Successive maps in exact sequence leads to $0$. Celluar Homology
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This comes from Hatcher's Algebraic Topology book on page 139
He says that the map in the diagram
$$0 to H_n(X^n) stackrel{j_n}to H_n(X^n, X^{n-1}) stackrel{partial_n}to H_{n-1}(X^{n-1})to?$$
The map $d_nd_{n+1}= j_{n+1}(partial _nj_n)partial_{n+1} = 0$ because of $partial _nj_n = 0$ due to exactness. I don't know which exact sequence he is referring to in his diagram
general-topology algebraic-topology proof-explanation homology-cohomology homological-algebra
asked Nov 19 at 8:20
Hawk
5,4381138102
|
show 10 more comments
up vote
1
down vote
favorite
This comes from Hatcher's Algebraic Topology book on page 139
He says that the map in the diagram
$$0 to H_n(X^n) stackrel{j_n}to H_n(X^n, X^{n-1}) stackrel{partial_n}to H_{n-1}(X^{n-1})to?$$
The map $d_nd_{n+1}= j_{n+1}(partial _nj_n)partial_{n+1} = 0$ because of $partial _nj_n = 0$ due to exactness. I don't know which exact sequence he is referring to in his diagram
general-topology algebraic-topology proof-explanation homology-cohomology homological-algebra
asked Nov 19 at 8:20
Hawk
5,4381138102
He is referring to the exact sequence you have written on the 3rd line. E.g. the one of the pair $(X^n,X^{n-1})$.
– Tyrone
Nov 19 at 9:46
@Tyrone, yeah but that is where I am not 100% sure why that sequence ends in $0$.
– Hawk
Nov 19 at 9:47
Recall that $X^{n-1}$ is the $(n-1)$-skeleton of $X$, so without $n$-cells you have $H_nX^{n-1}=0$, giving that $j_n$ is injective. The group you have labelled $?$ will be $H_{n-1}X^{n}$.
– Tyrone
Nov 19 at 9:48
@Tyrone you are referring to $H_n(X^{n},X^{n-1})$ right? Because $H_n(X^{n-1}) approx H_n(X)$
– Hawk
Nov 19 at 9:50
@Tyrone, the group before $?$ is $H_{n-1}(X^{n-1})$
– Hawk
Nov 19 at 9:51
|
show 10 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
This comes from Hatcher's Algebraic Topology book on page 139
He says that the map in the diagram
$$0 to H_n(X^n) stackrel{j_n}to H_n(X^n, X^{n-1}) stackrel{partial_n}to H_{n-1}(X^{n-1})to?$$
The map $d_nd_{n+1}= j_{n+1}(partial _nj_n)partial_{n+1} = 0$ because of $partial _nj_n = 0$ due to exactness. I don't know which exact sequence he is referring to in his diagram
general-topology algebraic-topology proof-explanation homology-cohomology homological-algebra
asked Nov 19 at 8:20
Hawk
5,4381138102
This comes from Hatcher's Algebraic Topology book on page 139
He says that the map in the diagram
$$0 to H_n(X^n) stackrel{j_n}to H_n(X^n, X^{n-1}) stackrel{partial_n}to H_{n-1}(X^{n-1})to?$$
The map $d_nd_{n+1}= j_{n+1}(partial _nj_n)partial_{n+1} = 0$ because of $partial _nj_n = 0$ due to exactness. I don't know which exact sequence he is referring to in his diagram
general-topology algebraic-topology proof-explanation homology-cohomology homological-algebra
general-topology algebraic-topology proof-explanation homology-cohomology homological-algebra
asked Nov 19 at 8:20
Hawk
5,4381138102
asked Nov 19 at 8:20
Hawk
5,4381138102
asked Nov 19 at 8:20
Hawk
5,4381138102
asked Nov 19 at 8:20
Hawk
5,4381138102
asked Nov 19 at 8:20
Hawk
5,4381138102
5,4381138102
He is referring to the exact sequence you have written on the 3rd line. E.g. the one of the pair $(X^n,X^{n-1})$.
– Tyrone
Nov 19 at 9:46
@Tyrone, yeah but that is where I am not 100% sure why that sequence ends in $0$.
– Hawk
Nov 19 at 9:47
Recall that $X^{n-1}$ is the $(n-1)$-skeleton of $X$, so without $n$-cells you have $H_nX^{n-1}=0$, giving that $j_n$ is injective. The group you have labelled $?$ will be $H_{n-1}X^{n}$.
– Tyrone
Nov 19 at 9:48
@Tyrone you are referring to $H_n(X^{n},X^{n-1})$ right? Because $H_n(X^{n-1}) approx H_n(X)$
– Hawk
Nov 19 at 9:50
@Tyrone, the group before $?$ is $H_{n-1}(X^{n-1})$
– Hawk
Nov 19 at 9:51
|
show 10 more comments
He is referring to the exact sequence you have written on the 3rd line. E.g. the one of the pair $(X^n,X^{n-1})$.
– Tyrone
Nov 19 at 9:46
@Tyrone, yeah but that is where I am not 100% sure why that sequence ends in $0$.
– Hawk
Nov 19 at 9:47
Recall that $X^{n-1}$ is the $(n-1)$-skeleton of $X$, so without $n$-cells you have $H_nX^{n-1}=0$, giving that $j_n$ is injective. The group you have labelled $?$ will be $H_{n-1}X^{n}$.
