Mixing
$Utimes W cong Vtimes W $ does this imply that $Ucong V$
$begingroup$
Let $U, V,$ and $W$ be smooth manifolds such that $Utimes W$ is diffeomorphic to $Vtimes W $ does that imply that $U$ is diffeomorphic to $V$.?
Is this result also true/False in other categories like Groups, Topological spaces?
In the case of smooth manifolds, I think this is true by using rank theorem since by rank theorem the map in local coordinates is of the form $(x^1,x^2,...,x^n)to (x^1,x^2,...,x^n)$ and we can think of $U$ as embedded in $U times W$ as $Utimes {y}$ for some $yin W$ fixed. and similarly, think of $V$ as embedded in $V times W$ as $Vtimes {y}$.
Using rank theorem we can prove that $Utimes {y}$ and $Vtimes {y}$ are locally diffeomorphic and use the fact that $Utimes W cong Vtimes W $ to prove that they are in fact diffeomorphic?
Is my approach right? If not where does it fail?
For groups, this
implies that the statement is true for finite groups.
I have a small question here if $Htimes G cong Ktimes G $ then why cant we quotient on both sides ${e}times G$ to get $Hcong K$?
Is the result true for topological spaces?
Is there any classification of Categories in which the result is true?
general-topology group-theory proof-verification differential-geometry differential-topology
edited Jan 22 at 15:07
Shaun
9,334113684
asked Jan 22 at 13:39
user345777user345777
428311
$endgroup$
|
show 5 more comments
$begingroup$
Let $U, V,$ and $W$ be smooth manifolds such that $Utimes W$ is diffeomorphic to $Vtimes W $ does that imply that $U$ is diffeomorphic to $V$.?
Is this result also true/False in other categories like Groups, Topological spaces?
In the case of smooth manifolds, I think this is true by using rank theorem since by rank theorem the map in local coordinates is of the form $(x^1,x^2,...,x^n)to (x^1,x^2,...,x^n)$ and we can think of $U$ as embedded in $U times W$ as $Utimes {y}$ for some $yin W$ fixed. and similarly, think of $V$ as embedded in $V times W$ as $Vtimes {y}$.
Using rank theorem we can prove that $Utimes {y}$ and $Vtimes {y}$ are locally diffeomorphic and use the fact that $Utimes W cong Vtimes W $ to prove that they are in fact diffeomorphic?
Is my approach right? If not where does it fail?
For groups, this
implies that the statement is true for finite groups.
I have a small question here if $Htimes G cong Ktimes G $ then why cant we quotient on both sides ${e}times G$ to get $Hcong K$?
Is the result true for topological spaces?
Is there any classification of Categories in which the result is true?
general-topology group-theory proof-verification differential-geometry differential-topology
edited Jan 22 at 15:07
Shaun
9,334113684
asked Jan 22 at 13:39
user345777user345777
428311
$endgroup$
2
$begingroup$
I have little to no background in differential geometry, but if $W=emptyset$ is a smooth manifold, then that might give a counterexample.
$endgroup$
– SmileyCraft
Jan 22 at 13:43
$begingroup$
I think it's a good idea to quotient by ${e}times G$, why shouldn't it work ?
$endgroup$
– Surb
Jan 22 at 13:58
8
$begingroup$
Look up the Whitehead manifold. Its product with $Bbb R$ is Euclidean space again. There are closed manifold examples as well.
$endgroup$
– user98602
Jan 22 at 14:22
1
$begingroup$
Similar questions for groups have been asked here a few times. For example, 1, 2,3, 4. See also 5, on free products.
$endgroup$
– user1729
Jan 22 at 15:18
1
$begingroup$
Also, I think this question is interesting, but is far too broad! I can see at least 5 questions: The first question (on smooth manifolds), "Is my approach right?", "Is it true for groups?", "Is it true for topological spaces?", and "Can we classify the categories for which it is true?"
$endgroup$
– user1729
Jan 22 at 15:27
|
show 5 more comments
$begingroup$
Let $U, V,$ and $W$ be smooth manifolds such that $Utimes W$ is diffeomorphic to $Vtimes W $ does that imply that $U$ is diffeomorphic to $V$.?
