Find an example of K, A, B, and I to show $beta_ineq beta_i(A)+beta_i(B)-beta_i(Acap B)$












0














I’ve been trying to find a simplicial complex $K$, where $A$ and $B$ are subcomplexes, and $K=Acup B$ to show that the Betti number



$beta_i(K)neq beta_i(A)+beta_i(B)-beta_i(Acap B)$



but every complex I come up with contradicts this for $i=0$, what would be a good example for this?






topological-data-analysis







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  • What is "= / ="? Do you mean "not equal"? Use "neq" in MathJax to get "$neq$"
    – MPW
    Nov 30 at 16:56












  • I already fixed it, thanks
    – LexyFidds
    Nov 30 at 16:57












  • Look at the Mayer--Vietoris sequence, this will give you the precise relation between these numbers.
    – Pedro Tamaroff♦
    Nov 30 at 17:18
















0














I’ve been trying to find a simplicial complex $K$, where $A$ and $B$ are subcomplexes, and $K=Acup B$ to show that the Betti number



$beta_i(K)neq beta_i(A)+beta_i(B)-beta_i(Acap B)$



but every complex I come up with contradicts this for $i=0$, what would be a good example for this?






topological-data-analysis







share|cite|improve this question
























  • What is "= / ="? Do you mean "not equal"? Use "neq" in MathJax to get "$neq$"
    – MPW
    Nov 30 at 16:56












  • I already fixed it, thanks
    – LexyFidds
    Nov 30 at 16:57












  • Look at the Mayer--Vietoris sequence, this will give you the precise relation between these numbers.
    – Pedro Tamaroff♦
    Nov 30 at 17:18














0












0








0







I’ve been trying to find a simplicial complex $K$, where $A$ and $B$ are subcomplexes, and $K=Acup B$ to show that the Betti number



$beta_i(K)neq beta_i(A)+beta_i(B)-beta_i(Acap B)$



but every complex I come up with contradicts this for $i=0$, what would be a good example for this?






topological-data-analysis







share|cite|improve this question















I’ve been trying to find a simplicial complex $K$, where $A$ and $B$ are subcomplexes, and $K=Acup B$ to show that the Betti number



$beta_i(K)neq beta_i(A)+beta_i(B)-beta_i(Acap B)$



but every complex I come up with contradicts this for $i=0$, what would be a good example for this?






topological-data-analysis




topological-data-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 30 at 17:15

























asked Nov 30 at 16:54









LexyFidds

226




226












  • What is "= / ="? Do you mean "not equal"? Use "neq" in MathJax to get "$neq$"
    – MPW
    Nov 30 at 16:56












  • I already fixed it, thanks
    – LexyFidds
    Nov 30 at 16:57












  • Look at the Mayer--Vietoris sequence, this will give you the precise relation between these numbers.
    – Pedro Tamaroff♦
    Nov 30 at 17:18


















  • What is "= / ="? Do you mean "not equal"? Use "neq" in MathJax to get "$neq$"
    – MPW
    Nov 30 at 16:56












  • I already fixed it, thanks
    – LexyFidds
    Nov 30 at 16:57












  • Look at the Mayer--Vietoris sequence, this will give you the precise relation between these numbers.
    – Pedro Tamaroff♦
    Nov 30 at 17:18
















What is "= / ="? Do you mean "not equal"? Use "neq" in MathJax to get "$neq$"
– MPW
Nov 30 at 16:56






What is "= / ="? Do you mean "not equal"? Use "neq" in MathJax to get "$neq$"
– MPW
Nov 30 at 16:56














I already fixed it, thanks
– LexyFidds
Nov 30 at 16:57






I already fixed it, thanks
– LexyFidds
Nov 30 at 16:57














Look at the Mayer--Vietoris sequence, this will give you the precise relation between these numbers.
– Pedro Tamaroff♦
Nov 30 at 17:18




Look at the Mayer--Vietoris sequence, this will give you the precise relation between these numbers.
– Pedro Tamaroff♦
Nov 30 at 17:18










1 Answer
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Here's an example (for $i=0$) where $beta_0(K)neqbeta_0(A)+beta_0(B)-beta_0(Acap B)$. Let simplicial complex $K$ be the boundary of a triangle, with three vertices $[0]$, $[1]$, $[2]$, and three edges $[0,1]$, $[1,2]$, $[2,0]$. Let subcomplex $A$ be the union of two edges, containing all three vertices $[0]$, $[1]$, $[2]$, and only two edges $[0,1]$, $[1,2]$. Let subcomplex $B$ be the third remaining edge, containing vertices $[0]$, $[2]$ and only one edge $[2,0]$. We have $K=Acup B$.



Note that $K$, $A$, and $B$ are each connected, giving $beta_0(K)=beta_0(A)=beta_0(B)=1$. However, $Acap B$ consists of two connected components (two vertices $[0]$ and $[2]$), giving $beta_0(Acap B)=2$. So we have
$$beta_0(K)=1neq 0=1+1-2=beta_0(A)+beta_0(B)-beta_0(Acap B),$$
as desired. Note in this example that $beta_1(K)=1$ is nonzero, agreeing with the Mayer-Vietoris long exact sequence (https://en.wikipedia.org/wiki/Mayer%E2%80%93Vietoris_sequence).






