Confusion in using Mayer-Vietoris theorem
$begingroup$
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
asked Dec 16 '18 at 13:22
user132770user132770
496
$endgroup$
add a comment |
$begingroup$
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
asked Dec 16 '18 at 13:22
user132770user132770
496
$endgroup$
add a comment |
$begingroup$
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
asked Dec 16 '18 at 13:22
user132770user132770
496
$endgroup$
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
asked Dec 16 '18 at 13:22
user132770user132770
496
asked Dec 16 '18 at 13:22
user132770user132770
496
asked Dec 16 '18 at 13:22
user132770user132770
496
asked Dec 16 '18 at 13:22
user132770user132770
496
asked Dec 16 '18 at 13:22
user132770user132770
496
496
add a comment |
add a comment |
1 Answer
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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...
answered Dec 16 '18 at 13:35
Kenny WongKenny Wong
18.5k21438
$endgroup$
$begingroup$
Arghh it's just that then. Many thanks.
$endgroup$
– user132770
Dec 16 '18 at 13:39
add a comment |
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1 Answer
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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...
answered Dec 16 '18 at 13:35
Kenny WongKenny Wong
18.5k21438
$endgroup$
$begingroup$
Arghh it's just that then. Many thanks.
$endgroup$
– user132770
Dec 16 '18 at 13:39
add a comment |
$begingroup$
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...
answered Dec 16 '18 at 13:35
Kenny WongKenny Wong
18.5k21438
$endgroup$
$begingroup$
Arghh it's just that then. Many thanks.
$endgroup$
– user132770
Dec 16 '18 at 13:39
add a comment |
$begingroup$
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...
answered Dec 16 '18 at 13:35
Kenny WongKenny Wong
18.5k21438
$endgroup$
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...
answered Dec 16 '18 at 13:35
Kenny WongKenny Wong
18.5k21438
answered Dec 16 '18 at 13:35
Kenny WongKenny Wong
18.5k21438
answered Dec 16 '18 at 13:35
Kenny WongKenny Wong
18.5k21438
answered Dec 16 '18 at 13:35
Kenny WongKenny Wong
18.5k21438
18.5k21438
$begingroup$
Arghh it's just that then. Many thanks.
$endgroup$
– user132770
Dec 16 '18 at 13:39
add a comment |
$begingroup$
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
add a comment |
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