Is this result also true/False in other categories like Groups, Topological spaces?
In the case of smooth manifolds, I think this is true by using rank theorem since by rank theorem the map in local coordinates is of the form $(x^1,x^2,...,x^n)to (x^1,x^2,...,x^n)$ and we can think of $U$ as embedded in $U times W$ as $Utimes {y}$ for some $yin W$ fixed. and similarly, think of $V$ as embedded in $V times W$ as $Vtimes {y}$.
Using rank theorem we can prove that $Utimes {y}$ and $Vtimes {y}$ are locally diffeomorphic and use the fact that $Utimes W cong Vtimes W $ to prove that they are in fact diffeomorphic?
Is my approach right? If not where does it fail?
For groups, this
implies that the statement is true for finite groups.
I have a small question here if $Htimes G cong Ktimes G $ then why cant we quotient on both sides ${e}times G$ to get $Hcong K$?
Is the result true for topological spaces?
Is there any classification of Categories in which the result is true?
general-topology group-theory proof-verification differential-geometry differential-topology
edited Jan 22 at 15:07
Shaun
9,334113684
asked Jan 22 at 13:39
user345777user345777
428311
$endgroup$
Let $U, V,$ and $W$ be smooth manifolds such that $Utimes W$ is diffeomorphic to $Vtimes W $ does that imply that $U$ is diffeomorphic to $V$.?
Is this result also true/False in other categories like Groups, Topological spaces?
In the case of smooth manifolds, I think this is true by using rank theorem since by rank theorem the map in local coordinates is of the form $(x^1,x^2,...,x^n)to (x^1,x^2,...,x^n)$ and we can think of $U$ as embedded in $U times W$ as $Utimes {y}$ for some $yin W$ fixed. and similarly, think of $V$ as embedded in $V times W$ as $Vtimes {y}$.
Using rank theorem we can prove that $Utimes {y}$ and $Vtimes {y}$ are locally diffeomorphic and use the fact that $Utimes W cong Vtimes W $ to prove that they are in fact diffeomorphic?
Is my approach right? If not where does it fail?
For groups, this
implies that the statement is true for finite groups.
I have a small question here if $Htimes G cong Ktimes G $ then why cant we quotient on both sides ${e}times G$ to get $Hcong K$?
Is the result true for topological spaces?
Is there any classification of Categories in which the result is true?
general-topology group-theory proof-verification differential-geometry differential-topology
general-topology group-theory proof-verification differential-geometry differential-topology
edited Jan 22 at 15:07
Shaun
9,334113684
asked Jan 22 at 13:39
user345777user345777
428311
edited Jan 22 at 15:07
Shaun
9,334113684
asked Jan 22 at 13:39
user345777user345777
428311
edited Jan 22 at 15:07
Shaun
9,334113684
edited Jan 22 at 15:07
Shaun
9,334113684
edited Jan 22 at 15:07
Shaun
9,334113684
9,334113684
asked Jan 22 at 13:39
user345777user345777
428311
asked Jan 22 at 13:39
user345777user345777
428311
asked Jan 22 at 13:39
user345777user345777
428311
428311
2
$begingroup$
I have little to no background in differential geometry, but if $W=emptyset$ is a smooth manifold, then that might give a counterexample.
$endgroup$
– SmileyCraft
Jan 22 at 13:43
$begingroup$
I think it's a good idea to quotient by ${e}times G$, why shouldn't it work ?
$endgroup$
– Surb
Jan 22 at 13:58
8
$begingroup$
Look up the Whitehead manifold. Its product with $Bbb R$ is Euclidean space again. There are closed manifold examples as well.
$endgroup$
– user98602
Jan 22 at 14:22
1
$begingroup$
Similar questions for groups have been asked here a few times. For example, 1, 2,3, 4. See also 5, on free products.
$endgroup$
– user1729
Jan 22 at 15:18
1
$begingroup$
Also, I think this question is interesting, but is far too broad! I can see at least 5 questions: The first question (on smooth manifolds), "Is my approach right?", "Is it true for groups?", "Is it true for topological spaces?", and "Can we classify the categories for which it is true?"
