Confusion in using Mayer-Vietoris theorem












2












$begingroup$


For two topological spaces $A$ and $B$, in order to show that $H(A sqcup B) cong H(A) oplus H(B) $
in this question and in general I believe one can use Mayer-Vietoris to obtain the result easily. However, I don't quite understand how it applies in this case - what if $A sqcup B$ is not contained in the interior of $A$ and $B$?



Also in the link above the OP claims $A cap B= emptyset $ to conclude that homology group is trivial but $A cap B$ need not be empty or am I missing something?



Would greatly appreciated if someone could shed a light on these.






algebraic-topology







share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    For two topological spaces $A$ and $B$, in order to show that $H(A sqcup B) cong H(A) oplus H(B) $
    in this question and in general I believe one can use Mayer-Vietoris to obtain the result easily. However, I don't quite understand how it applies in this case - what if $A sqcup B$ is not contained in the interior of $A$ and $B$?



    Also in the link above the OP claims $A cap B= emptyset $ to conclude that homology group is trivial but $A cap B$ need not be empty or am I missing something?



    Would greatly appreciated if someone could shed a light on these.






    algebraic-topology







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      For two topological spaces $A$ and $B$, in order to show that $H(A sqcup B) cong H(A) oplus H(B) $
      in this question and in general I believe one can use Mayer-Vietoris to obtain the result easily. However, I don't quite understand how it applies in this case - what if $A sqcup B$ is not contained in the interior of $A$ and $B$?



      Also in the link above the OP claims $A cap B= emptyset $ to conclude that homology group is trivial but $A cap B$ need not be empty or am I missing something?



      Would greatly appreciated if someone could shed a light on these.






      algebraic-topology







      share|cite|improve this question









      $endgroup$




      For two topological spaces $A$ and $B$, in order to show that $H(A sqcup B) cong H(A) oplus H(B) $
      in this question and in general I believe one can use Mayer-Vietoris to obtain the result easily. However, I don't quite understand how it applies in this case - what if $A sqcup B$ is not contained in the interior of $A$ and $B$?



      Also in the link above the OP claims $A cap B= emptyset $ to conclude that homology group is trivial but $A cap B$ need not be empty or am I missing something?



      Would greatly appreciated if someone could shed a light on these.






      algebraic-topology




      algebraic-topology






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 16 '18 at 13:22









      user132770user132770

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          $begingroup$

          Maybe the notation caused some confusion: $A sqcup B$ is the disjoint union of $A$ and $B$.



          This implies that $A cap B = emptyset$, which immediately addresses your second point.



          Also, this implies that both $A$ and $B$ are open in $A sqcup B$, which means that the interior of $A$ (resp. of $B$) in $A sqcup B$ is $A$ itself (resp. $B$ itself). Hence the interiors of $A$ and $B$ cover $A sqcup B$, which means that the conditions for Mayer-Vietoris are satisfied.



          But I must say that you don't really need Mayer-Vietoris to compute $H^star (A sqcup B)$ - it's much better to work directly from the definition of homology...






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Arghh it's just that then. Many thanks.
            $endgroup$
            – user132770
            Dec 16 '18 at 13:39











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          $begingroup$

          Maybe the notation caused some confusion: $A sqcup B$ is the disjoint union of $A$ and $B$.



          This implies that $A cap B = emptyset$, which immediately addresses your second point.



          Also, this implies that both $A$ and $B$ are open in $A sqcup B$, which means that the interior of $A$ (resp. of $B$) in $A sqcup B$ is $A$ itself (resp. $B$ itself). Hence the interiors of $A$ and $B$ cover $A sqcup B$, which means that the conditions for Mayer-Vietoris are satisfied.



          But I must say that you don't really need Mayer-Vietoris to compute $H^star (A sqcup B)$ - it's much better to work directly from the definition of homology...






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Arghh it's just that then. Many thanks.
            $endgroup$
            – user132770
            Dec 16 '18 at 13:39
















          1












          $begingroup$

          Maybe the notation caused some confusion: $A sqcup B$ is the disjoint union of $A$ and $B$.



          This implies that $A cap B = emptyset$, which immediately addresses your second point.



          Also, this implies that both $A$ and $B$ are open in $A sqcup B$, which means that the interior of $A$ (resp. of $B$) in $A sqcup B$ is $A$ itself (resp. $B$ itself). Hence the interiors of $A$ and $B$ cover $A sqcup B$, which means that the conditions for Mayer-Vietoris are satisfied.



          But I must say that you don't really need Mayer-Vietoris to compute $H^star (A sqcup B)$ - it's much better to work directly from the definition of homology...






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Arghh it's just that then. Many thanks.
            $endgroup$
            – user132770
            Dec 16 '18 at 13:39














          1












          1








          1





          $begingroup$

          Maybe the notation caused some confusion: $A sqcup B$ is the disjoint union of $A$ and $B$.



          This implies that $A cap B = emptyset$, which immediately addresses your second point.



          Also, this implies that both $A$ and $B$ are open in $A sqcup B$, which means that the interior of $A$ (resp. of $B$) in $A sqcup B$ is $A$ itself (resp. $B$ itself). Hence the interiors of $A$ and $B$ cover $A sqcup B$, which means that the conditions for Mayer-Vietoris are satisfied.



          But I must say that you don't really need Mayer-Vietoris to compute $H^star (A sqcup B)$ - it's much better to work directly from the definition of homology...






          share|cite|improve this answer









          $endgroup$



          Maybe the notation caused some confusion: $A sqcup B$ is the disjoint union of $A$ and $B$.



          This implies that $A cap B = emptyset$, which immediately addresses your second point.



          Also, this implies that both $A$ and $B$ are open in $A sqcup B$, which means that the interior of $A$ (resp. of $B$) in $A sqcup B$ is $A$ itself (resp. $B$ itself). Hence the interiors of $A$ and $B$ cover $A sqcup B$, which means that the conditions for Mayer-Vietoris are satisfied.



          But I must say that you don't really need Mayer-Vietoris to compute $H^star (A sqcup B)$ - it's much better to work directly from the definition of homology...







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 16 '18 at 13:35









          Kenny WongKenny Wong

          19k21440




          19k21440












          • $begingroup$
            Arghh it's just that then. Many thanks.
            $endgroup$
            – user132770
            Dec 16 '18 at 13:39


















          • $begingroup$
            Arghh it's just that then. Many thanks.
            $endgroup$
            – user132770
            Dec 16 '18 at 13:39
















          $begingroup$
          Arghh it's just that then. Many thanks.
          $endgroup$
          – user132770
          Dec 16 '18 at 13:39




          $begingroup$
          Arghh it's just that then. Many thanks.
          $endgroup$
          – user132770
          Dec 16 '18 at 13:39


















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