$Xsubset S^n$ is a neighborhood retract, then $Xto S^n$ inclusion map is nullhomotopic
$begingroup$
Set $ngeq 1$. Suppose $Xsubset S^n$ is a neighborhood retract where $S^n$ is $n-$sphere. $Xsubset Y$ is neighborhood retract if there is an open subset/neighborhood of $Y$ containing $X$ with $X$ as its retraction.
Denote $U$ as the neighborhood of $X$ as above for retraction. Then $H(U)=H(X)oplus H(U,X)$. The claim is that $H_n(U)=0$. It suffices to look at $H_n(U,p)=tilde{H}_n(U)=H_n(U)$ where $tilde{H}$ is reduced homology and last step is by couple $(U,p)$ long exact sequence. WLOG $U$ is connected. Since $U$ is open, there is a even smaller open $B_p$ ball around $p$. Now $H_n(U,p)=H_n(U,B_p)$ where first step is by deformation retraction. How do I see $H_n(U,B_p)=0$?
$textbf{Q:}$ I do not see how to proceed further as I do not have anything to cut off. I am being sloppy here. A lot of $=$ should be $cong$.
$textbf{Q':}$ The book says "because $Xneq S^n$, inclusion map $Xto S^n$ is nullhomotopic". How is so obvious? Am I missing something here?
Ref. Dold, Algebraic Topology Chpt IV, Sec 6, Cor 6.5 pg 73.
abstract-algebra general-topology algebraic-topology
asked Dec 26 '18 at 18:08
user45765user45765
2,6702724
$endgroup$
add a comment |
$begingroup$
Set $ngeq 1$. Suppose $Xsubset S^n$ is a neighborhood retract where $S^n$ is $n-$sphere. $Xsubset Y$ is neighborhood retract if there is an open subset/neighborhood of $Y$ containing $X$ with $X$ as its retraction.
Denote $U$ as the neighborhood of $X$ as above for retraction. Then $H(U)=H(X)oplus H(U,X)$. The claim is that $H_n(U)=0$. It suffices to look at $H_n(U,p)=tilde{H}_n(U)=H_n(U)$ where $tilde{H}$ is reduced homology and last step is by couple $(U,p)$ long exact sequence. WLOG $U$ is connected. Since $U$ is open, there is a even smaller open $B_p$ ball around $p$. Now $H_n(U,p)=H_n(U,B_p)$ where first step is by deformation retraction. How do I see $H_n(U,B_p)=0$?
$textbf{Q:}$ I do not see how to proceed further as I do not have anything to cut off. I am being sloppy here. A lot of $=$ should be $cong$.
$textbf{Q':}$ The book says "because $Xneq S^n$, inclusion map $Xto S^n$ is nullhomotopic". How is so obvious? Am I missing something here?
Ref. Dold, Algebraic Topology Chpt IV, Sec 6, Cor 6.5 pg 73.
abstract-algebra general-topology algebraic-topology
asked Dec 26 '18 at 18:08
user45765user45765
2,6702724
$endgroup$
add a comment |
$begingroup$
Set $ngeq 1$. Suppose $Xsubset S^n$ is a neighborhood retract where $S^n$ is $n-$sphere. $Xsubset Y$ is neighborhood retract if there is an open subset/neighborhood of $Y$ containing $X$ with $X$ as its retraction.
Denote $U$ as the neighborhood of $X$ as above for retraction. Then $H(U)=H(X)oplus H(U,X)$. The claim is that $H_n(U)=0$. It suffices to look at $H_n(U,p)=tilde{H}_n(U)=H_n(U)$ where $tilde{H}$ is reduced homology and last step is by couple $(U,p)$ long exact sequence. WLOG $U$ is connected. Since $U$ is open, there is a even smaller open $B_p$ ball around $p$. Now $H_n(U,p)=H_n(U,B_p)$ where first step is by deformation retraction. How do I see $H_n(U,B_p)=0$?
