Proving that hopf map from $S^3 to S^2 $ is not null homotopic
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I want to prove that hopf map from $S^3 to S^2 $ is not null homotopic. Is there some elementary proof of this fact?
algebraic-topology homotopy-theory
asked Oct 1 '16 at 9:36
happymath
3,80511439
add a comment |
up vote
1
down vote
favorite
I want to prove that hopf map from $S^3 to S^2 $ is not null homotopic. Is there some elementary proof of this fact?
algebraic-topology homotopy-theory
asked Oct 1 '16 at 9:36
happymath
3,80511439
What facts do you know? It's hard to know what tools to use without a context. This result is quite overdetermined.
– Justin Young
Oct 9 '16 at 15:57
@Justin Young I am looking for a proof which only requires basic algebraic topology (by this I mean content of Hatcher's Algebraic topology book)
– happymath
Oct 11 '16 at 4:50
In that case, the answer below is best. The key point is that the cup product in $mathbb CP^2$ is non-trivial, as opposed to $S^2 vee S^4$.
– Justin Young
Oct 11 '16 at 14:05
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I want to prove that hopf map from $S^3 to S^2 $ is not null homotopic. Is there some elementary proof of this fact?
algebraic-topology homotopy-theory
asked Oct 1 '16 at 9:36
happymath
3,80511439
I want to prove that hopf map from $S^3 to S^2 $ is not null homotopic. Is there some elementary proof of this fact?
algebraic-topology homotopy-theory
algebraic-topology homotopy-theory
asked Oct 1 '16 at 9:36
happymath
3,80511439
asked Oct 1 '16 at 9:36
happymath
3,80511439
asked Oct 1 '16 at 9:36
happymath
3,80511439
asked Oct 1 '16 at 9:36
happymath
3,80511439
asked Oct 1 '16 at 9:36
happymath
3,80511439
3,80511439
What facts do you know? It's hard to know what tools to use without a context. This result is quite overdetermined.
– Justin Young
Oct 9 '16 at 15:57
@Justin Young I am looking for a proof which only requires basic algebraic topology (by this I mean content of Hatcher's Algebraic topology book)
– happymath
Oct 11 '16 at 4:50
In that case, the answer below is best. The key point is that the cup product in $mathbb CP^2$ is non-trivial, as opposed to $S^2 vee S^4$.
– Justin Young
Oct 11 '16 at 14:05
add a comment |
What facts do you know? It's hard to know what tools to use without a context. This result is quite overdetermined.
– Justin Young
Oct 9 '16 at 15:57
@Justin Young I am looking for a proof which only requires basic algebraic topology (by this I mean content of Hatcher's Algebraic topology book)
– happymath
Oct 11 '16 at 4:50
In that case, the answer below is best. The key point is that the cup product in $mathbb CP^2$ is non-trivial, as opposed to $S^2 vee S^4$.
– Justin Young
Oct 11 '16 at 14:05
What facts do you know? It's hard to know what tools to use without a context. This result is quite overdetermined.
– Justin Young
Oct 9 '16 at 15:57
What facts do you know? It's hard to know what tools to use without a context. This result is quite overdetermined.
– Justin Young
Oct 9 '16 at 15:57
@Justin Young I am looking for a proof which only requires basic algebraic topology (by this I mean content of Hatcher's Algebraic topology book)
– happymath
Oct 11 '16 at 4:50
@Justin Young I am looking for a proof which only requires basic algebraic topology (by this I mean content of Hatcher's Algebraic topology book)
– happymath
Oct 11 '16 at 4:50
In that case, the answer below is best. The key point is that the cup product in $mathbb CP^2$ is non-trivial, as opposed to $S^2 vee S^4$.
– Justin Young
Oct 11 '16 at 14:05
In that case, the answer below is best. The key point is that the cup product in $mathbb CP^2$ is non-trivial, as opposed to $S^2 vee S^4$.
– Justin Young
Oct 11 '16 at 14:05
add a comment |
3 Answers
3
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oldest
votes
up vote
4
down vote
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If it were nullhomotopic, what do you know about the homotopy type of its mapping cone?