$endgroup$
– user1729
Jan 22 at 15:27
|
show 5 more comments
2
$begingroup$
I have little to no background in differential geometry, but if $W=emptyset$ is a smooth manifold, then that might give a counterexample.
$endgroup$
– SmileyCraft
Jan 22 at 13:43
$begingroup$
I think it's a good idea to quotient by ${e}times G$, why shouldn't it work ?
$endgroup$
– Surb
Jan 22 at 13:58
8
$begingroup$
Look up the Whitehead manifold. Its product with $Bbb R$ is Euclidean space again. There are closed manifold examples as well.
$endgroup$
– user98602
Jan 22 at 14:22
1
$begingroup$
Similar questions for groups have been asked here a few times. For example, 1, 2,3, 4. See also 5, on free products.
$endgroup$
– user1729
Jan 22 at 15:18
1
$begingroup$
Also, I think this question is interesting, but is far too broad! I can see at least 5 questions: The first question (on smooth manifolds), "Is my approach right?", "Is it true for groups?", "Is it true for topological spaces?", and "Can we classify the categories for which it is true?"
$endgroup$
– user1729
Jan 22 at 15:27
2
2
$begingroup$
I have little to no background in differential geometry, but if $W=emptyset$ is a smooth manifold, then that might give a counterexample.
$endgroup$
– SmileyCraft
Jan 22 at 13:43
$begingroup$
I have little to no background in differential geometry, but if $W=emptyset$ is a smooth manifold, then that might give a counterexample.
$endgroup$
– SmileyCraft
Jan 22 at 13:43
$begingroup$
I think it's a good idea to quotient by ${e}times G$, why shouldn't it work ?
$endgroup$
– Surb
Jan 22 at 13:58
$begingroup$
I think it's a good idea to quotient by ${e}times G$, why shouldn't it work ?
$endgroup$
– Surb
Jan 22 at 13:58
8
8
$begingroup$
Look up the Whitehead manifold. Its product with $Bbb R$ is Euclidean space again. There are closed manifold examples as well.
$endgroup$
– user98602
Jan 22 at 14:22
$begingroup$
Look up the Whitehead manifold. Its product with $Bbb R$ is Euclidean space again. There are closed manifold examples as well.
$endgroup$
– user98602
Jan 22 at 14:22
1
1
$begingroup$
Similar questions for groups have been asked here a few times. For example, 1, 2,3, 4. See also 5, on free products.
$endgroup$
– user1729
Jan 22 at 15:18
$begingroup$
Similar questions for groups have been asked here a few times. For example, 1, 2,3, 4. See also 5, on free products.
$endgroup$
– user1729
Jan 22 at 15:18
1
1
$begingroup$
Also, I think this question is interesting, but is far too broad! I can see at least 5 questions: The first question (on smooth manifolds), "Is my approach right?", "Is it true for groups?", "Is it true for topological spaces?", and "Can we classify the categories for which it is true?"
$endgroup$
– user1729
Jan 22 at 15:27
$begingroup$
Also, I think this question is interesting, but is far too broad! I can see at least 5 questions: The first question (on smooth manifolds), "Is my approach right?", "Is it true for groups?", "Is it true for topological spaces?", and "Can we classify the categories for which it is true?"
$endgroup$
– user1729
Jan 22 at 15:27
|
show 5 more comments
3 Answers
3
active
oldest
votes
$begingroup$
This doesn't hold if $U,V,W$ are allowed to be disconnected. You can create silly counterexamples in the same way as with infinite groups: take $U = {0}$, $V = {1,2}$, and $W = mathbb{Z}$. Then $Utimes W cong mathbb{Z} cong V times W$ as manifolds, but of course $Unotcong V$. There are obvious higher-dimensional counterexamples of the same nature. (This is similar to $H = mathbb{Z}$, $K = mathbb{Z}^2$, $G = mathbb{Z}^{mathbb{Z}}$ in groups, or maybe more accurately in free abelian groups.)
As you noted, this question cites this paper which proves the result is true for finite groups, but also that it fails even for $G = mathbb{Z}$ and $H,K$ finitely presented! It does also hold for finitely generated (but not necessarily finite) abelian groups by the structure theorem, but it does not hold in general for (abelian) groups (see my example above).
