Mixing
Lack of homotopy equivalence between given spaces
$begingroup$
I am trying to show that spaces $mathbb{S}^3times mathbb{CP}^infty$ and $mathbb{S}^2$ are not homotopy equivalent.
(My goal is to use then as examples of non-homotopy equivalent spaces of the same homotopy groups).
I still do not clearly see simple argument why it is the case. Maybe it is obvious, but after some time I need at least a hint.
algebraic-topology homotopy-theory
asked Jan 9 at 17:57
mikismikis
1,4231821
$endgroup$
|
show 1 more comment
$begingroup$
I am trying to show that spaces $mathbb{S}^3times mathbb{CP}^infty$ and $mathbb{S}^2$ are not homotopy equivalent.
(My goal is to use then as examples of non-homotopy equivalent spaces of the same homotopy groups).
I still do not clearly see simple argument why it is the case. Maybe it is obvious, but after some time I need at least a hint.
algebraic-topology homotopy-theory
asked Jan 9 at 17:57
mikismikis
1,4231821
$endgroup$
2
$begingroup$
I think you have to switch your spheres there. The easiest argument is homology I think.
$endgroup$
– ThorbenK
Jan 9 at 17:59
$begingroup$
Yeah, you are right about the switching. I was thinking about that, but I would be greatful if some non-homological argument appear. (It will be interesting to see "purely topological" argument)
$endgroup$
– mikis
Jan 9 at 18:06
3
$begingroup$
How is an argument using homology not "purely topological"?
$endgroup$
– Eric Wofsey
Jan 9 at 20:13
$begingroup$
You are using some algebraic relations etc. That is the reason why I used quotation marks in "purely topological". It is a slightly philosophical question what is purely topological and I would not like to concentrate on this ;) More specifically, I would like to see some argument that do not use the machinery from homology/cohomology theory, but it is only an additional question - It will be interesting really to see some argument from general topology, maybe using some retractions etc.
$endgroup$
– mikis
Jan 9 at 21:15
2
$begingroup$
Good luck proving something about homotopy equivalences without using the tools of... homotopy theory.
$endgroup$
– Mike Miller
Jan 9 at 23:47
|
show 1 more comment
$begingroup$
I am trying to show that spaces $mathbb{S}^3times mathbb{CP}^infty$ and $mathbb{S}^2$ are not homotopy equivalent.
(My goal is to use then as examples of non-homotopy equivalent spaces of the same homotopy groups).
I still do not clearly see simple argument why it is the case. Maybe it is obvious, but after some time I need at least a hint.
algebraic-topology homotopy-theory
asked Jan 9 at 17:57
mikismikis
1,4231821
$endgroup$
I am trying to show that spaces $mathbb{S}^3times mathbb{CP}^infty$ and $mathbb{S}^2$ are not homotopy equivalent.
(My goal is to use then as examples of non-homotopy equivalent spaces of the same homotopy groups).
I still do not clearly see simple argument why it is the case. Maybe it is obvious, but after some time I need at least a hint.
algebraic-topology homotopy-theory
algebraic-topology homotopy-theory
asked Jan 9 at 17:57
mikismikis
1,4231821
asked Jan 9 at 17:57
mikismikis
1,4231821
edited Jan 9 at 18:02
mikis
asked Jan 9 at 17:57
mikismikis
1,4231821
asked Jan 9 at 17:57
mikismikis
1,4231821
asked Jan 9 at 17:57
mikismikis
1,4231821
1,4231821
2
$begingroup$
I think you have to switch your spheres there. The easiest argument is homology I think.
$endgroup$
– ThorbenK
Jan 9 at 17:59
$begingroup$
Yeah, you are right about the switching. I was thinking about that, but I would be greatful if some non-homological argument appear. (It will be interesting to see "purely topological" argument)
$endgroup$
– mikis
Jan 9 at 18:06
3
$begingroup$
How is an argument using homology not "purely topological"?
$endgroup$
– Eric Wofsey
Jan 9 at 20:13
$begingroup$
You are using some algebraic relations etc. That is the reason why I used quotation marks in "purely topological". It is a slightly philosophical question what is purely topological and I would not like to concentrate on this ;) More specifically, I would like to see some argument that do not use the machinery from homology/cohomology theory, but it is only an additional question - It will be interesting really to see some argument from general topology, maybe using some retractions etc.
$endgroup$
– mikis
Jan 9 at 21:15
2
$begingroup$
Good luck proving something about homotopy equivalences without using the tools of... homotopy theory.
$endgroup$
– Mike Miller
Jan 9 at 23:47
|
show 1 more comment
2
$begingroup$
I think you have to switch your spheres there. The easiest argument is homology I think.