On the other hand, the Hopf map is the attaching map of the $4$-cell in $Bbb CP^2$, so its mapping cone is just $Bbb CP^2$.
answered Oct 1 '16 at 9:45
iwriteonbananas
2,9521532
add a comment |
up vote
2
down vote
Briefly: This follows since the Hopf map $pi:S^3to S^2$ is surjective and satisfies the homotopy lifting property: if $pi$ were nullhomotopic, we could use the homotopy lifting property to construct a homotopy of $text{id}_{S^3}$ to a non-surjective map, which is impossible.
In more detail: The Hopf map $picolon S^3to S^2$ is a surjective submersion. Hence since $S^3$ is compact, Ehresmann's Lemma implies that $pi$ is a fiber bundle, and in particular a Hurewicz fibration. Therefore, $pi$ satisfies the homotopy lifting property.
Assume (to obtain a contradiction) that there is a nullhomotopy $pi_t:S^3to S^2$ with $pi_0 = pi$ and $pi_{1}$ a constant map.
By the homotopy lifting property, there exists a homotopy $h_tcolon S^3 to S^3$ satisfying $h_0 = text{id}_{S^3}$ and $picirc h_t = pi_t$ for all $t$. Since $pi$ is surjective and $pi_1$ is not, $pi circ h_1 = pi_1$ implies that $h_1$ is not surjective. Hence the Brouwer degree $text{deg}(h_1) = 0$. But by homotopy invariance of the degree, $text{deg}(h_1) = text{deg}(h_0) = text{deg}(text{id}_{S^3}) = 1$, so we have obtained a contradiction.
answered Nov 26 at 23:31
Matthew Kvalheim
651416
add a comment |
up vote
1
down vote
The exercises in the last chapter of Milnor's book, Topology from the Differentiable Viewpoint, may provide an elementary proof by the use of the idea of Linking number. You prove in a sequence of exercises that the linking number is a homotopy invariant. The only part, honestly, that I have not completely written down in detail is proving that for the hopf fibration the linking number is non zero - this would prove that the hopf map is not homotopic to the constant map.
answered Feb 11 '17 at 22:54
Behnam Esmayli
1,972515
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
If it were nullhomotopic, what do you know about the homotopy type of its mapping cone?
On the other hand, the Hopf map is the attaching map of the $4$-cell in $Bbb CP^2$, so its mapping cone is just $Bbb CP^2$.
answered Oct 1 '16 at 9:45
iwriteonbananas
2,9521532
add a comment |
up vote
4
down vote
accepted
If it were nullhomotopic, what do you know about the homotopy type of its mapping cone?
On the other hand, the Hopf map is the attaching map of the $4$-cell in $Bbb CP^2$, so its mapping cone is just $Bbb CP^2$.
answered Oct 1 '16 at 9:45
iwriteonbananas
2,9521532
add a comment |
up vote
4
down vote
accepted
up vote
4
down vote
accepted
If it were nullhomotopic, what do you know about the homotopy type of its mapping cone?
On the other hand, the Hopf map is the attaching map of the $4$-cell in $Bbb CP^2$, so its mapping cone is just $Bbb CP^2$.
answered Oct 1 '16 at 9:45
iwriteonbananas
2,9521532
If it were nullhomotopic, what do you know about the homotopy type of its mapping cone?
On the other hand, the Hopf map is the attaching map of the $4$-cell in $Bbb CP^2$, so its mapping cone is just $Bbb CP^2$.
answered Oct 1 '16 at 9:45
iwriteonbananas
2,9521532
answered Oct 1 '16 at 9:45
iwriteonbananas
2,9521532
answered Oct 1 '16 at 9:45
iwriteonbananas
2,9521532
answered Oct 1 '16 at 9:45
iwriteonbananas
2,9521532
2,9521532
add a comment |
add a comment |
up vote
2
down vote
Briefly: This follows since the Hopf map $pi:S^3to S^2$ is surjective and satisfies the homotopy lifting property: if $pi$ were nullhomotopic, we could use the homotopy lifting property to construct a homotopy of $text{id}_{S^3}$ to a non-surjective map, which is impossible.