$textbf{Q:}$ I do not see how to proceed further as I do not have anything to cut off. I am being sloppy here. A lot of $=$ should be $cong$.
$textbf{Q':}$ The book says "because $Xneq S^n$, inclusion map $Xto S^n$ is nullhomotopic". How is so obvious? Am I missing something here?
Ref. Dold, Algebraic Topology Chpt IV, Sec 6, Cor 6.5 pg 73.
abstract-algebra general-topology algebraic-topology
asked Dec 26 '18 at 18:08
user45765user45765
2,6702724
$endgroup$
Set $ngeq 1$. Suppose $Xsubset S^n$ is a neighborhood retract where $S^n$ is $n-$sphere. $Xsubset Y$ is neighborhood retract if there is an open subset/neighborhood of $Y$ containing $X$ with $X$ as its retraction.
Denote $U$ as the neighborhood of $X$ as above for retraction. Then $H(U)=H(X)oplus H(U,X)$. The claim is that $H_n(U)=0$. It suffices to look at $H_n(U,p)=tilde{H}_n(U)=H_n(U)$ where $tilde{H}$ is reduced homology and last step is by couple $(U,p)$ long exact sequence. WLOG $U$ is connected. Since $U$ is open, there is a even smaller open $B_p$ ball around $p$. Now $H_n(U,p)=H_n(U,B_p)$ where first step is by deformation retraction. How do I see $H_n(U,B_p)=0$?
$textbf{Q:}$ I do not see how to proceed further as I do not have anything to cut off. I am being sloppy here. A lot of $=$ should be $cong$.
$textbf{Q':}$ The book says "because $Xneq S^n$, inclusion map $Xto S^n$ is nullhomotopic". How is so obvious? Am I missing something here?
Ref. Dold, Algebraic Topology Chpt IV, Sec 6, Cor 6.5 pg 73.
abstract-algebra general-topology algebraic-topology
abstract-algebra general-topology algebraic-topology
asked Dec 26 '18 at 18:08
user45765user45765
2,6702724
asked Dec 26 '18 at 18:08
user45765user45765
2,6702724
asked Dec 26 '18 at 18:08
user45765user45765
2,6702724
asked Dec 26 '18 at 18:08
user45765user45765
2,6702724
asked Dec 26 '18 at 18:08
user45765user45765
2,6702724
2,6702724
add a comment |
add a comment |
2 Answers
2
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oldest
votes
$begingroup$
It is wrong if $X = S^n$ (it is well-known that the identity on $S^n$ is not nullhomotopic).
If $X subsetneqq S^n$, choose any point $x in S^n setminus X$. Then $R = S^n setminus { x }$ is homeomorphic to $mathbb{R}^n$ which is contractible, therefore the inclusion $X to R$ is nullhomotopic. Hence also the inclusion $X to S^n$ is nullhomotopic.
Note that this is true for any $X subsetneqq S^n$.
Added:
For any open $U subset S^n$ we have $H_i(S^n,U) = 0$ for $i > n$. This is is a special case of Proposition 6.4 in Dold. The proof is not at all trivial. It begins on p. 73 (Lemma 6.7) and ends on p.77.
Accepting $H_i(S^n,U) = 0$ for $i > n$, we see that the long exact sequence of $(S^n,U)$ contains the part $H_{n+1}(S^n,U) to H_n(U) to H_n(S^n)$. Since the first group is $0$, we see that $i_* : H_n(U) to H_n(S^n)$ is a monomorphism. In particular $H_n(U) approx text{im}(i_*)$. But if $U subsetneqq S^n$, then the inclusion $i : U to S^n$ is nullhomotopic. Hence $i_*$ is the zero homomorphism and therefore $text{im}(i_*) = 0$. We conclude $H_n(U) = 0$.
Note that the idea sketched in your question does not work. It is really hard work to prove $H_n(U) = 0$.
answered Dec 26 '18 at 18:25
Paul FrostPaul Frost
11.6k3934
$endgroup$
$begingroup$
I see. I was thinking about using excision or something to see the map factors through homology of something with $0$ top homology. I forgot it means that one can move the map in the target.
