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Computing Cohomology of $S^2/sim$, not Hausdorff
Clash Royale CLAN TAG#URR8PPP
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Fix the north and south poles $N, S$ of $S^2$. Given $G$, I want to compute the singular cohomology of $S^2/sim$, where $xsim y$ iff they lie on the same meridian between $N$ and $S$. We are then left with a space $X$ whose open sets are $X, Xbackslash N = U_S,Xbackslash S = U_N,$ and the open sets of $U_Ncap U_S simeq S^1$.
To calculate the cohomology I use Mayer-Vietoris:
$$
0 rightarrow H^0(X,G_X) simeq G rightarrow H^0(U_N,G_U_N) oplus H^0(U_S,G_U_S) simeq G oplus G rightarrow H^0(S^1, G_S^1) simeq G \
rightarrow H^1(X,G_X) rightarrow H^1(U_N,G_U_N) oplus H^1(U_S,G_U_S) rightarrow H^1(S^1, G_S^1) simeq G \
rightarrow H^2(X,G_X) rightarrow H^2(U_N,G_U_N) oplus H^2(U_S,G_U_S) rightarrow H^2(S^1, G_S^1) simeq 0
$$
I still need to show that the middle terms are all zero, i.e. that $ forall i>0, H^i(U_N,G_U_N) = H^i(U_S,G_U_S) = 0$.
Any hints? It doesn't seem obvious to me that $U_N$ is contractible.
algebraic-topology homology-cohomology
asked Jul 29 at 6:44
foaly
447414
add a comment |Â
up vote
1
down vote
favorite
Fix the north and south poles $N, S$ of $S^2$. Given $G$, I want to compute the singular cohomology of $S^2/sim$, where $xsim y$ iff they lie on the same meridian between $N$ and $S$. We are then left with a space $X$ whose open sets are $X, Xbackslash N = U_S,Xbackslash S = U_N,$ and the open sets of $U_Ncap U_S simeq S^1$.
To calculate the cohomology I use Mayer-Vietoris:
$$
0 rightarrow H^0(X,G_X) simeq G rightarrow H^0(U_N,G_U_N) oplus H^0(U_S,G_U_S) simeq G oplus G rightarrow H^0(S^1, G_S^1) simeq G \
rightarrow H^1(X,G_X) rightarrow H^1(U_N,G_U_N) oplus H^1(U_S,G_U_S) rightarrow H^1(S^1, G_S^1) simeq G \
rightarrow H^2(X,G_X) rightarrow H^2(U_N,G_U_N) oplus H^2(U_S,G_U_S) rightarrow H^2(S^1, G_S^1) simeq 0
$$
I still need to show that the middle terms are all zero, i.e. that $ forall i>0, H^i(U_N,G_U_N) = H^i(U_S,G_U_S) = 0$.
Any hints? It doesn't seem obvious to me that $U_N$ is contractible.
algebraic-topology homology-cohomology
asked Jul 29 at 6:44
foaly
447414
What's $G$?
– Lord Shark the Unknown
Jul 29 at 7:19
@LordSharktheUnknown an abelian group
– foaly
Jul 29 at 7:24
What's $G_U_N$ then?
– Lord Shark the Unknown
Jul 29 at 7:26
@LordSharktheUnknown I suppose I actually don't want the subscripts, but it doesn't make any difference since I found that $X$ is contractible, hence sheaf cohomology = singular cohomology.
– foaly
Jul 29 at 7:31
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Fix the north and south poles $N, S$ of $S^2$. Given $G$, I want to compute the singular cohomology of $S^2/sim$, where $xsim y$ iff they lie on the same meridian between $N$ and $S$. We are then left with a space $X$ whose open sets are $X, Xbackslash N = U_S,Xbackslash S = U_N,$ and the open sets of $U_Ncap U_S simeq S^1$.
To calculate the cohomology I use Mayer-Vietoris:
$$
0 rightarrow H^0(X,G_X) simeq G rightarrow H^0(U_N,G_U_N) oplus H^0(U_S,G_U_S) simeq G oplus G rightarrow H^0(S^1, G_S^1) simeq G \
rightarrow H^1(X,G_X) rightarrow H^1(U_N,G_U_N) oplus H^1(U_S,G_U_S) rightarrow H^1(S^1, G_S^1) simeq G \
rightarrow H^2(X,G_X) rightarrow H^2(U_N,G_U_N) oplus H^2(U_S,G_U_S) rightarrow H^2(S^1, G_S^1) simeq 0
$$
I still need to show that the middle terms are all zero, i.e. that $ forall i>0, H^i(U_N,G_U_N) = H^i(U_S,G_U_S) = 0$.
Any hints? It doesn't seem obvious to me that $U_N$ is contractible.
algebraic-topology homology-cohomology
asked Jul 29 at 6:44
foaly
447414
Fix the north and south poles $N, S$ of $S^2$. Given $G$, I want to compute the singular cohomology of $S^2/sim$, where $xsim y$ iff they lie on the same meridian between $N$ and $S$. We are then left with a space $X$ whose open sets are $X, Xbackslash N = U_S,Xbackslash S = U_N,$ and the open sets of $U_Ncap U_S simeq S^1$.
