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I am reading Lee's ''Introduction to Smooth Manifolds'' and it states that the tangent bundle TM has a natural topology and smooth structure that makes it a 2n-dimensional manifold.
I get most of the proof, my only problem is to check the Hausdorff condition. And it is because it does not specify which topology they use, so I thought of one that works but I do not know if it is the usual one.
Having the specified topology does not bother most of the proof, since you have a proyection: $pi: TM to M$ defined as $pi(p,X)=p$
And just take open sets as $pi^-1(U)$ where $U$ is open in $M$. This works well for most of the proof, but when you want to check the Hausdorff condition in two points in the same fiber $(p,X),(p,Y)$.
So to avoid this I can define the open sets as the collection of sets:
$(U,V_x_x in U):= (x,v) : x in U, v in V_x $ Where $U$ is an open set of $M$ and $V_x_x in U$ a collection of open sets $V_x subset T_xM$ where many but not all can be the empty set.
Is this right? I checked and it does not change the smooth structure since the local charts $(U,phi^sim)$ can be seen as $((U, T_xM_xin M),phi^sim)$ in my topology. Thanks in advanced.
differential-geometry differential-topology smooth-manifolds tangent-bundle
edited Jul 24 at 19:20
Sou
2,7062820
asked Jul 24 at 14:55
Bajo Fondo
376213
add a comment |Â
up vote
1
down vote
favorite
I am reading Lee's ''Introduction to Smooth Manifolds'' and it states that the tangent bundle TM has a natural topology and smooth structure that makes it a 2n-dimensional manifold.
I get most of the proof, my only problem is to check the Hausdorff condition. And it is because it does not specify which topology they use, so I thought of one that works but I do not know if it is the usual one.
Having the specified topology does not bother most of the proof, since you have a proyection: $pi: TM to M$ defined as $pi(p,X)=p$
And just take open sets as $pi^-1(U)$ where $U$ is open in $M$. This works well for most of the proof, but when you want to check the Hausdorff condition in two points in the same fiber $(p,X),(p,Y)$.
So to avoid this I can define the open sets as the collection of sets:
$(U,V_x_x in U):= (x,v) : x in U, v in V_x $ Where $U$ is an open set of $M$ and $V_x_x in U$ a collection of open sets $V_x subset T_xM$ where many but not all can be the empty set.
Is this right? I checked and it does not change the smooth structure since the local charts $(U,phi^sim)$ can be seen as $((U, T_xM_xin M),phi^sim)$ in my topology. Thanks in advanced.
differential-geometry differential-topology smooth-manifolds tangent-bundle
edited Jul 24 at 19:20
Sou
2,7062820
asked Jul 24 at 14:55
Bajo Fondo
376213
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am reading Lee's ''Introduction to Smooth Manifolds'' and it states that the tangent bundle TM has a natural topology and smooth structure that makes it a 2n-dimensional manifold.
I get most of the proof, my only problem is to check the Hausdorff condition. And it is because it does not specify which topology they use, so I thought of one that works but I do not know if it is the usual one.
Having the specified topology does not bother most of the proof, since you have a proyection: $pi: TM to M$ defined as $pi(p,X)=p$
And just take open sets as $pi^-1(U)$ where $U$ is open in $M$. This works well for most of the proof, but when you want to check the Hausdorff condition in two points in the same fiber $(p,X),(p,Y)$.
So to avoid this I can define the open sets as the collection of sets:
$(U,V_x_x in U):= (x,v) : x in U, v in V_x $ Where $U$ is an open set of $M$ and $V_x_x in U$ a collection of open sets $V_x subset T_xM$ where many but not all can be the empty set.