$endgroup$
– ThorbenK
Jan 9 at 17:59
$begingroup$
Yeah, you are right about the switching. I was thinking about that, but I would be greatful if some non-homological argument appear. (It will be interesting to see "purely topological" argument)
$endgroup$
– mikis
Jan 9 at 18:06
3
$begingroup$
How is an argument using homology not "purely topological"?
$endgroup$
– Eric Wofsey
Jan 9 at 20:13
$begingroup$
You are using some algebraic relations etc. That is the reason why I used quotation marks in "purely topological". It is a slightly philosophical question what is purely topological and I would not like to concentrate on this ;) More specifically, I would like to see some argument that do not use the machinery from homology/cohomology theory, but it is only an additional question - It will be interesting really to see some argument from general topology, maybe using some retractions etc.
$endgroup$
– mikis
Jan 9 at 21:15
2
$begingroup$
Good luck proving something about homotopy equivalences without using the tools of... homotopy theory.
$endgroup$
– Mike Miller
Jan 9 at 23:47
2
2
$begingroup$
I think you have to switch your spheres there. The easiest argument is homology I think.
$endgroup$
– ThorbenK
Jan 9 at 17:59
$begingroup$
I think you have to switch your spheres there. The easiest argument is homology I think.
$endgroup$
– ThorbenK
Jan 9 at 17:59
$begingroup$
Yeah, you are right about the switching. I was thinking about that, but I would be greatful if some non-homological argument appear. (It will be interesting to see "purely topological" argument)
$endgroup$
– mikis
Jan 9 at 18:06
$begingroup$
Yeah, you are right about the switching. I was thinking about that, but I would be greatful if some non-homological argument appear. (It will be interesting to see "purely topological" argument)
$endgroup$
– mikis
Jan 9 at 18:06
3
3
$begingroup$
How is an argument using homology not "purely topological"?
$endgroup$
– Eric Wofsey
Jan 9 at 20:13
$begingroup$
How is an argument using homology not "purely topological"?
$endgroup$
– Eric Wofsey
Jan 9 at 20:13
$begingroup$
You are using some algebraic relations etc. That is the reason why I used quotation marks in "purely topological". It is a slightly philosophical question what is purely topological and I would not like to concentrate on this ;) More specifically, I would like to see some argument that do not use the machinery from homology/cohomology theory, but it is only an additional question - It will be interesting really to see some argument from general topology, maybe using some retractions etc.
$endgroup$
– mikis
Jan 9 at 21:15
$begingroup$
You are using some algebraic relations etc. That is the reason why I used quotation marks in "purely topological". It is a slightly philosophical question what is purely topological and I would not like to concentrate on this ;) More specifically, I would like to see some argument that do not use the machinery from homology/cohomology theory, but it is only an additional question - It will be interesting really to see some argument from general topology, maybe using some retractions etc.
$endgroup$
– mikis
Jan 9 at 21:15
2
2
$begingroup$
Good luck proving something about homotopy equivalences without using the tools of... homotopy theory.
$endgroup$
– Mike Miller
Jan 9 at 23:47
$begingroup$
Good luck proving something about homotopy equivalences without using the tools of... homotopy theory.
$endgroup$
– Mike Miller
Jan 9 at 23:47
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
Here is an argument using only homotopy theory, with no homological considerations. In fact it uses nothing more than homotopy groups. It ends up being far more complicated than a homological argument, however, so I would suggest broadening your critera for accepting such an argument as an answer.
Assume there is a homotopy equivalence $Phi:S^3timesmathbb{C}P^inftyxrightarrowsimeq S^2$. Then for each $n$ we have a map $mathbb{C}P^nhookrightarrow mathbb{C}P^inftyhookrightarrow S^3timesmathbb{C}P^inftyxrightarrowsimeq S^2$ which induces an isomorphism on $pi_2$ and the trivial map on all other homotopy groups. In particular we have a map
$$phi:S^2congmathbb{C}P^1rightarrow S^2$$
which induces an isomorphism on $pi_2$ and the trivial map on all other homotopy groups. But this is absurd, since $pi_2S^2congmathbb{Z}$ is generated by the identity map, so we must have $phi=pm id_{S^2}$, and this leads to a contradiction. For instance, if $phi=+id_{S^2}$ then by functorality it induces the identity map on all homotopy groups. Therefore the only possibility is that $phi=-id_{S^2}inpi_2S^2$. But $-id_{S^2}=A$ is the antipodal map, from which it follows that $Acircphi$ is a map which induces an isomorphism on $pi_2$ and the trivial map on all other homotopy groups. Since then $Acirc Phi=Acirc A=id_{S^2}$ we are back in the last situation.