– Tyrone
Nov 19 at 9:48
@Tyrone you are referring to $H_n(X^{n},X^{n-1})$ right? Because $H_n(X^{n-1}) approx H_n(X)$
– Hawk
Nov 19 at 9:50
@Tyrone, the group before $?$ is $H_{n-1}(X^{n-1})$
– Hawk
Nov 19 at 9:51
He is referring to the exact sequence you have written on the 3rd line. E.g. the one of the pair $(X^n,X^{n-1})$.
– Tyrone
Nov 19 at 9:46
He is referring to the exact sequence you have written on the 3rd line. E.g. the one of the pair $(X^n,X^{n-1})$.
– Tyrone
Nov 19 at 9:46
@Tyrone, yeah but that is where I am not 100% sure why that sequence ends in $0$.
– Hawk
Nov 19 at 9:47
@Tyrone, yeah but that is where I am not 100% sure why that sequence ends in $0$.
– Hawk
Nov 19 at 9:47
Recall that $X^{n-1}$ is the $(n-1)$-skeleton of $X$, so without $n$-cells you have $H_nX^{n-1}=0$, giving that $j_n$ is injective. The group you have labelled $?$ will be $H_{n-1}X^{n}$.
– Tyrone
Nov 19 at 9:48
Recall that $X^{n-1}$ is the $(n-1)$-skeleton of $X$, so without $n$-cells you have $H_nX^{n-1}=0$, giving that $j_n$ is injective. The group you have labelled $?$ will be $H_{n-1}X^{n}$.
– Tyrone
Nov 19 at 9:48
@Tyrone you are referring to $H_n(X^{n},X^{n-1})$ right? Because $H_n(X^{n-1}) approx H_n(X)$
– Hawk
Nov 19 at 9:50
@Tyrone you are referring to $H_n(X^{n},X^{n-1})$ right? Because $H_n(X^{n-1}) approx H_n(X)$
– Hawk
Nov 19 at 9:50
@Tyrone, the group before $?$ is $H_{n-1}(X^{n-1})$
– Hawk
Nov 19 at 9:51
@Tyrone, the group before $?$ is $H_{n-1}(X^{n-1})$
– Hawk
Nov 19 at 9:51
|
show 10 more comments
1 Answer
1
active
oldest
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up vote
1
down vote
Since $(X^n, X^{n-1})$ is a good pair, it induces the long exact sequence
$0 to H_n(X^n) to H_n(X^n, X^{n-1}) to H_{n-1}(X^{n-1}) to H_{n-1}(X^n) to dots$
Definition of exactness leads to $img j_n = ker partial_n$, so $partial_n(img j_n) = partial_n(ker partial_n) = 0 $
answered Nov 22 at 0:48
Hawk
5,4381138102
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Since $(X^n, X^{n-1})$ is a good pair, it induces the long exact sequence
$0 to H_n(X^n) to H_n(X^n, X^{n-1}) to H_{n-1}(X^{n-1}) to H_{n-1}(X^n) to dots$
Definition of exactness leads to $img j_n = ker partial_n$, so $partial_n(img j_n) = partial_n(ker partial_n) = 0 $
answered Nov 22 at 0:48
Hawk
5,4381138102
add a comment |
up vote
1
down vote
Since $(X^n, X^{n-1})$ is a good pair, it induces the long exact sequence
$0 to H_n(X^n) to H_n(X^n, X^{n-1}) to H_{n-1}(X^{n-1}) to H_{n-1}(X^n) to dots$
Definition of exactness leads to $img j_n = ker partial_n$, so $partial_n(img j_n) = partial_n(ker partial_n) = 0 $
answered Nov 22 at 0:48
Hawk
5,4381138102
add a comment |
up vote
1
down vote
up vote
1
down vote
Since $(X^n, X^{n-1})$ is a good pair, it induces the long exact sequence
$0 to H_n(X^n) to H_n(X^n, X^{n-1}) to H_{n-1}(X^{n-1}) to H_{n-1}(X^n) to dots$
Definition of exactness leads to $img j_n = ker partial_n$, so $partial_n(img j_n) = partial_n(ker partial_n) = 0 $
answered Nov 22 at 0:48
Hawk
5,4381138102
Since $(X^n, X^{n-1})$ is a good pair, it induces the long exact sequence
$0 to H_n(X^n) to H_n(X^n, X^{n-1}) to H_{n-1}(X^{n-1}) to H_{n-1}(X^n) to dots$
Definition of exactness leads to $img j_n = ker partial_n$, so $partial_n(img j_n) = partial_n(ker partial_n) = 0 $
answered Nov 22 at 0:48
Hawk
5,4381138102
answered Nov 22 at 0:48
Hawk
5,4381138102
answered Nov 22 at 0:48
Hawk
5,4381138102
answered Nov 22 at 0:48
Hawk
5,4381138102
5,4381138102
add a comment |
add a comment |
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He is referring to the exact sequence you have written on the 3rd line. E.g. the one of the pair $(X^n,X^{n-1})$.
– Tyrone
Nov 19 at 9:46
@Tyrone, yeah but that is where I am not 100% sure why that sequence ends in $0$.
– Hawk
Nov 19 at 9:47
Recall that $X^{n-1}$ is the $(n-1)$-skeleton of $X$, so without $n$-cells you have $H_nX^{n-1}=0$, giving that $j_n$ is injective. The group you have labelled $?$ will be $H_{n-1}X^{n}$.
– Tyrone
Nov 19 at 9:48
@Tyrone you are referring to $H_n(X^{n},X^{n-1})$ right? Because $H_n(X^{n-1}) approx H_n(X)$
– Hawk
Nov 19 at 9:50
@Tyrone, the group before $?$ is $H_{n-1}(X^{n-1})$
– Hawk
Nov 19 at 9:51