As you can see in the paper cited above, it's a little tricky to get a counterexample for $G=mathbb{Z}$, and I suppose it's also a bit tricky for (closed) connected manifolds, but it looks like Mike Miller knows of one in the comments!
answered Jan 22 at 14:08
cspruncsprun
1,95829
$endgroup$
$begingroup$
Sure, but in groups where does the quoting argument fails?
$endgroup$
– user345777
Jan 22 at 14:13
1
$begingroup$
The set ${e}times G$ need not be preserved by the isomorphism, by which I mean the isomorphism need not map ${e_H}times G$ to ${e_K}times G$. "Quotienting on both sides by the same thing" only necessarily works when the normal subgroups are in correspondence under the isomorphism.
$endgroup$
– csprun
Jan 22 at 14:15
$begingroup$
Sure I get your point
$endgroup$
– user345777
Jan 22 at 14:25
add a comment |
$begingroup$
(Making this CW since the interesting part was done by George Lowther)
First, George Lowther gave a fantastic answer to the smooth manifold question here.
He proves:
There exist connected closed manifolds $X$ and $Y$ with different homotopy types but for which $Xtimes S^1$ is diffeomorphic to $Ytimes S^1$.
(This example also answers the question in the category of groups: the fundamental groups of $X$ and $Y$ are not isomorphic, but $pi_1(X)times mathbb{Z}cong pi_1(Y)times mathbb{Z}$.)
So, in short, NO, you cannot conclude from $Utimes Wcong Vtimes W$ that $Ucong W$. The problem with your argument is a diffeomorphism from $Utimes W$ to $Vtimes W$ does not have to map a subset of the form $Utimes {w}$ to a subset of some $Vtimes {w'}$, so there is no way to take a diffeomorphism of the larger spaces and reduce it to a diffeomorhpism of the smaller spaces.
$endgroup$
add a comment |
$begingroup$
The answer is no, but sometimes is yes.
You should see something about the Krull-Remak-Smith theorem in modules and groups and this question on topological spaces.
answered Jan 22 at 14:24
Ivan Di LibertiIvan Di Liberti
2,59811123
$endgroup$
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
$begingroup$
This doesn't hold if $U,V,W$ are allowed to be disconnected. You can create silly counterexamples in the same way as with infinite groups: take $U = {0}$, $V = {1,2}$, and $W = mathbb{Z}$. Then $Utimes W cong mathbb{Z} cong V times W$ as manifolds, but of course $Unotcong V$. There are obvious higher-dimensional counterexamples of the same nature. (This is similar to $H = mathbb{Z}$, $K = mathbb{Z}^2$, $G = mathbb{Z}^{mathbb{Z}}$ in groups, or maybe more accurately in free abelian groups.)
As you noted, this question cites this paper which proves the result is true for finite groups, but also that it fails even for $G = mathbb{Z}$ and $H,K$ finitely presented! It does also hold for finitely generated (but not necessarily finite) abelian groups by the structure theorem, but it does not hold in general for (abelian) groups (see my example above).
As you can see in the paper cited above, it's a little tricky to get a counterexample for $G=mathbb{Z}$, and I suppose it's also a bit tricky for (closed) connected manifolds, but it looks like Mike Miller knows of one in the comments!
answered Jan 22 at 14:08
cspruncsprun
1,95829
$endgroup$
$begingroup$
Sure, but in groups where does the quoting argument fails?
$endgroup$
– user345777
Jan 22 at 14:13
1
$begingroup$
The set ${e}times G$ need not be preserved by the isomorphism, by which I mean the isomorphism need not map ${e_H}times G$ to ${e_K}times G$. "Quotienting on both sides by the same thing" only necessarily works when the normal subgroups are in correspondence under the isomorphism.
$endgroup$
– csprun
Jan 22 at 14:15
$begingroup$
Sure I get your point
$endgroup$
– user345777
Jan 22 at 14:25
add a comment |
$begingroup$
This doesn't hold if $U,V,W$ are allowed to be disconnected. You can create silly counterexamples in the same way as with infinite groups: take $U = {0}$, $V = {1,2}$, and $W = mathbb{Z}$. Then $Utimes W cong mathbb{Z} cong V times W$ as manifolds, but of course $Unotcong V$. There are obvious higher-dimensional counterexamples of the same nature. (This is similar to $H = mathbb{Z}$, $K = mathbb{Z}^2$, $G = mathbb{Z}^{mathbb{Z}}$ in groups, or maybe more accurately in free abelian groups.)
