Why is this map $S^2to S^1$ nullhomotopic?Distinguishing homeomorphic from equal in identification spacesa map with contractible domain must be nullhomotopicpath space is contractibleContractible iff every map $f :X to Y$, for arbitrary $Y$, is nullhomotopic. <Proof Verification>Homologous to zero but not contractibleA map $h:S^1to X$ Induces a Trivial Homomorphism of Fundamental Groups Iff it is Nullhomotopic.Unit sphere without a point is contractibleClassifying continuous maps from closed 2-manifolds to various closed manifoldsWhat is the boundary map in the long exact sequence in homotopy groups? (Especially for $SO(n)$)On the homotopy group of a mapping cylinder
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Why is this map $S^2to S^1$ nullhomotopic?
Distinguishing homeomorphic from equal in identification spacesa map with contractible domain must be nullhomotopicpath space is contractibleContractible iff every map $f :X to Y$, for arbitrary $Y$, is nullhomotopic. <Proof Verification>Homologous to zero but not contractibleA map $h:S^1to X$ Induces a Trivial Homomorphism of Fundamental Groups Iff it is Nullhomotopic.Unit sphere without a point is contractibleClassifying continuous maps from closed 2-manifolds to various closed manifoldsWhat is the boundary map in the long exact sequence in homotopy groups? (Especially for $SO(n)$)On the homotopy group of a mapping cylinder
$begingroup$
I know that $pi_2(S^1)=0$ since $S^1$ has $mathbbR$ as universal cover, which is contractible. However, I have a map $S^2to S^1$ that I can't intuitively see why it is nullhomotopic (it has to be, since otherwise it would represent a nontrivial element of $pi_2(S^1)$).
Take $S^2$ to be the standard sphere centered at the origin of $mathbbR^3$. Project onto the $XY$-plane by $p_1: (x,y,z)mapsto (x,y,0)$. Then, do the same with the disc obtained, $p_2:(x,y,0)mapsto (x,0,0)$. Now we have an interval, that we can send homeomorphically (say by $h$) to $[0,1]$. Now, I can choose non-nullhomotopic maps $[0,1]to S^1$, such as the quotient map $q$ (not nullhomotopic because it represents a generator of singular homology) or $phi(t)=e^2pi i t$ (not nullohomotpic because it represents a generator of the fundamental group).
All the maps are continuous. The resulting map $f=qcirc hcirc p_2circ p_1$ (or $g=phicirc hcirc p_2circ p_1$) should be nullhomotopic, but I can't figure out a homotopy or a geometric intuition of how is that possible, since what I see is that in the end we're just performing the classical loop around $S^1$, which is not nullhomotopic.
general-topology algebraic-topology homotopy-theory spheres
asked Mar 29 at 10:49
JaviJavi
3,1432932
$endgroup$
add a comment |
$begingroup$
I know that $pi_2(S^1)=0$ since $S^1$ has $mathbbR$ as universal cover, which is contractible. However, I have a map $S^2to S^1$ that I can't intuitively see why it is nullhomotopic (it has to be, since otherwise it would represent a nontrivial element of $pi_2(S^1)$).
Take $S^2$ to be the standard sphere centered at the origin of $mathbbR^3$. Project onto the $XY$-plane by $p_1: (x,y,z)mapsto (x,y,0)$. Then, do the same with the disc obtained, $p_2:(x,y,0)mapsto (x,0,0)$. Now we have an interval, that we can send homeomorphically (say by $h$) to $[0,1]$. Now, I can choose non-nullhomotopic maps $[0,1]to S^1$, such as the quotient map $q$ (not nullhomotopic because it represents a generator of singular homology) or $phi(t)=e^2pi i t$ (not nullohomotpic because it represents a generator of the fundamental group).
All the maps are continuous. The resulting map $f=qcirc hcirc p_2circ p_1$ (or $g=phicirc hcirc p_2circ p_1$) should be nullhomotopic, but I can't figure out a homotopy or a geometric intuition of how is that possible, since what I see is that in the end we're just performing the classical loop around $S^1$, which is not nullhomotopic.
general-topology algebraic-topology homotopy-theory spheres
asked Mar 29 at 10:49
JaviJavi
3,1432932
$endgroup$
2
$begingroup$
So your problem is simpler. Any map $[0,1] to mathbbS^2$ is nullhomotopic. If you quotient the endpoints of $[0,1]$, then you can get a non-nullhomotopic map, but then the composition doesn't work out.