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    Here's an example (for $i=0$) where $beta_0(K)neqbeta_0(A)+beta_0(B)-beta_0(Acap B)$. Let simplicial complex $K$ be the boundary of a triangle, with three vertices $[0]$, $[1]$, $[2]$, and three edges $[0,1]$, $[1,2]$, $[2,0]$. Let subcomplex $A$ be the union of two edges, containing all three vertices $[0]$, $[1]$, $[2]$, and only two edges $[0,1]$, $[1,2]$. Let subcomplex $B$ be the third remaining edge, containing vertices $[0]$, $[2]$ and only one edge $[2,0]$. We have $K=Acup B$.



    Note that $K$, $A$, and $B$ are each connected, giving $beta_0(K)=beta_0(A)=beta_0(B)=1$. However, $Acap B$ consists of two connected components (two vertices $[0]$ and $[2]$), giving $beta_0(Acap B)=2$. So we have
    $$beta_0(K)=1neq 0=1+1-2=beta_0(A)+beta_0(B)-beta_0(Acap B),$$
    as desired. Note in this example that $beta_1(K)=1$ is nonzero, agreeing with the Mayer-Vietoris long exact sequence (https://en.wikipedia.org/wiki/Mayer%E2%80%93Vietoris_sequence).






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      Here's an example (for $i=0$) where $beta_0(K)neqbeta_0(A)+beta_0(B)-beta_0(Acap B)$. Let simplicial complex $K$ be the boundary of a triangle, with three vertices $[0]$, $[1]$, $[2]$, and three edges $[0,1]$, $[1,2]$, $[2,0]$. Let subcomplex $A$ be the union of two edges, containing all three vertices $[0]$, $[1]$, $[2]$, and only two edges $[0,1]$, $[1,2]$. Let subcomplex $B$ be the third remaining edge, containing vertices $[0]$, $[2]$ and only one edge $[2,0]$. We have $K=Acup B$.



      Note that $K$, $A$, and $B$ are each connected, giving $beta_0(K)=beta_0(A)=beta_0(B)=1$. However, $Acap B$ consists of two connected components (two vertices $[0]$ and $[2]$), giving $beta_0(Acap B)=2$. So we have
      $$beta_0(K)=1neq 0=1+1-2=beta_0(A)+beta_0(B)-beta_0(Acap B),$$
      as desired. Note in this example that $beta_1(K)=1$ is nonzero, agreeing with the Mayer-Vietoris long exact sequence (https://en.wikipedia.org/wiki/Mayer%E2%80%93Vietoris_sequence).






      share|cite|improve this answer
























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        Here's an example (for $i=0$) where $beta_0(K)neqbeta_0(A)+beta_0(B)-beta_0(Acap B)$. Let simplicial complex $K$ be the boundary of a triangle, with three vertices $[0]$, $[1]$, $[2]$, and three edges $[0,1]$, $[1,2]$, $[2,0]$. Let subcomplex $A$ be the union of two edges, containing all three vertices $[0]$, $[1]$, $[2]$, and only two edges $[0,1]$, $[1,2]$. Let subcomplex $B$ be the third remaining edge, containing vertices $[0]$, $[2]$ and only one edge $[2,0]$. We have $K=Acup B$.



        Note that $K$, $A$, and $B$ are each connected, giving $beta_0(K)=beta_0(A)=beta_0(B)=1$. However, $Acap B$ consists of two connected components (two vertices $[0]$ and $[2]$), giving $beta_0(Acap B)=2$. So we have
        $$beta_0(K)=1neq 0=1+1-2=beta_0(A)+beta_0(B)-beta_0(Acap B),$$
        as desired. Note in this example that $beta_1(K)=1$ is nonzero, agreeing with the Mayer-Vietoris long exact sequence (https://en.wikipedia.org/wiki/Mayer%E2%80%93Vietoris_sequence).






        share|cite|improve this answer












        Here's an example (for $i=0$) where $beta_0(K)neqbeta_0(A)+beta_0(B)-beta_0(Acap B)$. Let simplicial complex $K$ be the boundary of a triangle, with three vertices $[0]$, $[1]$, $[2]$, and three edges $[0,1]$, $[1,2]$, $[2,0]$. Let subcomplex $A$ be the union of two edges, containing all three vertices $[0]$, $[1]$, $[2]$, and only two edges $[0,1]$, $[1,2]$. Let subcomplex $B$ be the third remaining edge, containing vertices $[0]$, $[2]$ and only one edge $[2,0]$. We have $K=Acup B$.



        Note that $K$, $A$, and $B$ are each connected, giving $beta_0(K)=beta_0(A)=beta_0(B)=1$. However, $Acap B$ consists of two connected components (two vertices $[0]$ and $[2]$), giving $beta_0(Acap B)=2$. So we have
        $$beta_0(K)=1neq 0=1+1-2=beta_0(A)+beta_0(B)-beta_0(Acap B),$$
        as desired. Note in this example that $beta_1(K)=1$ is nonzero, agreeing with the Mayer-Vietoris long exact sequence (https://en.wikipedia.org/wiki/Mayer%E2%80%93Vietoris_sequence).







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 8 at 14:56









        Henry Adams

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