As you noted, this question cites this paper which proves the result is true for finite groups, but also that it fails even for $G = mathbb{Z}$ and $H,K$ finitely presented! It does also hold for finitely generated (but not necessarily finite) abelian groups by the structure theorem, but it does not hold in general for (abelian) groups (see my example above).
As you can see in the paper cited above, it's a little tricky to get a counterexample for $G=mathbb{Z}$, and I suppose it's also a bit tricky for (closed) connected manifolds, but it looks like Mike Miller knows of one in the comments!
answered Jan 22 at 14:08
cspruncsprun
1,95829
$endgroup$
$begingroup$
Sure, but in groups where does the quoting argument fails?
$endgroup$
– user345777
Jan 22 at 14:13
1
$begingroup$
The set ${e}times G$ need not be preserved by the isomorphism, by which I mean the isomorphism need not map ${e_H}times G$ to ${e_K}times G$. "Quotienting on both sides by the same thing" only necessarily works when the normal subgroups are in correspondence under the isomorphism.
$endgroup$
– csprun
Jan 22 at 14:15
$begingroup$
Sure I get your point
$endgroup$
– user345777
Jan 22 at 14:25
add a comment |
$begingroup$
This doesn't hold if $U,V,W$ are allowed to be disconnected. You can create silly counterexamples in the same way as with infinite groups: take $U = {0}$, $V = {1,2}$, and $W = mathbb{Z}$. Then $Utimes W cong mathbb{Z} cong V times W$ as manifolds, but of course $Unotcong V$. There are obvious higher-dimensional counterexamples of the same nature. (This is similar to $H = mathbb{Z}$, $K = mathbb{Z}^2$, $G = mathbb{Z}^{mathbb{Z}}$ in groups, or maybe more accurately in free abelian groups.)
As you noted, this question cites this paper which proves the result is true for finite groups, but also that it fails even for $G = mathbb{Z}$ and $H,K$ finitely presented! It does also hold for finitely generated (but not necessarily finite) abelian groups by the structure theorem, but it does not hold in general for (abelian) groups (see my example above).
As you can see in the paper cited above, it's a little tricky to get a counterexample for $G=mathbb{Z}$, and I suppose it's also a bit tricky for (closed) connected manifolds, but it looks like Mike Miller knows of one in the comments!
answered Jan 22 at 14:08
cspruncsprun
1,95829
$endgroup$
This doesn't hold if $U,V,W$ are allowed to be disconnected. You can create silly counterexamples in the same way as with infinite groups: take $U = {0}$, $V = {1,2}$, and $W = mathbb{Z}$. Then $Utimes W cong mathbb{Z} cong V times W$ as manifolds, but of course $Unotcong V$. There are obvious higher-dimensional counterexamples of the same nature. (This is similar to $H = mathbb{Z}$, $K = mathbb{Z}^2$, $G = mathbb{Z}^{mathbb{Z}}$ in groups, or maybe more accurately in free abelian groups.)
As you noted, this question cites this paper which proves the result is true for finite groups, but also that it fails even for $G = mathbb{Z}$ and $H,K$ finitely presented! It does also hold for finitely generated (but not necessarily finite) abelian groups by the structure theorem, but it does not hold in general for (abelian) groups (see my example above).
As you can see in the paper cited above, it's a little tricky to get a counterexample for $G=mathbb{Z}$, and I suppose it's also a bit tricky for (closed) connected manifolds, but it looks like Mike Miller knows of one in the comments!
answered Jan 22 at 14:08
cspruncsprun
1,95829
edited Jan 23 at 18:10
answered Jan 22 at 14:08
cspruncsprun
1,95829
answered Jan 22 at 14:08
cspruncsprun
1,95829
answered Jan 22 at 14:08
cspruncsprun
1,95829
1,95829
$begingroup$
Sure, but in groups where does the quoting argument fails?