$endgroup$
– D. Thomine
Mar 29 at 10:56
1
$begingroup$
So, essentially, what you've proved is that, if you identify (quotient) $(1, 0, 0)$ with $(-1,0,0)$ on the sphere, then there is a non-nullhomotopic map from the resulting space $M = mathbbS^2_/sim$ to $mathbbS_1$. Which is true. But $M$ is not a sphere.
$endgroup$
– D. Thomine
Mar 29 at 10:58
$begingroup$
@D.Thomine I don't understand what you mean by "then the composition doesn't work". I defined a map $S^2to S^1$. In fact, my first idea was defining it passing first to $M$, but it was more difficult to described. But, in that case, it is just a map $S^2to S^1$ which factors through $M$.
$endgroup$
– Javi
Mar 29 at 11:45
1
$begingroup$
There are a great many answers here- it shows that there are an incredible number of ways to show maps into $S^1$ are nullhomotopic. There is a generalization of the covering space argument that shows that if $pi_1 (X)$ has only the trivial homomorphism into $mathbbZ$, then any map into $S^1$ is nullhomotopic. If you are familiar with cohomology. this also implies that this happens if and only if $H^1(X;mathbbZ)$ is trivial. You might wonder then if this can be proved with cohomology, and it can! This is because $S^1$ "represents" $H^1(-;mathbbZ)$.
$endgroup$
– Connor Malin
Mar 29 at 16:26
$begingroup$
Thanks for your comment @ConnorMalin That's very intersting!
$endgroup$
– Javi
Mar 29 at 17:30
add a comment |
$begingroup$
I know that $pi_2(S^1)=0$ since $S^1$ has $mathbbR$ as universal cover, which is contractible. However, I have a map $S^2to S^1$ that I can't intuitively see why it is nullhomotopic (it has to be, since otherwise it would represent a nontrivial element of $pi_2(S^1)$).
Take $S^2$ to be the standard sphere centered at the origin of $mathbbR^3$. Project onto the $XY$-plane by $p_1: (x,y,z)mapsto (x,y,0)$. Then, do the same with the disc obtained, $p_2:(x,y,0)mapsto (x,0,0)$. Now we have an interval, that we can send homeomorphically (say by $h$) to $[0,1]$. Now, I can choose non-nullhomotopic maps $[0,1]to S^1$, such as the quotient map $q$ (not nullhomotopic because it represents a generator of singular homology) or $phi(t)=e^2pi i t$ (not nullohomotpic because it represents a generator of the fundamental group).
All the maps are continuous. The resulting map $f=qcirc hcirc p_2circ p_1$ (or $g=phicirc hcirc p_2circ p_1$) should be nullhomotopic, but I can't figure out a homotopy or a geometric intuition of how is that possible, since what I see is that in the end we're just performing the classical loop around $S^1$, which is not nullhomotopic.
general-topology algebraic-topology homotopy-theory spheres
asked Mar 29 at 10:49
JaviJavi
3,1432932
$endgroup$
I know that $pi_2(S^1)=0$ since $S^1$ has $mathbbR$ as universal cover, which is contractible. However, I have a map $S^2to S^1$ that I can't intuitively see why it is nullhomotopic (it has to be, since otherwise it would represent a nontrivial element of $pi_2(S^1)$).
Take $S^2$ to be the standard sphere centered at the origin of $mathbbR^3$. Project onto the $XY$-plane by $p_1: (x,y,z)mapsto (x,y,0)$. Then, do the same with the disc obtained, $p_2:(x,y,0)mapsto (x,0,0)$. Now we have an interval, that we can send homeomorphically (say by $h$) to $[0,1]$. Now, I can choose non-nullhomotopic maps $[0,1]to S^1$, such as the quotient map $q$ (not nullhomotopic because it represents a generator of singular homology) or $phi(t)=e^2pi i t$ (not nullohomotpic because it represents a generator of the fundamental group).
All the maps are continuous. The resulting map $f=qcirc hcirc p_2circ p_1$ (or $g=phicirc hcirc p_2circ p_1$) should be nullhomotopic, but I can't figure out a homotopy or a geometric intuition of how is that possible, since what I see is that in the end we're just performing the classical loop around $S^1$, which is not nullhomotopic.
general-topology algebraic-topology homotopy-theory spheres
general-topology algebraic-topology homotopy-theory spheres
asked Mar 29 at 10:49
JaviJavi
3,1432932
asked Mar 29 at 10:49
JaviJavi
3,1432932
asked Mar 29 at 10:49
JaviJavi
3,1432932
asked Mar 29 at 10:49
JaviJavi
3,1432932
asked Mar 29 at 10:49
JaviJavi
3,1432932
3,1432932
2
$begingroup$
So your problem is simpler. Any map $[0,1] to mathbbS^2$ is nullhomotopic. If you quotient the endpoints of $[0,1]$, then you can get a non-nullhomotopic map, but then the composition doesn't work out.