Hence $phi$ cannot exist, and this prohibits the existence of the homotopy equivalence $Phi$.
answered Jan 10 at 12:00
TyroneTyrone
4,62011225
$endgroup$
$begingroup$
That is a nice argument. I did not say I will not accept homological argument - I would only like to see some other argument ;)
$endgroup$
– mikis
Jan 10 at 20:50
add a comment |
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$begingroup$
Here is an argument using only homotopy theory, with no homological considerations. In fact it uses nothing more than homotopy groups. It ends up being far more complicated than a homological argument, however, so I would suggest broadening your critera for accepting such an argument as an answer.
Assume there is a homotopy equivalence $Phi:S^3timesmathbb{C}P^inftyxrightarrowsimeq S^2$. Then for each $n$ we have a map $mathbb{C}P^nhookrightarrow mathbb{C}P^inftyhookrightarrow S^3timesmathbb{C}P^inftyxrightarrowsimeq S^2$ which induces an isomorphism on $pi_2$ and the trivial map on all other homotopy groups. In particular we have a map
$$phi:S^2congmathbb{C}P^1rightarrow S^2$$
which induces an isomorphism on $pi_2$ and the trivial map on all other homotopy groups. But this is absurd, since $pi_2S^2congmathbb{Z}$ is generated by the identity map, so we must have $phi=pm id_{S^2}$, and this leads to a contradiction. For instance, if $phi=+id_{S^2}$ then by functorality it induces the identity map on all homotopy groups. Therefore the only possibility is that $phi=-id_{S^2}inpi_2S^2$. But $-id_{S^2}=A$ is the antipodal map, from which it follows that $Acircphi$ is a map which induces an isomorphism on $pi_2$ and the trivial map on all other homotopy groups. Since then $Acirc Phi=Acirc A=id_{S^2}$ we are back in the last situation.
Hence $phi$ cannot exist, and this prohibits the existence of the homotopy equivalence $Phi$.
answered Jan 10 at 12:00
TyroneTyrone
4,62011225
$endgroup$
$begingroup$
That is a nice argument. I did not say I will not accept homological argument - I would only like to see some other argument ;)
$endgroup$
– mikis
Jan 10 at 20:50
add a comment |
$begingroup$
Here is an argument using only homotopy theory, with no homological considerations. In fact it uses nothing more than homotopy groups. It ends up being far more complicated than a homological argument, however, so I would suggest broadening your critera for accepting such an argument as an answer.
Assume there is a homotopy equivalence $Phi:S^3timesmathbb{C}P^inftyxrightarrowsimeq S^2$. Then for each $n$ we have a map $mathbb{C}P^nhookrightarrow mathbb{C}P^inftyhookrightarrow S^3timesmathbb{C}P^inftyxrightarrowsimeq S^2$ which induces an isomorphism on $pi_2$ and the trivial map on all other homotopy groups. In particular we have a map
$$phi:S^2congmathbb{C}P^1rightarrow S^2$$
which induces an isomorphism on $pi_2$ and the trivial map on all other homotopy groups. But this is absurd, since $pi_2S^2congmathbb{Z}$ is generated by the identity map, so we must have $phi=pm id_{S^2}$, and this leads to a contradiction. For instance, if $phi=+id_{S^2}$ then by functorality it induces the identity map on all homotopy groups. Therefore the only possibility is that $phi=-id_{S^2}inpi_2S^2$. But $-id_{S^2}=A$ is the antipodal map, from which it follows that $Acircphi$ is a map which induces an isomorphism on $pi_2$ and the trivial map on all other homotopy groups. Since then $Acirc Phi=Acirc A=id_{S^2}$ we are back in the last situation.
Hence $phi$ cannot exist, and this prohibits the existence of the homotopy equivalence $Phi$.
answered Jan 10 at 12:00
TyroneTyrone
4,62011225
$endgroup$
$begingroup$
That is a nice argument. I did not say I will not accept homological argument - I would only like to see some other argument ;)
$endgroup$
– mikis
Jan 10 at 20:50
add a comment |
$begingroup$
Here is an argument using only homotopy theory, with no homological considerations. In fact it uses nothing more than homotopy groups. It ends up being far more complicated than a homological argument, however, so I would suggest broadening your critera for accepting such an argument as an answer.