$endgroup$
– user45765
Dec 26 '18 at 18:28
$begingroup$
So I cannot say top homology of open sets trivial in $S^n$.
$endgroup$
– user45765
Dec 26 '18 at 18:28
$begingroup$
You can prove this. I shall edit my answer..
$endgroup$
– Paul Frost
Dec 26 '18 at 18:40
add a comment |
$begingroup$
Image of the inclusion of X misses atleast one point of the sphere. And sphere - {one point} is just Euclidean space which is contractible. Now any map in a contractible space is null homotopic ( check this). And you are there!
Ps: You have not mentioned X is proper subset though.
answered Dec 26 '18 at 18:23
NeelNeel
567314
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
It is wrong if $X = S^n$ (it is well-known that the identity on $S^n$ is not nullhomotopic).
If $X subsetneqq S^n$, choose any point $x in S^n setminus X$. Then $R = S^n setminus { x }$ is homeomorphic to $mathbb{R}^n$ which is contractible, therefore the inclusion $X to R$ is nullhomotopic. Hence also the inclusion $X to S^n$ is nullhomotopic.
Note that this is true for any $X subsetneqq S^n$.
Added:
For any open $U subset S^n$ we have $H_i(S^n,U) = 0$ for $i > n$. This is is a special case of Proposition 6.4 in Dold. The proof is not at all trivial. It begins on p. 73 (Lemma 6.7) and ends on p.77.
Accepting $H_i(S^n,U) = 0$ for $i > n$, we see that the long exact sequence of $(S^n,U)$ contains the part $H_{n+1}(S^n,U) to H_n(U) to H_n(S^n)$. Since the first group is $0$, we see that $i_* : H_n(U) to H_n(S^n)$ is a monomorphism. In particular $H_n(U) approx text{im}(i_*)$. But if $U subsetneqq S^n$, then the inclusion $i : U to S^n$ is nullhomotopic. Hence $i_*$ is the zero homomorphism and therefore $text{im}(i_*) = 0$. We conclude $H_n(U) = 0$.
Note that the idea sketched in your question does not work. It is really hard work to prove $H_n(U) = 0$.
answered Dec 26 '18 at 18:25
Paul FrostPaul Frost
11.6k3934
$endgroup$
$begingroup$
I see. I was thinking about using excision or something to see the map factors through homology of something with $0$ top homology. I forgot it means that one can move the map in the target.
$endgroup$
– user45765
Dec 26 '18 at 18:28
$begingroup$
So I cannot say top homology of open sets trivial in $S^n$.
$endgroup$
– user45765
Dec 26 '18 at 18:28
$begingroup$
You can prove this. I shall edit my answer..
$endgroup$
– Paul Frost
Dec 26 '18 at 18:40
add a comment |
$begingroup$
It is wrong if $X = S^n$ (it is well-known that the identity on $S^n$ is not nullhomotopic).
If $X subsetneqq S^n$, choose any point $x in S^n setminus X$. Then $R = S^n setminus { x }$ is homeomorphic to $mathbb{R}^n$ which is contractible, therefore the inclusion $X to R$ is nullhomotopic. Hence also the inclusion $X to S^n$ is nullhomotopic.
Note that this is true for any $X subsetneqq S^n$.
Added:
For any open $U subset S^n$ we have $H_i(S^n,U) = 0$ for $i > n$. This is is a special case of Proposition 6.4 in Dold. The proof is not at all trivial. It begins on p. 73 (Lemma 6.7) and ends on p.77.
Accepting $H_i(S^n,U) = 0$ for $i > n$, we see that the long exact sequence of $(S^n,U)$ contains the part $H_{n+1}(S^n,U) to H_n(U) to H_n(S^n)$. Since the first group is $0$, we see that $i_* : H_n(U) to H_n(S^n)$ is a monomorphism. In particular $H_n(U) approx text{im}(i_*)$. But if $U subsetneqq S^n$, then the inclusion $i : U to S^n$ is nullhomotopic. Hence $i_*$ is the zero homomorphism and therefore $text{im}(i_*) = 0$. We conclude $H_n(U) = 0$.