To calculate the cohomology I use Mayer-Vietoris:
$$
0 rightarrow H^0(X,G_X) simeq G rightarrow H^0(U_N,G_U_N) oplus H^0(U_S,G_U_S) simeq G oplus G rightarrow H^0(S^1, G_S^1) simeq G \
rightarrow H^1(X,G_X) rightarrow H^1(U_N,G_U_N) oplus H^1(U_S,G_U_S) rightarrow H^1(S^1, G_S^1) simeq G \
rightarrow H^2(X,G_X) rightarrow H^2(U_N,G_U_N) oplus H^2(U_S,G_U_S) rightarrow H^2(S^1, G_S^1) simeq 0
$$
I still need to show that the middle terms are all zero, i.e. that $ forall i>0, H^i(U_N,G_U_N) = H^i(U_S,G_U_S) = 0$.
Any hints? It doesn't seem obvious to me that $U_N$ is contractible.
algebraic-topology homology-cohomology
asked Jul 29 at 6:44
foaly
447414
edited Jul 29 at 6:58
asked Jul 29 at 6:44
foaly
447414
asked Jul 29 at 6:44
foaly
447414
asked Jul 29 at 6:44
foaly
447414
447414
What's $G$?
– Lord Shark the Unknown
Jul 29 at 7:19
@LordSharktheUnknown an abelian group
– foaly
Jul 29 at 7:24
What's $G_U_N$ then?
– Lord Shark the Unknown
Jul 29 at 7:26
@LordSharktheUnknown I suppose I actually don't want the subscripts, but it doesn't make any difference since I found that $X$ is contractible, hence sheaf cohomology = singular cohomology.
– foaly
Jul 29 at 7:31
add a comment |Â
What's $G$?
– Lord Shark the Unknown
Jul 29 at 7:19
@LordSharktheUnknown an abelian group
– foaly
Jul 29 at 7:24
What's $G_U_N$ then?
– Lord Shark the Unknown
Jul 29 at 7:26
@LordSharktheUnknown I suppose I actually don't want the subscripts, but it doesn't make any difference since I found that $X$ is contractible, hence sheaf cohomology = singular cohomology.
– foaly
Jul 29 at 7:31
What's $G$?
– Lord Shark the Unknown
Jul 29 at 7:19
What's $G$?
– Lord Shark the Unknown
Jul 29 at 7:19
@LordSharktheUnknown an abelian group
– foaly
Jul 29 at 7:24
@LordSharktheUnknown an abelian group
– foaly
Jul 29 at 7:24
What's $G_U_N$ then?
– Lord Shark the Unknown
Jul 29 at 7:26
What's $G_U_N$ then?
– Lord Shark the Unknown
Jul 29 at 7:26
@LordSharktheUnknown I suppose I actually don't want the subscripts, but it doesn't make any difference since I found that $X$ is contractible, hence sheaf cohomology = singular cohomology.
– foaly
Jul 29 at 7:31
@LordSharktheUnknown I suppose I actually don't want the subscripts, but it doesn't make any difference since I found that $X$ is contractible, hence sheaf cohomology = singular cohomology.
– foaly
Jul 29 at 7:31
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
NVM -- induced by the null-homotopy on Euclidean space we have a null-homotopy on $U_N$. (Check that the induced map is continuous.)
answered Jul 29 at 7:24
foaly
447414
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
NVM -- induced by the null-homotopy on Euclidean space we have a null-homotopy on $U_N$. (Check that the induced map is continuous.)
answered Jul 29 at 7:24
foaly
447414
add a comment |Â
up vote
1
down vote
accepted
NVM -- induced by the null-homotopy on Euclidean space we have a null-homotopy on $U_N$. (Check that the induced map is continuous.)
answered Jul 29 at 7:24
foaly
447414
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
NVM -- induced by the null-homotopy on Euclidean space we have a null-homotopy on $U_N$. (Check that the induced map is continuous.)
answered Jul 29 at 7:24
foaly
447414
NVM -- induced by the null-homotopy on Euclidean space we have a null-homotopy on $U_N$. (Check that the induced map is continuous.)
answered Jul 29 at 7:24
foaly
447414
answered Jul 29 at 7:24
foaly
447414
answered Jul 29 at 7:24
foaly
447414
answered Jul 29 at 7:24
foaly
447414
447414
add a comment |Â
add a comment |Â
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What's $G$?
– Lord Shark the Unknown
Jul 29 at 7:19
@LordSharktheUnknown an abelian group
– foaly
Jul 29 at 7:24
What's $G_U_N$ then?
– Lord Shark the Unknown
Jul 29 at 7:26
@LordSharktheUnknown I suppose I actually don't want the subscripts, but it doesn't make any difference since I found that $X$ is contractible, hence sheaf cohomology = singular cohomology.
– foaly
Jul 29 at 7:31