Is this right? I checked and it does not change the smooth structure since the local charts $(U,phi^sim)$ can be seen as $((U, T_xM_xin M),phi^sim)$ in my topology. Thanks in advanced.
differential-geometry differential-topology smooth-manifolds tangent-bundle
edited Jul 24 at 19:20
Sou
2,7062820
asked Jul 24 at 14:55
Bajo Fondo
376213
I am reading Lee's ''Introduction to Smooth Manifolds'' and it states that the tangent bundle TM has a natural topology and smooth structure that makes it a 2n-dimensional manifold.
I get most of the proof, my only problem is to check the Hausdorff condition. And it is because it does not specify which topology they use, so I thought of one that works but I do not know if it is the usual one.
Having the specified topology does not bother most of the proof, since you have a proyection: $pi: TM to M$ defined as $pi(p,X)=p$
And just take open sets as $pi^-1(U)$ where $U$ is open in $M$. This works well for most of the proof, but when you want to check the Hausdorff condition in two points in the same fiber $(p,X),(p,Y)$.
So to avoid this I can define the open sets as the collection of sets:
$(U,V_x_x in U):= (x,v) : x in U, v in V_x $ Where $U$ is an open set of $M$ and $V_x_x in U$ a collection of open sets $V_x subset T_xM$ where many but not all can be the empty set.
Is this right? I checked and it does not change the smooth structure since the local charts $(U,phi^sim)$ can be seen as $((U, T_xM_xin M),phi^sim)$ in my topology. Thanks in advanced.
differential-geometry differential-topology smooth-manifolds tangent-bundle
edited Jul 24 at 19:20
Sou
2,7062820
asked Jul 24 at 14:55
Bajo Fondo
376213
edited Jul 24 at 19:20
Sou
2,7062820
edited Jul 24 at 19:20
Sou
2,7062820
edited Jul 24 at 19:20
Sou
2,7062820
2,7062820
asked Jul 24 at 14:55
Bajo Fondo
376213
asked Jul 24 at 14:55
Bajo Fondo
376213
asked Jul 24 at 14:55
Bajo Fondo
376213
376213
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
You know that each chart $phi : U to V subset mathbbR^n$ for $M$ induces a canonical bijection $tildephi : TM mid_U = p^-1(U) to V times mathbbR^n$. We define $W subset TM$ to be open if $tildephi(W cap p^-1(U))$ is open in $V times mathbbR^n$ for all $phi$.
It is an easy exercise to show that this is in fact a Hausdorff topology.
answered Jul 24 at 16:39
Paul Frost
3,623420
Ok.. I get it, thanks. I think you can also take $V_n_n in mathbbN$ a countable basis of $R^n$ and $U_n_n in mathbbN$ a countable basis of $M$, then define a subbasis where the elements are exacly $tildephi(V_n cap p^-1(U_m))$ for positive intergers $n,m$ This is mostly to assure asmooth manifold structure.
– Bajo Fondo
Jul 25 at 2:18
add a comment |Â
up vote
1
down vote
I guess the theorem that you mention is Proposition 3.18 on Lee's book.
For any smooth chart $(U_alpha,varphi_alpha)$ of $M$, we have a map $widetildevarphi_alpha : pi^-1(U_alpha) to BbbR^2n$ defined as
$$
widetildevarphi_alpha (v_p) = (x^1(p),dots,x^n(p),v^1,dots,v^n) .
$$
This map is a bijection onto its image $varphi_alpha(U) times BbbR^n$.
The topology defined on $TM$ is the one that generated by $widetildevarphi_alpha^-1(V) : forall alpha in A, Vsubseteq BbbR^2n text is open$. You can check directly that the collection above indeed generate a topology.
To show the Hausdorff property, let $p in U subseteq M$, where $U$ is the domain of some smooth chart $(U,varphi)$ of $M$. And $v_p,w_p in pi^-1(p) subseteq pi^-1(U)subseteq TM$, be a pair of points that lie in the same fiber.