$endgroup$
– user345777
Jan 22 at 14:13
1
$begingroup$
The set ${e}times G$ need not be preserved by the isomorphism, by which I mean the isomorphism need not map ${e_H}times G$ to ${e_K}times G$. "Quotienting on both sides by the same thing" only necessarily works when the normal subgroups are in correspondence under the isomorphism.
$endgroup$
– csprun
Jan 22 at 14:15
$begingroup$
Sure I get your point
$endgroup$
– user345777
Jan 22 at 14:25
add a comment |
$begingroup$
Sure, but in groups where does the quoting argument fails?
$endgroup$
– user345777
Jan 22 at 14:13
1
$begingroup$
The set ${e}times G$ need not be preserved by the isomorphism, by which I mean the isomorphism need not map ${e_H}times G$ to ${e_K}times G$. "Quotienting on both sides by the same thing" only necessarily works when the normal subgroups are in correspondence under the isomorphism.
$endgroup$
– csprun
Jan 22 at 14:15
$begingroup$
Sure I get your point
$endgroup$
– user345777
Jan 22 at 14:25
$begingroup$
Sure, but in groups where does the quoting argument fails?
$endgroup$
– user345777
Jan 22 at 14:13
$begingroup$
Sure, but in groups where does the quoting argument fails?
$endgroup$
– user345777
Jan 22 at 14:13
1
1
$begingroup$
The set ${e}times G$ need not be preserved by the isomorphism, by which I mean the isomorphism need not map ${e_H}times G$ to ${e_K}times G$. "Quotienting on both sides by the same thing" only necessarily works when the normal subgroups are in correspondence under the isomorphism.
$endgroup$
– csprun
Jan 22 at 14:15
$begingroup$
The set ${e}times G$ need not be preserved by the isomorphism, by which I mean the isomorphism need not map ${e_H}times G$ to ${e_K}times G$. "Quotienting on both sides by the same thing" only necessarily works when the normal subgroups are in correspondence under the isomorphism.
$endgroup$
– csprun
Jan 22 at 14:15
$begingroup$
Sure I get your point
$endgroup$
– user345777
Jan 22 at 14:25
$begingroup$
Sure I get your point
$endgroup$
– user345777
Jan 22 at 14:25
add a comment |
$begingroup$
(Making this CW since the interesting part was done by George Lowther)
First, George Lowther gave a fantastic answer to the smooth manifold question here.
He proves:
There exist connected closed manifolds $X$ and $Y$ with different homotopy types but for which $Xtimes S^1$ is diffeomorphic to $Ytimes S^1$.
(This example also answers the question in the category of groups: the fundamental groups of $X$ and $Y$ are not isomorphic, but $pi_1(X)times mathbb{Z}cong pi_1(Y)times mathbb{Z}$.)
So, in short, NO, you cannot conclude from $Utimes Wcong Vtimes W$ that $Ucong W$. The problem with your argument is a diffeomorphism from $Utimes W$ to $Vtimes W$ does not have to map a subset of the form $Utimes {w}$ to a subset of some $Vtimes {w'}$, so there is no way to take a diffeomorphism of the larger spaces and reduce it to a diffeomorhpism of the smaller spaces.
$endgroup$
add a comment |
$begingroup$
(Making this CW since the interesting part was done by George Lowther)
First, George Lowther gave a fantastic answer to the smooth manifold question here.
He proves:
There exist connected closed manifolds $X$ and $Y$ with different homotopy types but for which $Xtimes S^1$ is diffeomorphic to $Ytimes S^1$.
(This example also answers the question in the category of groups: the fundamental groups of $X$ and $Y$ are not isomorphic, but $pi_1(X)times mathbb{Z}cong pi_1(Y)times mathbb{Z}$.)
So, in short, NO, you cannot conclude from $Utimes Wcong Vtimes W$ that $Ucong W$. The problem with your argument is a diffeomorphism from $Utimes W$ to $Vtimes W$ does not have to map a subset of the form $Utimes {w}$ to a subset of some $Vtimes {w'}$, so there is no way to take a diffeomorphism of the larger spaces and reduce it to a diffeomorhpism of the smaller spaces.
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add a comment |
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(Making this CW since the interesting part was done by George Lowther)
First, George Lowther gave a fantastic answer to the smooth manifold question here.