$endgroup$
– D. Thomine
Mar 29 at 10:56
1
$begingroup$
So, essentially, what you've proved is that, if you identify (quotient) $(1, 0, 0)$ with $(-1,0,0)$ on the sphere, then there is a non-nullhomotopic map from the resulting space $M = mathbbS^2_/sim$ to $mathbbS_1$. Which is true. But $M$ is not a sphere.
$endgroup$
– D. Thomine
Mar 29 at 10:58
$begingroup$
@D.Thomine I don't understand what you mean by "then the composition doesn't work". I defined a map $S^2to S^1$. In fact, my first idea was defining it passing first to $M$, but it was more difficult to described. But, in that case, it is just a map $S^2to S^1$ which factors through $M$.
$endgroup$
– Javi
Mar 29 at 11:45
1
$begingroup$
There are a great many answers here- it shows that there are an incredible number of ways to show maps into $S^1$ are nullhomotopic. There is a generalization of the covering space argument that shows that if $pi_1 (X)$ has only the trivial homomorphism into $mathbbZ$, then any map into $S^1$ is nullhomotopic. If you are familiar with cohomology. this also implies that this happens if and only if $H^1(X;mathbbZ)$ is trivial. You might wonder then if this can be proved with cohomology, and it can! This is because $S^1$ "represents" $H^1(-;mathbbZ)$.
$endgroup$
– Connor Malin
Mar 29 at 16:26
$begingroup$
Thanks for your comment @ConnorMalin That's very intersting!
$endgroup$
– Javi
Mar 29 at 17:30
add a comment |
2
$begingroup$
So your problem is simpler. Any map $[0,1] to mathbbS^2$ is nullhomotopic. If you quotient the endpoints of $[0,1]$, then you can get a non-nullhomotopic map, but then the composition doesn't work out.
$endgroup$
– D. Thomine
Mar 29 at 10:56
1
$begingroup$
So, essentially, what you've proved is that, if you identify (quotient) $(1, 0, 0)$ with $(-1,0,0)$ on the sphere, then there is a non-nullhomotopic map from the resulting space $M = mathbbS^2_/sim$ to $mathbbS_1$. Which is true. But $M$ is not a sphere.
$endgroup$
– D. Thomine
Mar 29 at 10:58
$begingroup$
@D.Thomine I don't understand what you mean by "then the composition doesn't work". I defined a map $S^2to S^1$. In fact, my first idea was defining it passing first to $M$, but it was more difficult to described. But, in that case, it is just a map $S^2to S^1$ which factors through $M$.
$endgroup$
– Javi
Mar 29 at 11:45
1
$begingroup$
There are a great many answers here- it shows that there are an incredible number of ways to show maps into $S^1$ are nullhomotopic. There is a generalization of the covering space argument that shows that if $pi_1 (X)$ has only the trivial homomorphism into $mathbbZ$, then any map into $S^1$ is nullhomotopic. If you are familiar with cohomology. this also implies that this happens if and only if $H^1(X;mathbbZ)$ is trivial. You might wonder then if this can be proved with cohomology, and it can! This is because $S^1$ "represents" $H^1(-;mathbbZ)$.
$endgroup$
– Connor Malin
Mar 29 at 16:26
$begingroup$
Thanks for your comment @ConnorMalin That's very intersting!
$endgroup$
– Javi
Mar 29 at 17:30
2
2
$begingroup$
So your problem is simpler. Any map $[0,1] to mathbbS^2$ is nullhomotopic. If you quotient the endpoints of $[0,1]$, then you can get a non-nullhomotopic map, but then the composition doesn't work out.
$endgroup$
– D. Thomine
Mar 29 at 10:56
$begingroup$
So your problem is simpler. Any map $[0,1] to mathbbS^2$ is nullhomotopic. If you quotient the endpoints of $[0,1]$, then you can get a non-nullhomotopic map, but then the composition doesn't work out.
$endgroup$
– D. Thomine
Mar 29 at 10:56
1
1
$begingroup$
So, essentially, what you've proved is that, if you identify (quotient) $(1, 0, 0)$ with $(-1,0,0)$ on the sphere, then there is a non-nullhomotopic map from the resulting space $M = mathbbS^2_/sim$ to $mathbbS_1$. Which is true. But $M$ is not a sphere.