With this topology, the map $widetildevarphi : pi^-1(U) to BbbR^2n$ is a homeomorphism onto its image. We can write $widetildevarphi(v_p) = (hatp,v)$, with $hatp = varphi(p)$. So let $widetildevarphi(v_p)=(hatp,v)$ and $widetildevarphi(w_p) = (hatp,w)$ be their images and $v,w in BbbR^n$ be two distinct vectors. To obtain disjoint neighbourhoods of $v_p$ and $w_p$, just take disjoint neighbourhoods $B_1,B_2 subseteq BbbR^n$ of $v$ and $w$ resp. and maped back the disjoint open sets $varphi(U)times B_1$ and $varphi(U)times B_2$ to $pi^-1(U)$. So $widetildevarphi^-1(varphi(U) times B_1)$ and $widetildevarphi^-1(varphi(U) times B_2)$ are the desired neighbourhoods for $v_p$ and $w_p$.
answered Jul 24 at 19:19
Sou
2,7062820
Thanks, in my book is Lemma 4.1 (There is also, no prop 3.18), there is a little subtletly here since you are asserting an homeomorphism, that might not be true with your defined topology. Maybe define a subbasis with those pre-images, taking a countable basis of $mathbbR^n$. I mixed your answer with Frost's and I think it works, thanks.
– Bajo Fondo
Jul 25 at 2:20
@BajoFondo : Maybe you use different edition (mine was 2ed). Why you say that $widetildevarphi$ might not be a homeomorphism ? I think its clearly stated in the proposition (by exhibit its continous inverse).
– Sou
Jul 25 at 2:35
Take $U times V subset mathbbR^n times mathbbR^n$ open set. Then $widetildevarphi^-1(U times V)$ is not $pi^-1(V_alpha)$ for any $V_alpha$. Fix $p in U$, then $p times T_pM subset pi^-1(U)$ but take a $v_p$ that is not in $V$, then $(p,v_p) in p times T_pM subset pi^-1(U)$, but it is not in $widetildevarphi^-1(U times V)$, hence $widetildevarphi^-1(U times V)$ cannot be of the form $pi^-1(U)$. The idea is that with your topology your open sets are the union of the elements of a class $p times T_pM; p in U_alpha$.
– Bajo Fondo
Jul 25 at 19:30
@BajoFondo You're right i made a mistake about the topology. Let me amend it.
– Sou
Jul 25 at 20:06
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
You know that each chart $phi : U to V subset mathbbR^n$ for $M$ induces a canonical bijection $tildephi : TM mid_U = p^-1(U) to V times mathbbR^n$. We define $W subset TM$ to be open if $tildephi(W cap p^-1(U))$ is open in $V times mathbbR^n$ for all $phi$.
It is an easy exercise to show that this is in fact a Hausdorff topology.
answered Jul 24 at 16:39
Paul Frost
3,623420
Ok.. I get it, thanks. I think you can also take $V_n_n in mathbbN$ a countable basis of $R^n$ and $U_n_n in mathbbN$ a countable basis of $M$, then define a subbasis where the elements are exacly $tildephi(V_n cap p^-1(U_m))$ for positive intergers $n,m$ This is mostly to assure asmooth manifold structure.
– Bajo Fondo
Jul 25 at 2:18
add a comment |Â
up vote
2
down vote
accepted
You know that each chart $phi : U to V subset mathbbR^n$ for $M$ induces a canonical bijection $tildephi : TM mid_U = p^-1(U) to V times mathbbR^n$. We define $W subset TM$ to be open if $tildephi(W cap p^-1(U))$ is open in $V times mathbbR^n$ for all $phi$.
It is an easy exercise to show that this is in fact a Hausdorff topology.
answered Jul 24 at 16:39
Paul Frost
3,623420
Ok.. I get it, thanks. I think you can also take $V_n_n in mathbbN$ a countable basis of $R^n$ and $U_n_n in mathbbN$ a countable basis of $M$, then define a subbasis where the elements are exacly $tildephi(V_n cap p^-1(U_m))$ for positive intergers $n,m$ This is mostly to assure asmooth manifold structure.
