complexes of simplicial chains $C(K)$ and $C(L)$. Then there is an exact sequence of complexes
          $$0to C(K)to C(L)to Yto 0tag{*}$$
          where the complex $Y$ consists only of a $Bbb Z$ in dimension $n$. Then (*)
          has an associated long exact sequence of homology. But the homology of $Y$
          consists of just a $Bbb Z$ is dimension $n$. We conclude that $H_m(K)cong H_m(L)$
          for $mnotin{n-1,n}$ and that there is an exact sequence
          $$0to H_n(K)to H_n(L)toBbb Zto H_{n-1}(K)to H_{n-1}(L)to0.$$
          We can conclude several things from this, if we know something about the
          homology groups of $K$. So adding $sigma$ kills the homology of the cycle
          $partialsigma$ in $H_{n-1}(K)$. Moreover $H_n(L)$ will be either $H_n(K)$
          or the sum of $H_n(K)$ with a copy of $Bbb Z$ etc.






          share|cite|improve this answer












          Suppose that $sigma$ has dimension $n$. Consider the complexes of simplicial chains $C(K)$ and $C(L)$. Then there is an exact sequence of complexes
          $$0to C(K)to C(L)to Yto 0tag{*}$$
          where the complex $Y$ consists only of a $Bbb Z$ in dimension $n$. Then (*)
          has an associated long exact sequence of homology. But the homology of $Y$
          consists of just a $Bbb Z$ is dimension $n$. We conclude that $H_m(K)cong H_m(L)$
          for $mnotin{n-1,n}$ and that there is an exact sequence
          $$0to H_n(K)to H_n(L)toBbb Zto H_{n-1}(K)to H_{n-1}(L)to0.$$
          We can conclude several things from this, if we know something about the
          homology groups of $K$. So adding $sigma$ kills the homology of the cycle
          $partialsigma$ in $H_{n-1}(K)$. Moreover $H_n(L)$ will be either $H_n(K)$
          or the sum of $H_n(K)$ with a copy of $Bbb Z$ etc.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 18 at 16:56









          Lord Shark the Unknown

          101k958132




          101k958132












          • Thanks. For the last sentence, can we know (possibly with some added assumptions or extra conditions) whether $H_n(L)=H_n(K)$ or sum of $H_n(K)$ with a copy of Z? Is there any condition that can tell us which case it falls into?
            – yoyostein
            Nov 18 at 17:06






          • 2




            It depends on whether the homology class of $partialsigma$ in $H_{n-1}(K)$ is torsion.
            – Lord Shark the Unknown
            Nov 18 at 17:13


















          • Thanks. For the last sentence, can we know (possibly with some added assumptions or extra conditions) whether $H_n(L)=H_n(K)$ or sum of $H_n(K)$ with a copy of Z? Is there any condition that can tell us which case it falls into?
            – yoyostein
            Nov 18 at 17:06






          • 2




            It depends on whether the homology class of $partialsigma$ in $H_{n-1}(K)$ is torsion.
            – Lord Shark the Unknown
            Nov 18 at 17:13
















          Thanks. For the last sentence, can we know (possibly with some added assumptions or extra conditions) whether $H_n(L)=H_n(K)$ or sum of $H_n(K)$ with a copy of Z? Is there any condition that can tell us which case it falls into?
          – yoyostein
          Nov 18 at 17:06




          Thanks. For the last sentence, can we know (possibly with some added assumptions or extra conditions) whether $H_n(L)=H_n(K)$ or sum of $H_n(K)$ with a copy of Z? Is there any condition that can tell us which case it falls into?
          – yoyostein
          Nov 18 at 17:06




          2




          2




          It depends on whether the homology class of $partialsigma$ in $H_{n-1}(K)$ is torsion.
          – Lord Shark the Unknown
          Nov 18 at 17:13




          It depends on whether the homology class of $partialsigma$ in $H_{n-1}(K)$ is torsion.
          – Lord Shark the Unknown
          Nov 18 at 17:13


















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