$endgroup$
– D. Thomine
Mar 29 at 10:58
$begingroup$
So, essentially, what you've proved is that, if you identify (quotient) $(1, 0, 0)$ with $(-1,0,0)$ on the sphere, then there is a non-nullhomotopic map from the resulting space $M = mathbbS^2_/sim$ to $mathbbS_1$. Which is true. But $M$ is not a sphere.
$endgroup$
– D. Thomine
Mar 29 at 10:58
$begingroup$
@D.Thomine I don't understand what you mean by "then the composition doesn't work". I defined a map $S^2to S^1$. In fact, my first idea was defining it passing first to $M$, but it was more difficult to described. But, in that case, it is just a map $S^2to S^1$ which factors through $M$.
$endgroup$
– Javi
Mar 29 at 11:45
$begingroup$
@D.Thomine I don't understand what you mean by "then the composition doesn't work". I defined a map $S^2to S^1$. In fact, my first idea was defining it passing first to $M$, but it was more difficult to described. But, in that case, it is just a map $S^2to S^1$ which factors through $M$.
$endgroup$
– Javi
Mar 29 at 11:45
1
1
$begingroup$
There are a great many answers here- it shows that there are an incredible number of ways to show maps into $S^1$ are nullhomotopic. There is a generalization of the covering space argument that shows that if $pi_1 (X)$ has only the trivial homomorphism into $mathbbZ$, then any map into $S^1$ is nullhomotopic. If you are familiar with cohomology. this also implies that this happens if and only if $H^1(X;mathbbZ)$ is trivial. You might wonder then if this can be proved with cohomology, and it can! This is because $S^1$ "represents" $H^1(-;mathbbZ)$.
$endgroup$
– Connor Malin
Mar 29 at 16:26
$begingroup$
There are a great many answers here- it shows that there are an incredible number of ways to show maps into $S^1$ are nullhomotopic. There is a generalization of the covering space argument that shows that if $pi_1 (X)$ has only the trivial homomorphism into $mathbbZ$, then any map into $S^1$ is nullhomotopic. If you are familiar with cohomology. this also implies that this happens if and only if $H^1(X;mathbbZ)$ is trivial. You might wonder then if this can be proved with cohomology, and it can! This is because $S^1$ "represents" $H^1(-;mathbbZ)$.
$endgroup$
– Connor Malin
Mar 29 at 16:26
$begingroup$
Thanks for your comment @ConnorMalin That's very intersting!
$endgroup$
– Javi
Mar 29 at 17:30
$begingroup$
Thanks for your comment @ConnorMalin That's very intersting!
$endgroup$
– Javi
Mar 29 at 17:30
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
You claim that $q : [0,1] to S^1$ is not nullhomotopic. But it is. Define $h : [0,1] times [0,1] to S^1, h(s,t) = q(st)$. Then $h(s,0) = q(0)$ which is constant and $q(s,1) = q(s)$.
You misunderstanding comes from the fact that the closed path $q$ is a generator of $pi_1(S^1)$. But for fundamental groups we consider homotopies of closed paths which keep the endpoints $ 0, 1 $ of $[0,1]$ fixed. This kind of homotopy is a very special one. For maps $S^2 to S^1$ we are allowed to use arbitary homotopies. Even if we require that some basepoint of $S^2$ is kept fixed under homotopies ("pointed homotopies" ), we get the same result: All pointed maps are pointed homotopic to the constant pointed map.
answered Mar 29 at 11:14
Paul FrostPaul Frost
12.2k3935
$endgroup$
$begingroup$
By the closed paht $q$ you mean the map $phi$ on my question?
$endgroup$
– Javi
Mar 29 at 11:36
1
$begingroup$
Yes. In fact, your map $phi$ is a quotient map $[0,1] to S^1$. I would even say it is the standard quotient map. But we can take any quotient map $q : [0,1] to S^1$.
$endgroup$
– Paul Frost
Mar 29 at 11:40
$begingroup$
So, in the same way we have the homotopy $h$, wouldn't we have $H:[0,1]times [0,1]to S^1$, $H(s,t)=phi(st)$? $phi(s,0)=phi(0)$ which is constant and $phi(s,1)=phi(s)$. Does it fail to be continuous?
$endgroup$
– Javi
Mar 29 at 11:42
$begingroup$
It is continuous because multplication $mu : [0,1] times [0,1] to [0,1], mu(s,t) =st$, is continuous.
$endgroup$
– Paul Frost
Mar 29 at 11:44
1
$begingroup$
As I explained in my answer, for fundamental groups we need homotopies keeping $ 0,1$ fixed. The above homotopy only keeps $0$ fixed, thus it is not a homotopy of paths. But nevertheless $phi$ is nullhomotopic.