– Bajo Fondo
Jul 25 at 2:18
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
You know that each chart $phi : U to V subset mathbbR^n$ for $M$ induces a canonical bijection $tildephi : TM mid_U = p^-1(U) to V times mathbbR^n$. We define $W subset TM$ to be open if $tildephi(W cap p^-1(U))$ is open in $V times mathbbR^n$ for all $phi$.
It is an easy exercise to show that this is in fact a Hausdorff topology.
answered Jul 24 at 16:39
Paul Frost
3,623420
You know that each chart $phi : U to V subset mathbbR^n$ for $M$ induces a canonical bijection $tildephi : TM mid_U = p^-1(U) to V times mathbbR^n$. We define $W subset TM$ to be open if $tildephi(W cap p^-1(U))$ is open in $V times mathbbR^n$ for all $phi$.
It is an easy exercise to show that this is in fact a Hausdorff topology.
answered Jul 24 at 16:39
Paul Frost
3,623420
answered Jul 24 at 16:39
Paul Frost
3,623420
answered Jul 24 at 16:39
Paul Frost
3,623420
answered Jul 24 at 16:39
Paul Frost
3,623420
3,623420
Ok.. I get it, thanks. I think you can also take $V_n_n in mathbbN$ a countable basis of $R^n$ and $U_n_n in mathbbN$ a countable basis of $M$, then define a subbasis where the elements are exacly $tildephi(V_n cap p^-1(U_m))$ for positive intergers $n,m$ This is mostly to assure asmooth manifold structure.
– Bajo Fondo
Jul 25 at 2:18
add a comment |Â
Ok.. I get it, thanks. I think you can also take $V_n_n in mathbbN$ a countable basis of $R^n$ and $U_n_n in mathbbN$ a countable basis of $M$, then define a subbasis where the elements are exacly $tildephi(V_n cap p^-1(U_m))$ for positive intergers $n,m$ This is mostly to assure asmooth manifold structure.
– Bajo Fondo
Jul 25 at 2:18
Ok.. I get it, thanks. I think you can also take $V_n_n in mathbbN$ a countable basis of $R^n$ and $U_n_n in mathbbN$ a countable basis of $M$, then define a subbasis where the elements are exacly $tildephi(V_n cap p^-1(U_m))$ for positive intergers $n,m$ This is mostly to assure asmooth manifold structure.
– Bajo Fondo
Jul 25 at 2:18
Ok.. I get it, thanks. I think you can also take $V_n_n in mathbbN$ a countable basis of $R^n$ and $U_n_n in mathbbN$ a countable basis of $M$, then define a subbasis where the elements are exacly $tildephi(V_n cap p^-1(U_m))$ for positive intergers $n,m$ This is mostly to assure asmooth manifold structure.
– Bajo Fondo
Jul 25 at 2:18
add a comment |Â
up vote
1
down vote
I guess the theorem that you mention is Proposition 3.18 on Lee's book.
For any smooth chart $(U_alpha,varphi_alpha)$ of $M$, we have a map $widetildevarphi_alpha : pi^-1(U_alpha) to BbbR^2n$ defined as
$$
widetildevarphi_alpha (v_p) = (x^1(p),dots,x^n(p),v^1,dots,v^n) .
$$
This map is a bijection onto its image $varphi_alpha(U) times BbbR^n$.
The topology defined on $TM$ is the one that generated by $widetildevarphi_alpha^-1(V) : forall alpha in A, Vsubseteq BbbR^2n text is open$. You can check directly that the collection above indeed generate a topology.
To show the Hausdorff property, let $p in U subseteq M$, where $U$ is the domain of some smooth chart $(U,varphi)$ of $M$. And $v_p,w_p in pi^-1(p) subseteq pi^-1(U)subseteq TM$, be a pair of points that lie in the same fiber.