Assume there is a homotopy equivalence $Phi:S^3timesmathbb{C}P^inftyxrightarrowsimeq S^2$. Then for each $n$ we have a map $mathbb{C}P^nhookrightarrow mathbb{C}P^inftyhookrightarrow S^3timesmathbb{C}P^inftyxrightarrowsimeq S^2$ which induces an isomorphism on $pi_2$ and the trivial map on all other homotopy groups. In particular we have a map
$$phi:S^2congmathbb{C}P^1rightarrow S^2$$
which induces an isomorphism on $pi_2$ and the trivial map on all other homotopy groups. But this is absurd, since $pi_2S^2congmathbb{Z}$ is generated by the identity map, so we must have $phi=pm id_{S^2}$, and this leads to a contradiction. For instance, if $phi=+id_{S^2}$ then by functorality it induces the identity map on all homotopy groups. Therefore the only possibility is that $phi=-id_{S^2}inpi_2S^2$. But $-id_{S^2}=A$ is the antipodal map, from which it follows that $Acircphi$ is a map which induces an isomorphism on $pi_2$ and the trivial map on all other homotopy groups. Since then $Acirc Phi=Acirc A=id_{S^2}$ we are back in the last situation.
Hence $phi$ cannot exist, and this prohibits the existence of the homotopy equivalence $Phi$.
answered Jan 10 at 12:00
TyroneTyrone
4,62011225
$endgroup$
Here is an argument using only homotopy theory, with no homological considerations. In fact it uses nothing more than homotopy groups. It ends up being far more complicated than a homological argument, however, so I would suggest broadening your critera for accepting such an argument as an answer.
Assume there is a homotopy equivalence $Phi:S^3timesmathbb{C}P^inftyxrightarrowsimeq S^2$. Then for each $n$ we have a map $mathbb{C}P^nhookrightarrow mathbb{C}P^inftyhookrightarrow S^3timesmathbb{C}P^inftyxrightarrowsimeq S^2$ which induces an isomorphism on $pi_2$ and the trivial map on all other homotopy groups. In particular we have a map
$$phi:S^2congmathbb{C}P^1rightarrow S^2$$
which induces an isomorphism on $pi_2$ and the trivial map on all other homotopy groups. But this is absurd, since $pi_2S^2congmathbb{Z}$ is generated by the identity map, so we must have $phi=pm id_{S^2}$, and this leads to a contradiction. For instance, if $phi=+id_{S^2}$ then by functorality it induces the identity map on all homotopy groups. Therefore the only possibility is that $phi=-id_{S^2}inpi_2S^2$. But $-id_{S^2}=A$ is the antipodal map, from which it follows that $Acircphi$ is a map which induces an isomorphism on $pi_2$ and the trivial map on all other homotopy groups. Since then $Acirc Phi=Acirc A=id_{S^2}$ we are back in the last situation.
Hence $phi$ cannot exist, and this prohibits the existence of the homotopy equivalence $Phi$.
answered Jan 10 at 12:00
TyroneTyrone
4,62011225
answered Jan 10 at 12:00
TyroneTyrone
4,62011225
answered Jan 10 at 12:00
TyroneTyrone
4,62011225
answered Jan 10 at 12:00
TyroneTyrone
4,62011225
4,62011225
$begingroup$
That is a nice argument. I did not say I will not accept homological argument - I would only like to see some other argument ;)
$endgroup$
– mikis
Jan 10 at 20:50
add a comment |
$begingroup$
That is a nice argument. I did not say I will not accept homological argument - I would only like to see some other argument ;)
$endgroup$
– mikis
Jan 10 at 20:50
$begingroup$
That is a nice argument. I did not say I will not accept homological argument - I would only like to see some other argument ;)
$endgroup$
– mikis
Jan 10 at 20:50
$begingroup$
That is a nice argument. I did not say I will not accept homological argument - I would only like to see some other argument ;)
$endgroup$
– mikis
Jan 10 at 20:50
add a comment |
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2
$begingroup$
I think you have to switch your spheres there. The easiest argument is homology I think.
$endgroup$
– ThorbenK
Jan 9 at 17:59
$begingroup$
Yeah, you are right about the switching. I was thinking about that, but I would be greatful if some non-homological argument appear. (It will be interesting to see "purely topological" argument)
$endgroup$
– mikis
Jan 9 at 18:06
3
$begingroup$
How is an argument using homology not "purely topological"?
$endgroup$
– Eric Wofsey
Jan 9 at 20:13
$begingroup$
You are using some algebraic relations etc. That is the reason why I used quotation marks in "purely topological". It is a slightly philosophical question what is purely topological and I would not like to concentrate on this ;) More specifically, I would like to see some argument that do not use the machinery from homology/cohomology theory, but it is only an additional question - It will be interesting really to see some argument from general topology, maybe using some retractions etc.
$endgroup$
– mikis
Jan 9 at 21:15
2
$begingroup$
Good luck proving something about homotopy equivalences without using the tools of... homotopy theory.
$endgroup$
– Mike Miller
Jan 9 at 23:47