$endgroup$
– Paul Frost
Mar 29 at 11:50
|
show 1 more comment
$begingroup$
It may not be easy to describe a nullhomotopy on the end result, but you have intermediate steps (a disc and a line segment) which are trivially nullhomotopic. Insert a homotopy at one of those points in your function chain.
The image of the resulting homotopy may look discontinuous, as you're tearing a hole in the circle. But the parts that are torn apart do not correspond to points close together on $S^2$, so it is, in fact, still continuous.
answered Mar 29 at 10:52
ArthurArthur
122k7122211
$endgroup$
add a comment |
$begingroup$
Imagine poking the sphere inwards from $x=0$ and $x=1$ so that under projection to the $x$-axis it doesn't reach all the way around the interval $[0,1]$. Performing this in $mathbb R^3$ exhibits a homotopy between two embeddings of $S^2$. From this point it should be clear that following this homotopy with the map you describe yields a nullhomotopic map.
answered Mar 29 at 11:05
Rylee LymanRylee Lyman
382210
$endgroup$
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
$begingroup$
You claim that $q : [0,1] to S^1$ is not nullhomotopic. But it is. Define $h : [0,1] times [0,1] to S^1, h(s,t) = q(st)$. Then $h(s,0) = q(0)$ which is constant and $q(s,1) = q(s)$.
You misunderstanding comes from the fact that the closed path $q$ is a generator of $pi_1(S^1)$. But for fundamental groups we consider homotopies of closed paths which keep the endpoints $ 0, 1 $ of $[0,1]$ fixed. This kind of homotopy is a very special one. For maps $S^2 to S^1$ we are allowed to use arbitary homotopies. Even if we require that some basepoint of $S^2$ is kept fixed under homotopies ("pointed homotopies" ), we get the same result: All pointed maps are pointed homotopic to the constant pointed map.
answered Mar 29 at 11:14
Paul FrostPaul Frost
12.2k3935
$endgroup$
$begingroup$
By the closed paht $q$ you mean the map $phi$ on my question?
$endgroup$
– Javi
Mar 29 at 11:36
1
$begingroup$
Yes. In fact, your map $phi$ is a quotient map $[0,1] to S^1$. I would even say it is the standard quotient map. But we can take any quotient map $q : [0,1] to S^1$.
$endgroup$
– Paul Frost
Mar 29 at 11:40
$begingroup$
So, in the same way we have the homotopy $h$, wouldn't we have $H:[0,1]times [0,1]to S^1$, $H(s,t)=phi(st)$? $phi(s,0)=phi(0)$ which is constant and $phi(s,1)=phi(s)$. Does it fail to be continuous?
$endgroup$
– Javi
Mar 29 at 11:42
$begingroup$
It is continuous because multplication $mu : [0,1] times [0,1] to [0,1], mu(s,t) =st$, is continuous.
$endgroup$
– Paul Frost
Mar 29 at 11:44
1
$begingroup$
As I explained in my answer, for fundamental groups we need homotopies keeping $ 0,1$ fixed. The above homotopy only keeps $0$ fixed, thus it is not a homotopy of paths. But nevertheless $phi$ is nullhomotopic.
$endgroup$
– Paul Frost
Mar 29 at 11:50
|
show 1 more comment
$begingroup$
You claim that $q : [0,1] to S^1$ is not nullhomotopic. But it is. Define $h : [0,1] times [0,1] to S^1, h(s,t) = q(st)$. Then $h(s,0) = q(0)$ which is constant and $q(s,1) = q(s)$.
You misunderstanding comes from the fact that the closed path $q$ is a generator of $pi_1(S^1)$. But for fundamental groups we consider homotopies of closed paths which keep the endpoints $ 0, 1 $ of $[0,1]$ fixed. This kind of homotopy is a very special one. For maps $S^2 to S^1$ we are allowed to use arbitary homotopies. Even if we require that some basepoint of $S^2$ is kept fixed under homotopies ("pointed homotopies" ), we get the same result: All pointed maps are pointed homotopic to the constant pointed map.
answered Mar 29 at 11:14
Paul FrostPaul Frost
12.2k3935
$endgroup$
$begingroup$
By the closed paht $q$ you mean the map $phi$ on my question?
$endgroup$
– Javi
Mar 29 at 11:36
1
$begingroup$
Yes. In fact, your map $phi$ is a quotient map $[0,1] to S^1$. I would even say it is the standard quotient map. But we can take any quotient map $q : [0,1] to S^1$.
$endgroup$
– Paul Frost
Mar 29 at 11:40
$begingroup$
So, in the same way we have the homotopy $h$, wouldn't we have $H:[0,1]times [0,1]to S^1$, $H(s,t)=phi(st)$? $phi(s,0)=phi(0)$ which is constant and $phi(s,1)=phi(s)$. Does it fail to be continuous?