With this topology, the map $widetildevarphi : pi^-1(U) to BbbR^2n$ is a homeomorphism onto its image. We can write $widetildevarphi(v_p) = (hatp,v)$, with $hatp = varphi(p)$. So let $widetildevarphi(v_p)=(hatp,v)$ and $widetildevarphi(w_p) = (hatp,w)$ be their images and $v,w in BbbR^n$ be two distinct vectors. To obtain disjoint neighbourhoods of $v_p$ and $w_p$, just take disjoint neighbourhoods $B_1,B_2 subseteq BbbR^n$ of $v$ and $w$ resp. and maped back the disjoint open sets $varphi(U)times B_1$ and $varphi(U)times B_2$ to $pi^-1(U)$. So $widetildevarphi^-1(varphi(U) times B_1)$ and $widetildevarphi^-1(varphi(U) times B_2)$ are the desired neighbourhoods for $v_p$ and $w_p$.
answered Jul 24 at 19:19
Sou
2,7062820
Thanks, in my book is Lemma 4.1 (There is also, no prop 3.18), there is a little subtletly here since you are asserting an homeomorphism, that might not be true with your defined topology. Maybe define a subbasis with those pre-images, taking a countable basis of $mathbbR^n$. I mixed your answer with Frost's and I think it works, thanks.
– Bajo Fondo
Jul 25 at 2:20
@BajoFondo : Maybe you use different edition (mine was 2ed). Why you say that $widetildevarphi$ might not be a homeomorphism ? I think its clearly stated in the proposition (by exhibit its continous inverse).
– Sou
Jul 25 at 2:35
Take $U times V subset mathbbR^n times mathbbR^n$ open set. Then $widetildevarphi^-1(U times V)$ is not $pi^-1(V_alpha)$ for any $V_alpha$. Fix $p in U$, then $p times T_pM subset pi^-1(U)$ but take a $v_p$ that is not in $V$, then $(p,v_p) in p times T_pM subset pi^-1(U)$, but it is not in $widetildevarphi^-1(U times V)$, hence $widetildevarphi^-1(U times V)$ cannot be of the form $pi^-1(U)$. The idea is that with your topology your open sets are the union of the elements of a class $p times T_pM; p in U_alpha$.
– Bajo Fondo
Jul 25 at 19:30
@BajoFondo You're right i made a mistake about the topology. Let me amend it.
– Sou
Jul 25 at 20:06
add a comment |Â
up vote
1
down vote
I guess the theorem that you mention is Proposition 3.18 on Lee's book.
For any smooth chart $(U_alpha,varphi_alpha)$ of $M$, we have a map $widetildevarphi_alpha : pi^-1(U_alpha) to BbbR^2n$ defined as
$$
widetildevarphi_alpha (v_p) = (x^1(p),dots,x^n(p),v^1,dots,v^n) .
$$
This map is a bijection onto its image $varphi_alpha(U) times BbbR^n$.
The topology defined on $TM$ is the one that generated by $widetildevarphi_alpha^-1(V) : forall alpha in A, Vsubseteq BbbR^2n text is open$. You can check directly that the collection above indeed generate a topology.
To show the Hausdorff property, let $p in U subseteq M$, where $U$ is the domain of some smooth chart $(U,varphi)$ of $M$. And $v_p,w_p in pi^-1(p) subseteq pi^-1(U)subseteq TM$, be a pair of points that lie in the same fiber.
With this topology, the map $widetildevarphi : pi^-1(U) to BbbR^2n$ is a homeomorphism onto its image. We can write $widetildevarphi(v_p) = (hatp,v)$, with $hatp = varphi(p)$. So let $widetildevarphi(v_p)=(hatp,v)$ and $widetildevarphi(w_p) = (hatp,w)$ be their images and $v,w in BbbR^n$ be two distinct vectors. To obtain disjoint neighbourhoods of $v_p$ and $w_p$, just take disjoint neighbourhoods $B_1,B_2 subseteq BbbR^n$ of $v$ and $w$ resp. and maped back the disjoint open sets $varphi(U)times B_1$ and $varphi(U)times B_2$ to $pi^-1(U)$. So $widetildevarphi^-1(varphi(U) times B_1)$ and $widetildevarphi^-1(varphi(U) times B_2)$ are the desired neighbourhoods for $v_p$ and $w_p$.
