Cohomology of $mathbb{S}^n$
2
$begingroup$
I have a corollary stating:
if X is contractible then $H_0(X) = mathbb{R}$ and $H_n(X) = {0}$ for $n>0$.
But $mathbb{S}^n$ is contractible and $H_n(mathbb{S}^n) = mathbb{R}$.
I must be missing something.
Thank you!
differential-geometry
asked Dec 19 '18 at 22:51
PerelManPerelMan
649313
$endgroup$
add a comment |
2
$begingroup$
I have a corollary stating:
if X is contractible then $H_0(X) = mathbb{R}$ and $H_n(X) = {0}$ for $n>0$.
But $mathbb{S}^n$ is contractible and $H_n(mathbb{S}^n) = mathbb{R}$.
I must be missing something.
Thank you!
differential-geometry
asked Dec 19 '18 at 22:51
PerelManPerelMan
649313
$endgroup$
8
$begingroup$
It’s not contractible!
$endgroup$
– Ted Shifrin
Dec 19 '18 at 22:52
$begingroup$
If contractible means any loop on, say $mathbb{S}^2$, is homotopic to a point, that is true for $mathbb{S}^2$ , isn't it ?
$endgroup$
– PerelMan
Dec 19 '18 at 22:55
5
$begingroup$
I think you are confusing contractible with simply connected.
$endgroup$
– positrón0802
Dec 19 '18 at 22:57
$begingroup$
Right! Thank you all, I will double check.
$endgroup$
– PerelMan
Dec 19 '18 at 22:58
add a comment |
2
2
2
$begingroup$
I have a corollary stating:
if X is contractible then $H_0(X) = mathbb{R}$ and $H_n(X) = {0}$ for $n>0$.
But $mathbb{S}^n$ is contractible and $H_n(mathbb{S}^n) = mathbb{R}$.
I must be missing something.
Thank you!
differential-geometry
asked Dec 19 '18 at 22:51
PerelManPerelMan
649313
$endgroup$
I have a corollary stating:
if X is contractible then $H_0(X) = mathbb{R}$ and $H_n(X) = {0}$ for $n>0$.
But $mathbb{S}^n$ is contractible and $H_n(mathbb{S}^n) = mathbb{R}$.
I must be missing something.
Thank you!
differential-geometry
differential-geometry
asked Dec 19 '18 at 22:51
PerelManPerelMan
649313
asked Dec 19 '18 at 22:51
PerelManPerelMan
649313
asked Dec 19 '18 at 22:51
PerelManPerelMan
649313
asked Dec 19 '18 at 22:51
PerelManPerelMan
649313
asked Dec 19 '18 at 22:51
PerelManPerelMan
649313
649313
8
$begingroup$
It’s not contractible!
$endgroup$
– Ted Shifrin
Dec 19 '18 at 22:52
$begingroup$
If contractible means any loop on, say $mathbb{S}^2$, is homotopic to a point, that is true for $mathbb{S}^2$ , isn't it ?
$endgroup$
– PerelMan
Dec 19 '18 at 22:55
5
$begingroup$
I think you are confusing contractible with simply connected.
$endgroup$
– positrón0802
Dec 19 '18 at 22:57
$begingroup$
Right! Thank you all, I will double check.
$endgroup$
– PerelMan
Dec 19 '18 at 22:58
add a comment |
8
$begingroup$
It’s not contractible!
$endgroup$
– Ted Shifrin
Dec 19 '18 at 22:52
$begingroup$
If contractible means any loop on, say $mathbb{S}^2$, is homotopic to a point, that is true for $mathbb{S}^2$ , isn't it ?
$endgroup$
– PerelMan
Dec 19 '18 at 22:55
5
$begingroup$
I think you are confusing contractible with simply connected.
$endgroup$
– positrón0802
Dec 19 '18 at 22:57
$begingroup$
Right! Thank you all, I will double check.
$endgroup$
– PerelMan
Dec 19 '18 at 22:58
8
8
$begingroup$
It’s not contractible!
$endgroup$
– Ted Shifrin
Dec 19 '18 at 22:52
$begingroup$
It’s not contractible!
$endgroup$
– Ted Shifrin
Dec 19 '18 at 22:52
$begingroup$
If contractible means any loop on, say $mathbb{S}^2$, is homotopic to a point, that is true for $mathbb{S}^2$ , isn't it ?
$endgroup$
– PerelMan
Dec 19 '18 at 22:55
$begingroup$
If contractible means any loop on, say $mathbb{S}^2$, is homotopic to a point, that is true for $mathbb{S}^2$ , isn't it ?
$endgroup$
– PerelMan
Dec 19 '18 at 22:55
5
5
$begingroup$
I think you are confusing contractible with simply connected.
$endgroup$
– positrón0802
Dec 19 '18 at 22:57
$begingroup$
I think you are confusing contractible with simply connected.
$endgroup$
– positrón0802
Dec 19 '18 at 22:57
$begingroup$
Right! Thank you all, I will double check.
$endgroup$
– PerelMan
Dec 19 '18 at 22:58
$begingroup$
Right! Thank you all, I will double check.
$endgroup$
– PerelMan
Dec 19 '18 at 22:58
add a comment |
1 Answer
1
active
oldest
votes
5
$begingroup$
$mathbb{S}^n$ is not contractible.
answered Dec 19 '18 at 22:52
positrón0802positrón0802
4,468520
$endgroup$
add a comment |
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1 Answer
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1 Answer
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5
$begingroup$
$mathbb{S}^n$ is not contractible.
answered Dec 19 '18 at 22:52
positrón0802positrón0802
4,468520
$endgroup$
add a comment |
5
$begingroup$
$mathbb{S}^n$ is not contractible.
answered Dec 19 '18 at 22:52
positrón0802positrón0802
4,468520
$endgroup$
add a comment |
5
5
5
$begingroup$
$mathbb{S}^n$ is not contractible.
answered Dec 19 '18 at 22:52
positrón0802positrón0802
4,468520
$endgroup$
$mathbb{S}^n$ is not contractible.
answered Dec 19 '18 at 22:52
positrón0802positrón0802
4,468520
answered Dec 19 '18 at 22:52
positrón0802positrón0802
4,468520
answered Dec 19 '18 at 22:52
positrón0802positrón0802
4,468520
answered Dec 19 '18 at 22:52
positrón0802positrón0802
4,468520
4,468520
add a comment |
add a comment |
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8
$begingroup$
It’s not contractible!
$endgroup$
– Ted Shifrin
Dec 19 '18 at 22:52
$begingroup$
If contractible means any loop on, say $mathbb{S}^2$, is homotopic to a point, that is true for $mathbb{S}^2$ , isn't it ?
$endgroup$
– PerelMan
Dec 19 '18 at 22:55
5
$begingroup$
I think you are confusing contractible with simply connected.
$endgroup$
– positrón0802
Dec 19 '18 at 22:57
$begingroup$
Right! Thank you all, I will double check.
$endgroup$
– PerelMan
Dec 19 '18 at 22:58