$endgroup$
– Javi
Mar 29 at 11:42
$begingroup$
It is continuous because multplication $mu : [0,1] times [0,1] to [0,1], mu(s,t) =st$, is continuous.
$endgroup$
– Paul Frost
Mar 29 at 11:44
1
$begingroup$
As I explained in my answer, for fundamental groups we need homotopies keeping $ 0,1$ fixed. The above homotopy only keeps $0$ fixed, thus it is not a homotopy of paths. But nevertheless $phi$ is nullhomotopic.
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– Paul Frost
Mar 29 at 11:50
|
show 1 more comment
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You claim that $q : [0,1] to S^1$ is not nullhomotopic. But it is. Define $h : [0,1] times [0,1] to S^1, h(s,t) = q(st)$. Then $h(s,0) = q(0)$ which is constant and $q(s,1) = q(s)$.
You misunderstanding comes from the fact that the closed path $q$ is a generator of $pi_1(S^1)$. But for fundamental groups we consider homotopies of closed paths which keep the endpoints $ 0, 1 $ of $[0,1]$ fixed. This kind of homotopy is a very special one. For maps $S^2 to S^1$ we are allowed to use arbitary homotopies. Even if we require that some basepoint of $S^2$ is kept fixed under homotopies ("pointed homotopies" ), we get the same result: All pointed maps are pointed homotopic to the constant pointed map.
answered Mar 29 at 11:14
Paul FrostPaul Frost
12.2k3935
$endgroup$
You claim that $q : [0,1] to S^1$ is not nullhomotopic. But it is. Define $h : [0,1] times [0,1] to S^1, h(s,t) = q(st)$. Then $h(s,0) = q(0)$ which is constant and $q(s,1) = q(s)$.
You misunderstanding comes from the fact that the closed path $q$ is a generator of $pi_1(S^1)$. But for fundamental groups we consider homotopies of closed paths which keep the endpoints $ 0, 1 $ of $[0,1]$ fixed. This kind of homotopy is a very special one. For maps $S^2 to S^1$ we are allowed to use arbitary homotopies. Even if we require that some basepoint of $S^2$ is kept fixed under homotopies ("pointed homotopies" ), we get the same result: All pointed maps are pointed homotopic to the constant pointed map.
answered Mar 29 at 11:14
Paul FrostPaul Frost
12.2k3935
answered Mar 29 at 11:14
Paul FrostPaul Frost
12.2k3935
answered Mar 29 at 11:14
Paul FrostPaul Frost
12.2k3935
answered Mar 29 at 11:14
Paul FrostPaul Frost
12.2k3935
12.2k3935
$begingroup$
By the closed paht $q$ you mean the map $phi$ on my question?
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– Javi
Mar 29 at 11:36
1
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Yes. In fact, your map $phi$ is a quotient map $[0,1] to S^1$. I would even say it is the standard quotient map. But we can take any quotient map $q : [0,1] to S^1$.
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– Paul Frost
Mar 29 at 11:40
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So, in the same way we have the homotopy $h$, wouldn't we have $H:[0,1]times [0,1]to S^1$, $H(s,t)=phi(st)$? $phi(s,0)=phi(0)$ which is constant and $phi(s,1)=phi(s)$. Does it fail to be continuous?
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– Javi
Mar 29 at 11:42
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It is continuous because multplication $mu : [0,1] times [0,1] to [0,1], mu(s,t) =st$, is continuous.
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– Paul Frost
Mar 29 at 11:44
1
$begingroup$
As I explained in my answer, for fundamental groups we need homotopies keeping $ 0,1$ fixed. The above homotopy only keeps $0$ fixed, thus it is not a homotopy of paths. But nevertheless $phi$ is nullhomotopic.
$endgroup$
– Paul Frost
Mar 29 at 11:50
|
show 1 more comment
$begingroup$
By the closed paht $q$ you mean the map $phi$ on my question?
$endgroup$
– Javi
Mar 29 at 11:36
1
$begingroup$
Yes. In fact, your map $phi$ is a quotient map $[0,1] to S^1$. I would even say it is the standard quotient map. But we can take any quotient map $q : [0,1] to S^1$.
$endgroup$
– Paul Frost
Mar 29 at 11:40
$begingroup$
So, in the same way we have the homotopy $h$, wouldn't we have $H:[0,1]times [0,1]to S^1$, $H(s,t)=phi(st)$? $phi(s,0)=phi(0)$ which is constant and $phi(s,1)=phi(s)$. Does it fail to be continuous?