Note that the idea sketched in your question does not work. It is really hard work to prove $H_n(U) = 0$.
answered Dec 26 '18 at 18:25
Paul FrostPaul Frost
11.6k3934
$endgroup$
$begingroup$
I see. I was thinking about using excision or something to see the map factors through homology of something with $0$ top homology. I forgot it means that one can move the map in the target.
$endgroup$
– user45765
Dec 26 '18 at 18:28
$begingroup$
So I cannot say top homology of open sets trivial in $S^n$.
$endgroup$
– user45765
Dec 26 '18 at 18:28
$begingroup$
You can prove this. I shall edit my answer..
$endgroup$
– Paul Frost
Dec 26 '18 at 18:40
add a comment |
$begingroup$
It is wrong if $X = S^n$ (it is well-known that the identity on $S^n$ is not nullhomotopic).
If $X subsetneqq S^n$, choose any point $x in S^n setminus X$. Then $R = S^n setminus { x }$ is homeomorphic to $mathbb{R}^n$ which is contractible, therefore the inclusion $X to R$ is nullhomotopic. Hence also the inclusion $X to S^n$ is nullhomotopic.
Note that this is true for any $X subsetneqq S^n$.
Added:
For any open $U subset S^n$ we have $H_i(S^n,U) = 0$ for $i > n$. This is is a special case of Proposition 6.4 in Dold. The proof is not at all trivial. It begins on p. 73 (Lemma 6.7) and ends on p.77.
Accepting $H_i(S^n,U) = 0$ for $i > n$, we see that the long exact sequence of $(S^n,U)$ contains the part $H_{n+1}(S^n,U) to H_n(U) to H_n(S^n)$. Since the first group is $0$, we see that $i_* : H_n(U) to H_n(S^n)$ is a monomorphism. In particular $H_n(U) approx text{im}(i_*)$. But if $U subsetneqq S^n$, then the inclusion $i : U to S^n$ is nullhomotopic. Hence $i_*$ is the zero homomorphism and therefore $text{im}(i_*) = 0$. We conclude $H_n(U) = 0$.
Note that the idea sketched in your question does not work. It is really hard work to prove $H_n(U) = 0$.
answered Dec 26 '18 at 18:25
Paul FrostPaul Frost
11.6k3934
$endgroup$
It is wrong if $X = S^n$ (it is well-known that the identity on $S^n$ is not nullhomotopic).
If $X subsetneqq S^n$, choose any point $x in S^n setminus X$. Then $R = S^n setminus { x }$ is homeomorphic to $mathbb{R}^n$ which is contractible, therefore the inclusion $X to R$ is nullhomotopic. Hence also the inclusion $X to S^n$ is nullhomotopic.
Note that this is true for any $X subsetneqq S^n$.
Added:
For any open $U subset S^n$ we have $H_i(S^n,U) = 0$ for $i > n$. This is is a special case of Proposition 6.4 in Dold. The proof is not at all trivial. It begins on p. 73 (Lemma 6.7) and ends on p.77.
Accepting $H_i(S^n,U) = 0$ for $i > n$, we see that the long exact sequence of $(S^n,U)$ contains the part $H_{n+1}(S^n,U) to H_n(U) to H_n(S^n)$. Since the first group is $0$, we see that $i_* : H_n(U) to H_n(S^n)$ is a monomorphism. In particular $H_n(U) approx text{im}(i_*)$. But if $U subsetneqq S^n$, then the inclusion $i : U to S^n$ is nullhomotopic. Hence $i_*$ is the zero homomorphism and therefore $text{im}(i_*) = 0$. We conclude $H_n(U) = 0$.
