Mixing
Successive maps in exact sequence leads to $0$. Celluar Homology
up vote
0
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 4 hours ago
Hawk
5,4161138102
|
show 10 more comments
up vote
0
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 4 hours ago
Hawk
5,4161138102
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
3 hours ago
@Tyrone, yeah but that is where I am not 100% sure why that sequence ends in $0$.
– Hawk
3 hours ago
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
3 hours ago
@Tyrone you are referring to $H_n(X^{n},X^{n-1})$ right? Because $H_n(X^{n-1}) approx H_n(X)$
– Hawk
3 hours ago
@Tyrone, the group before $?$ is $H_{n-1}(X^{n-1})$
– Hawk
3 hours ago
|
show 10 more comments
up vote
0
down vote
favorite
up vote
0
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 4 hours ago
Hawk
5,4161138102
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 4 hours ago
Hawk
5,4161138102
asked 4 hours ago
Hawk
5,4161138102
asked 4 hours ago
Hawk
5,4161138102
asked 4 hours ago
Hawk
5,4161138102
asked 4 hours ago
Hawk
5,4161138102
5,4161138102
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
3 hours ago
@Tyrone, yeah but that is where I am not 100% sure why that sequence ends in $0$.
– Hawk
3 hours ago
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
3 hours ago
@Tyrone you are referring to $H_n(X^{n},X^{n-1})$ right? Because $H_n(X^{n-1}) approx H_n(X)$
– Hawk
3 hours ago
@Tyrone, the group before $?$ is $H_{n-1}(X^{n-1})$
– Hawk
3 hours ago
|
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
3 hours ago
@Tyrone, yeah but that is where I am not 100% sure why that sequence ends in $0$.
– Hawk
3 hours ago
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
3 hours ago
@Tyrone you are referring to $H_n(X^{n},X^{n-1})$ right? Because $H_n(X^{n-1}) approx H_n(X)$
– Hawk
3 hours ago
@Tyrone, the group before $?$ is $H_{n-1}(X^{n-1})$
– Hawk
3 hours ago
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
3 hours ago
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
3 hours ago
@Tyrone, yeah but that is where I am not 100% sure why that sequence ends in $0$.
– Hawk
3 hours ago
@Tyrone, yeah but that is where I am not 100% sure why that sequence ends in $0$.
– Hawk
3 hours ago
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
3 hours ago
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
3 hours ago
@Tyrone you are referring to $H_n(X^{n},X^{n-1})$ right? Because $H_n(X^{n-1}) approx H_n(X)$
– Hawk
3 hours ago
@Tyrone you are referring to $H_n(X^{n},X^{n-1})$ right? Because $H_n(X^{n-1}) approx H_n(X)$
– Hawk
3 hours ago
@Tyrone, the group before $?$ is $H_{n-1}(X^{n-1})$
– Hawk
3 hours ago
@Tyrone, the group before $?$ is $H_{n-1}(X^{n-1})$
– Hawk
3 hours ago
|
show 10 more comments
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3004649%2fsuccessive-maps-in-exact-sequence-leads-to-0-celluar-homology%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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
3 hours ago
@Tyrone, yeah but that is where I am not 100% sure why that sequence ends in $0$.
– Hawk
3 hours ago
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
3 hours ago
@Tyrone you are referring to $H_n(X^{n},X^{n-1})$ right? Because $H_n(X^{n-1}) approx H_n(X)$
– Hawk
3 hours ago
@Tyrone, the group before $?$ is $H_{n-1}(X^{n-1})$
– Hawk
3 hours ago