$endgroup$
– Javi
Mar 29 at 11:42
$begingroup$
It is continuous because multplication $mu : [0,1] times [0,1] to [0,1], mu(s,t) =st$, is continuous.
$endgroup$
– Paul Frost
Mar 29 at 11:44
1
$begingroup$
As I explained in my answer, for fundamental groups we need homotopies keeping $ 0,1$ fixed. The above homotopy only keeps $0$ fixed, thus it is not a homotopy of paths. But nevertheless $phi$ is nullhomotopic.
$endgroup$
– Paul Frost
Mar 29 at 11:50
$begingroup$
By the closed paht $q$ you mean the map $phi$ on my question?
$endgroup$
– Javi
Mar 29 at 11:36
$begingroup$
By the closed paht $q$ you mean the map $phi$ on my question?
$endgroup$
– Javi
Mar 29 at 11:36
1
1
$begingroup$
Yes. In fact, your map $phi$ is a quotient map $[0,1] to S^1$. I would even say it is the standard quotient map. But we can take any quotient map $q : [0,1] to S^1$.
$endgroup$
– Paul Frost
Mar 29 at 11:40
$begingroup$
Yes. In fact, your map $phi$ is a quotient map $[0,1] to S^1$. I would even say it is the standard quotient map. But we can take any quotient map $q : [0,1] to S^1$.
$endgroup$
– Paul Frost
Mar 29 at 11:40
$begingroup$
So, in the same way we have the homotopy $h$, wouldn't we have $H:[0,1]times [0,1]to S^1$, $H(s,t)=phi(st)$? $phi(s,0)=phi(0)$ which is constant and $phi(s,1)=phi(s)$. Does it fail to be continuous?
$endgroup$
– Javi
Mar 29 at 11:42
$begingroup$
So, in the same way we have the homotopy $h$, wouldn't we have $H:[0,1]times [0,1]to S^1$, $H(s,t)=phi(st)$? $phi(s,0)=phi(0)$ which is constant and $phi(s,1)=phi(s)$. Does it fail to be continuous?
$endgroup$
– Javi
Mar 29 at 11:42
$begingroup$
It is continuous because multplication $mu : [0,1] times [0,1] to [0,1], mu(s,t) =st$, is continuous.
$endgroup$
– Paul Frost
Mar 29 at 11:44
$begingroup$
It is continuous because multplication $mu : [0,1] times [0,1] to [0,1], mu(s,t) =st$, is continuous.
$endgroup$
– Paul Frost
Mar 29 at 11:44
1
1
$begingroup$
As I explained in my answer, for fundamental groups we need homotopies keeping $ 0,1$ fixed. The above homotopy only keeps $0$ fixed, thus it is not a homotopy of paths. But nevertheless $phi$ is nullhomotopic.
$endgroup$
– Paul Frost
Mar 29 at 11:50
$begingroup$
As I explained in my answer, for fundamental groups we need homotopies keeping $ 0,1$ fixed. The above homotopy only keeps $0$ fixed, thus it is not a homotopy of paths. But nevertheless $phi$ is nullhomotopic.
$endgroup$
– Paul Frost
Mar 29 at 11:50
|
show 1 more comment
$begingroup$
It may not be easy to describe a nullhomotopy on the end result, but you have intermediate steps (a disc and a line segment) which are trivially nullhomotopic. Insert a homotopy at one of those points in your function chain.
The image of the resulting homotopy may look discontinuous, as you're tearing a hole in the circle. But the parts that are torn apart do not correspond to points close together on $S^2$, so it is, in fact, still continuous.
answered Mar 29 at 10:52
ArthurArthur
122k7122211
$endgroup$
add a comment |
$begingroup$
It may not be easy to describe a nullhomotopy on the end result, but you have intermediate steps (a disc and a line segment) which are trivially nullhomotopic. Insert a homotopy at one of those points in your function chain.
The image of the resulting homotopy may look discontinuous, as you're tearing a hole in the circle. But the parts that are torn apart do not correspond to points close together on $S^2$, so it is, in fact, still continuous.
answered Mar 29 at 10:52
ArthurArthur
122k7122211
$endgroup$
add a comment |
$begingroup$
It may not be easy to describe a nullhomotopy on the end result, but you have intermediate steps (a disc and a line segment) which are trivially nullhomotopic. Insert a homotopy at one of those points in your function chain.