He proves:
There exist connected closed manifolds $X$ and $Y$ with different homotopy types but for which $Xtimes S^1$ is diffeomorphic to $Ytimes S^1$.
(This example also answers the question in the category of groups: the fundamental groups of $X$ and $Y$ are not isomorphic, but $pi_1(X)times mathbb{Z}cong pi_1(Y)times mathbb{Z}$.)
So, in short, NO, you cannot conclude from $Utimes Wcong Vtimes W$ that $Ucong W$. The problem with your argument is a diffeomorphism from $Utimes W$ to $Vtimes W$ does not have to map a subset of the form $Utimes {w}$ to a subset of some $Vtimes {w'}$, so there is no way to take a diffeomorphism of the larger spaces and reduce it to a diffeomorhpism of the smaller spaces.
$endgroup$
(Making this CW since the interesting part was done by George Lowther)
First, George Lowther gave a fantastic answer to the smooth manifold question here.
He proves:
There exist connected closed manifolds $X$ and $Y$ with different homotopy types but for which $Xtimes S^1$ is diffeomorphic to $Ytimes S^1$.
(This example also answers the question in the category of groups: the fundamental groups of $X$ and $Y$ are not isomorphic, but $pi_1(X)times mathbb{Z}cong pi_1(Y)times mathbb{Z}$.)
So, in short, NO, you cannot conclude from $Utimes Wcong Vtimes W$ that $Ucong W$. The problem with your argument is a diffeomorphism from $Utimes W$ to $Vtimes W$ does not have to map a subset of the form $Utimes {w}$ to a subset of some $Vtimes {w'}$, so there is no way to take a diffeomorphism of the larger spaces and reduce it to a diffeomorhpism of the smaller spaces.
answered Jan 22 at 15:24
community wiki
Jason DeVito
add a comment |
add a comment |
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The answer is no, but sometimes is yes.
You should see something about the Krull-Remak-Smith theorem in modules and groups and this question on topological spaces.
answered Jan 22 at 14:24
Ivan Di LibertiIvan Di Liberti
2,59811123
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add a comment |
$begingroup$
The answer is no, but sometimes is yes.
You should see something about the Krull-Remak-Smith theorem in modules and groups and this question on topological spaces.
answered Jan 22 at 14:24
Ivan Di LibertiIvan Di Liberti
2,59811123
$endgroup$
add a comment |
$begingroup$
The answer is no, but sometimes is yes.
You should see something about the Krull-Remak-Smith theorem in modules and groups and this question on topological spaces.
answered Jan 22 at 14:24
Ivan Di LibertiIvan Di Liberti
2,59811123
$endgroup$
The answer is no, but sometimes is yes.
You should see something about the Krull-Remak-Smith theorem in modules and groups and this question on topological spaces.
answered Jan 22 at 14:24
Ivan Di LibertiIvan Di Liberti
2,59811123
answered Jan 22 at 14:24
Ivan Di LibertiIvan Di Liberti
2,59811123
answered Jan 22 at 14:24
Ivan Di LibertiIvan Di Liberti
2,59811123
answered Jan 22 at 14:24
Ivan Di LibertiIvan Di Liberti
2,59811123
2,59811123
add a comment |
add a comment |
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I have little to no background in differential geometry, but if $W=emptyset$ is a smooth manifold, then that might give a counterexample.
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– SmileyCraft
Jan 22 at 13:43
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I think it's a good idea to quotient by ${e}times G$, why shouldn't it work ?
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– Surb
Jan 22 at 13:58
8
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Look up the Whitehead manifold. Its product with $Bbb R$ is Euclidean space again. There are closed manifold examples as well.
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– user98602
Jan 22 at 14:22
1
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Similar questions for groups have been asked here a few times. For example, 1, 2,3, 4. See also 5, on free products.
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– user1729
Jan 22 at 15:18
1
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Also, I think this question is interesting, but is far too broad! I can see at least 5 questions: The first question (on smooth manifolds), "Is my approach right?", "Is it true for groups?", "Is it true for topological spaces?", and "Can we classify the categories for which it is true?"
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– user1729
Jan 22 at 15:27