In more detail: The Hopf map $picolon S^3to S^2$ is a surjective submersion. Hence since $S^3$ is compact, Ehresmann's Lemma implies that $pi$ is a fiber bundle, and in particular a Hurewicz fibration. Therefore, $pi$ satisfies the homotopy lifting property.
Assume (to obtain a contradiction) that there is a nullhomotopy $pi_t:S^3to S^2$ with $pi_0 = pi$ and $pi_{1}$ a constant map.
By the homotopy lifting property, there exists a homotopy $h_tcolon S^3 to S^3$ satisfying $h_0 = text{id}_{S^3}$ and $picirc h_t = pi_t$ for all $t$. Since $pi$ is surjective and $pi_1$ is not, $pi circ h_1 = pi_1$ implies that $h_1$ is not surjective. Hence the Brouwer degree $text{deg}(h_1) = 0$. But by homotopy invariance of the degree, $text{deg}(h_1) = text{deg}(h_0) = text{deg}(text{id}_{S^3}) = 1$, so we have obtained a contradiction.
answered Nov 26 at 23:31
Matthew Kvalheim
651416
add a comment |
up vote
2
down vote
Briefly: This follows since the Hopf map $pi:S^3to S^2$ is surjective and satisfies the homotopy lifting property: if $pi$ were nullhomotopic, we could use the homotopy lifting property to construct a homotopy of $text{id}_{S^3}$ to a non-surjective map, which is impossible.
In more detail: The Hopf map $picolon S^3to S^2$ is a surjective submersion. Hence since $S^3$ is compact, Ehresmann's Lemma implies that $pi$ is a fiber bundle, and in particular a Hurewicz fibration. Therefore, $pi$ satisfies the homotopy lifting property.
Assume (to obtain a contradiction) that there is a nullhomotopy $pi_t:S^3to S^2$ with $pi_0 = pi$ and $pi_{1}$ a constant map.
By the homotopy lifting property, there exists a homotopy $h_tcolon S^3 to S^3$ satisfying $h_0 = text{id}_{S^3}$ and $picirc h_t = pi_t$ for all $t$. Since $pi$ is surjective and $pi_1$ is not, $pi circ h_1 = pi_1$ implies that $h_1$ is not surjective. Hence the Brouwer degree $text{deg}(h_1) = 0$. But by homotopy invariance of the degree, $text{deg}(h_1) = text{deg}(h_0) = text{deg}(text{id}_{S^3}) = 1$, so we have obtained a contradiction.
answered Nov 26 at 23:31
Matthew Kvalheim
651416
add a comment |
up vote
2
down vote
up vote
2
down vote
Briefly: This follows since the Hopf map $pi:S^3to S^2$ is surjective and satisfies the homotopy lifting property: if $pi$ were nullhomotopic, we could use the homotopy lifting property to construct a homotopy of $text{id}_{S^3}$ to a non-surjective map, which is impossible.
In more detail: The Hopf map $picolon S^3to S^2$ is a surjective submersion. Hence since $S^3$ is compact, Ehresmann's Lemma implies that $pi$ is a fiber bundle, and in particular a Hurewicz fibration. Therefore, $pi$ satisfies the homotopy lifting property.
Assume (to obtain a contradiction) that there is a nullhomotopy $pi_t:S^3to S^2$ with $pi_0 = pi$ and $pi_{1}$ a constant map.
By the homotopy lifting property, there exists a homotopy $h_tcolon S^3 to S^3$ satisfying $h_0 = text{id}_{S^3}$ and $picirc h_t = pi_t$ for all $t$. Since $pi$ is surjective and $pi_1$ is not, $pi circ h_1 = pi_1$ implies that $h_1$ is not surjective. Hence the Brouwer degree $text{deg}(h_1) = 0$. But by homotopy invariance of the degree, $text{deg}(h_1) = text{deg}(h_0) = text{deg}(text{id}_{S^3}) = 1$, so we have obtained a contradiction.
answered Nov 26 at 23:31
Matthew Kvalheim
651416
Briefly: This follows since the Hopf map $pi:S^3to S^2$ is surjective and satisfies the homotopy lifting property: if $pi$ were nullhomotopic, we could use the homotopy lifting property to construct a homotopy of $text{id}_{S^3}$ to a non-surjective map, which is impossible.