Note that the idea sketched in your question does not work. It is really hard work to prove $H_n(U) = 0$.
answered Dec 26 '18 at 18:25
Paul FrostPaul Frost
11.6k3934
edited Dec 26 '18 at 23:06
answered Dec 26 '18 at 18:25
Paul FrostPaul Frost
11.6k3934
answered Dec 26 '18 at 18:25
Paul FrostPaul Frost
11.6k3934
answered Dec 26 '18 at 18:25
Paul FrostPaul Frost
11.6k3934
11.6k3934
$begingroup$
I see. I was thinking about using excision or something to see the map factors through homology of something with $0$ top homology. I forgot it means that one can move the map in the target.
$endgroup$
– user45765
Dec 26 '18 at 18:28
$begingroup$
So I cannot say top homology of open sets trivial in $S^n$.
$endgroup$
– user45765
Dec 26 '18 at 18:28
$begingroup$
You can prove this. I shall edit my answer..
$endgroup$
– Paul Frost
Dec 26 '18 at 18:40
add a comment |
$begingroup$
I see. I was thinking about using excision or something to see the map factors through homology of something with $0$ top homology. I forgot it means that one can move the map in the target.
$endgroup$
– user45765
Dec 26 '18 at 18:28
$begingroup$
So I cannot say top homology of open sets trivial in $S^n$.
$endgroup$
– user45765
Dec 26 '18 at 18:28
$begingroup$
You can prove this. I shall edit my answer..
$endgroup$
– Paul Frost
Dec 26 '18 at 18:40
$begingroup$
I see. I was thinking about using excision or something to see the map factors through homology of something with $0$ top homology. I forgot it means that one can move the map in the target.
$endgroup$
– user45765
Dec 26 '18 at 18:28
$begingroup$
I see. I was thinking about using excision or something to see the map factors through homology of something with $0$ top homology. I forgot it means that one can move the map in the target.
$endgroup$
– user45765
Dec 26 '18 at 18:28
$begingroup$
So I cannot say top homology of open sets trivial in $S^n$.
$endgroup$
– user45765
Dec 26 '18 at 18:28
$begingroup$
So I cannot say top homology of open sets trivial in $S^n$.
$endgroup$
– user45765
Dec 26 '18 at 18:28
$begingroup$
You can prove this. I shall edit my answer..
$endgroup$
– Paul Frost
Dec 26 '18 at 18:40
$begingroup$
You can prove this. I shall edit my answer..
$endgroup$
– Paul Frost
Dec 26 '18 at 18:40
add a comment |
$begingroup$
Image of the inclusion of X misses atleast one point of the sphere. And sphere - {one point} is just Euclidean space which is contractible. Now any map in a contractible space is null homotopic ( check this). And you are there!
Ps: You have not mentioned X is proper subset though.
answered Dec 26 '18 at 18:23
NeelNeel
567314
$endgroup$
add a comment |
$begingroup$
Image of the inclusion of X misses atleast one point of the sphere. And sphere - {one point} is just Euclidean space which is contractible. Now any map in a contractible space is null homotopic ( check this). And you are there!
Ps: You have not mentioned X is proper subset though.
answered Dec 26 '18 at 18:23
NeelNeel
567314
$endgroup$
add a comment |
$begingroup$
Image of the inclusion of X misses atleast one point of the sphere. And sphere - {one point} is just Euclidean space which is contractible. Now any map in a contractible space is null homotopic ( check this). And you are there!
Ps: You have not mentioned X is proper subset though.
answered Dec 26 '18 at 18:23
NeelNeel
567314
$endgroup$
Image of the inclusion of X misses atleast one point of the sphere. And sphere - {one point} is just Euclidean space which is contractible. Now any map in a contractible space is null homotopic ( check this). And you are there!
Ps: You have not mentioned X is proper subset though.
answered Dec 26 '18 at 18:23
NeelNeel
567314
edited Dec 26 '18 at 18:32
answered Dec 26 '18 at 18:23
NeelNeel
567314
answered Dec 26 '18 at 18:23
NeelNeel
567314
answered Dec 26 '18 at 18:23
NeelNeel
567314
567314
add a comment |
add a comment |
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