The image of the resulting homotopy may look discontinuous, as you're tearing a hole in the circle. But the parts that are torn apart do not correspond to points close together on $S^2$, so it is, in fact, still continuous.
answered Mar 29 at 10:52
ArthurArthur
122k7122211
$endgroup$
It may not be easy to describe a nullhomotopy on the end result, but you have intermediate steps (a disc and a line segment) which are trivially nullhomotopic. Insert a homotopy at one of those points in your function chain.
The image of the resulting homotopy may look discontinuous, as you're tearing a hole in the circle. But the parts that are torn apart do not correspond to points close together on $S^2$, so it is, in fact, still continuous.
answered Mar 29 at 10:52
ArthurArthur
122k7122211
edited Mar 29 at 10:57
answered Mar 29 at 10:52
ArthurArthur
122k7122211
answered Mar 29 at 10:52
ArthurArthur
122k7122211
answered Mar 29 at 10:52
ArthurArthur
122k7122211
122k7122211
add a comment |
add a comment |
$begingroup$
Imagine poking the sphere inwards from $x=0$ and $x=1$ so that under projection to the $x$-axis it doesn't reach all the way around the interval $[0,1]$. Performing this in $mathbb R^3$ exhibits a homotopy between two embeddings of $S^2$. From this point it should be clear that following this homotopy with the map you describe yields a nullhomotopic map.
answered Mar 29 at 11:05
Rylee LymanRylee Lyman
382210
$endgroup$
add a comment |
$begingroup$
Imagine poking the sphere inwards from $x=0$ and $x=1$ so that under projection to the $x$-axis it doesn't reach all the way around the interval $[0,1]$. Performing this in $mathbb R^3$ exhibits a homotopy between two embeddings of $S^2$. From this point it should be clear that following this homotopy with the map you describe yields a nullhomotopic map.
answered Mar 29 at 11:05
Rylee LymanRylee Lyman
382210
$endgroup$
add a comment |
$begingroup$
Imagine poking the sphere inwards from $x=0$ and $x=1$ so that under projection to the $x$-axis it doesn't reach all the way around the interval $[0,1]$. Performing this in $mathbb R^3$ exhibits a homotopy between two embeddings of $S^2$. From this point it should be clear that following this homotopy with the map you describe yields a nullhomotopic map.
answered Mar 29 at 11:05
Rylee LymanRylee Lyman
382210
$endgroup$
Imagine poking the sphere inwards from $x=0$ and $x=1$ so that under projection to the $x$-axis it doesn't reach all the way around the interval $[0,1]$. Performing this in $mathbb R^3$ exhibits a homotopy between two embeddings of $S^2$. From this point it should be clear that following this homotopy with the map you describe yields a nullhomotopic map.
answered Mar 29 at 11:05
Rylee LymanRylee Lyman
382210
answered Mar 29 at 11:05
Rylee LymanRylee Lyman
382210
answered Mar 29 at 11:05
Rylee LymanRylee Lyman
382210
answered Mar 29 at 11:05
Rylee LymanRylee Lyman
382210
382210
add a comment |
add a comment |
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So your problem is simpler. Any map $[0,1] to mathbbS^2$ is nullhomotopic. If you quotient the endpoints of $[0,1]$, then you can get a non-nullhomotopic map, but then the composition doesn't work out.
$endgroup$
– D. Thomine
Mar 29 at 10:56
1
$begingroup$
So, essentially, what you've proved is that, if you identify (quotient) $(1, 0, 0)$ with $(-1,0,0)$ on the sphere, then there is a non-nullhomotopic map from the resulting space $M = mathbbS^2_/sim$ to $mathbbS_1$. Which is true. But $M$ is not a sphere.
$endgroup$
– D. Thomine
Mar 29 at 10:58
$begingroup$
@D.Thomine I don't understand what you mean by "then the composition doesn't work". I defined a map $S^2to S^1$. In fact, my first idea was defining it passing first to $M$, but it was more difficult to described. But, in that case, it is just a map $S^2to S^1$ which factors through $M$.
$endgroup$
– Javi
Mar 29 at 11:45
1
$begingroup$
There are a great many answers here- it shows that there are an incredible number of ways to show maps into $S^1$ are nullhomotopic. There is a generalization of the covering space argument that shows that if $pi_1 (X)$ has only the trivial homomorphism into $mathbbZ$, then any map into $S^1$ is nullhomotopic. If you are familiar with cohomology. this also implies that this happens if and only if $H^1(X;mathbbZ)$ is trivial. You might wonder then if this can be proved with cohomology, and it can! This is because $S^1$ "represents" $H^1(-;mathbbZ)$.
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– Connor Malin
Mar 29 at 16:26
$begingroup$
Thanks for your comment @ConnorMalin That's very intersting!
$endgroup$
– Javi
Mar 29 at 17:30