In more detail: The Hopf map $picolon S^3to S^2$ is a surjective submersion. Hence since $S^3$ is compact, Ehresmann's Lemma implies that $pi$ is a fiber bundle, and in particular a Hurewicz fibration. Therefore, $pi$ satisfies the homotopy lifting property.
Assume (to obtain a contradiction) that there is a nullhomotopy $pi_t:S^3to S^2$ with $pi_0 = pi$ and $pi_{1}$ a constant map.
By the homotopy lifting property, there exists a homotopy $h_tcolon S^3 to S^3$ satisfying $h_0 = text{id}_{S^3}$ and $picirc h_t = pi_t$ for all $t$. Since $pi$ is surjective and $pi_1$ is not, $pi circ h_1 = pi_1$ implies that $h_1$ is not surjective. Hence the Brouwer degree $text{deg}(h_1) = 0$. But by homotopy invariance of the degree, $text{deg}(h_1) = text{deg}(h_0) = text{deg}(text{id}_{S^3}) = 1$, so we have obtained a contradiction.
answered Nov 26 at 23:31
Matthew Kvalheim
651416
answered Nov 26 at 23:31
Matthew Kvalheim
651416
answered Nov 26 at 23:31
Matthew Kvalheim
651416
answered Nov 26 at 23:31
Matthew Kvalheim
651416
651416
add a comment |
add a comment |
up vote
1
down vote
The exercises in the last chapter of Milnor's book, Topology from the Differentiable Viewpoint, may provide an elementary proof by the use of the idea of Linking number. You prove in a sequence of exercises that the linking number is a homotopy invariant. The only part, honestly, that I have not completely written down in detail is proving that for the hopf fibration the linking number is non zero - this would prove that the hopf map is not homotopic to the constant map.
answered Feb 11 '17 at 22:54
Behnam Esmayli
1,972515
add a comment |
up vote
1
down vote
The exercises in the last chapter of Milnor's book, Topology from the Differentiable Viewpoint, may provide an elementary proof by the use of the idea of Linking number. You prove in a sequence of exercises that the linking number is a homotopy invariant. The only part, honestly, that I have not completely written down in detail is proving that for the hopf fibration the linking number is non zero - this would prove that the hopf map is not homotopic to the constant map.
answered Feb 11 '17 at 22:54
Behnam Esmayli
1,972515
add a comment |
up vote
1
down vote
up vote
1
down vote
The exercises in the last chapter of Milnor's book, Topology from the Differentiable Viewpoint, may provide an elementary proof by the use of the idea of Linking number. You prove in a sequence of exercises that the linking number is a homotopy invariant. The only part, honestly, that I have not completely written down in detail is proving that for the hopf fibration the linking number is non zero - this would prove that the hopf map is not homotopic to the constant map.
answered Feb 11 '17 at 22:54
Behnam Esmayli
1,972515
The exercises in the last chapter of Milnor's book, Topology from the Differentiable Viewpoint, may provide an elementary proof by the use of the idea of Linking number. You prove in a sequence of exercises that the linking number is a homotopy invariant. The only part, honestly, that I have not completely written down in detail is proving that for the hopf fibration the linking number is non zero - this would prove that the hopf map is not homotopic to the constant map.
answered Feb 11 '17 at 22:54
Behnam Esmayli
1,972515
answered Feb 11 '17 at 22:54
Behnam Esmayli
1,972515
answered Feb 11 '17 at 22:54
Behnam Esmayli
1,972515
answered Feb 11 '17 at 22:54
Behnam Esmayli
1,972515
1,972515
add a comment |
add a comment |
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What facts do you know? It's hard to know what tools to use without a context. This result is quite overdetermined.
– Justin Young
Oct 9 '16 at 15:57
@Justin Young I am looking for a proof which only requires basic algebraic topology (by this I mean content of Hatcher's Algebraic topology book)
– happymath
Oct 11 '16 at 4:50
In that case, the answer below is best. The key point is that the cup product in $mathbb CP^2$ is non-trivial, as opposed to $S^2 vee S^4$.
– Justin Young
Oct 11 '16 at 14:05