answered Jul 24 at 19:19
Sou
2,7062820
Thanks, in my book is Lemma 4.1 (There is also, no prop 3.18), there is a little subtletly here since you are asserting an homeomorphism, that might not be true with your defined topology. Maybe define a subbasis with those pre-images, taking a countable basis of $mathbbR^n$. I mixed your answer with Frost's and I think it works, thanks.
– Bajo Fondo
Jul 25 at 2:20
@BajoFondo : Maybe you use different edition (mine was 2ed). Why you say that $widetildevarphi$ might not be a homeomorphism ? I think its clearly stated in the proposition (by exhibit its continous inverse).
– Sou
Jul 25 at 2:35
Take $U times V subset mathbbR^n times mathbbR^n$ open set. Then $widetildevarphi^-1(U times V)$ is not $pi^-1(V_alpha)$ for any $V_alpha$. Fix $p in U$, then $p times T_pM subset pi^-1(U)$ but take a $v_p$ that is not in $V$, then $(p,v_p) in p times T_pM subset pi^-1(U)$, but it is not in $widetildevarphi^-1(U times V)$, hence $widetildevarphi^-1(U times V)$ cannot be of the form $pi^-1(U)$. The idea is that with your topology your open sets are the union of the elements of a class $p times T_pM; p in U_alpha$.
– Bajo Fondo
Jul 25 at 19:30
@BajoFondo You're right i made a mistake about the topology. Let me amend it.
– Sou
Jul 25 at 20:06
add a comment |Â
up vote
1
down vote
up vote
1
down vote
I guess the theorem that you mention is Proposition 3.18 on Lee's book.
For any smooth chart $(U_alpha,varphi_alpha)$ of $M$, we have a map $widetildevarphi_alpha : pi^-1(U_alpha) to BbbR^2n$ defined as
$$
widetildevarphi_alpha (v_p) = (x^1(p),dots,x^n(p),v^1,dots,v^n) .
$$
This map is a bijection onto its image $varphi_alpha(U) times BbbR^n$.
The topology defined on $TM$ is the one that generated by $widetildevarphi_alpha^-1(V) : forall alpha in A, Vsubseteq BbbR^2n text is open$. You can check directly that the collection above indeed generate a topology.
To show the Hausdorff property, let $p in U subseteq M$, where $U$ is the domain of some smooth chart $(U,varphi)$ of $M$. And $v_p,w_p in pi^-1(p) subseteq pi^-1(U)subseteq TM$, be a pair of points that lie in the same fiber.
With this topology, the map $widetildevarphi : pi^-1(U) to BbbR^2n$ is a homeomorphism onto its image. We can write $widetildevarphi(v_p) = (hatp,v)$, with $hatp = varphi(p)$. So let $widetildevarphi(v_p)=(hatp,v)$ and $widetildevarphi(w_p) = (hatp,w)$ be their images and $v,w in BbbR^n$ be two distinct vectors. To obtain disjoint neighbourhoods of $v_p$ and $w_p$, just take disjoint neighbourhoods $B_1,B_2 subseteq BbbR^n$ of $v$ and $w$ resp. and maped back the disjoint open sets $varphi(U)times B_1$ and $varphi(U)times B_2$ to $pi^-1(U)$. So $widetildevarphi^-1(varphi(U) times B_1)$ and $widetildevarphi^-1(varphi(U) times B_2)$ are the desired neighbourhoods for $v_p$ and $w_p$.
answered Jul 24 at 19:19
Sou
2,7062820
I guess the theorem that you mention is Proposition 3.18 on Lee's book.
For any smooth chart $(U_alpha,varphi_alpha)$ of $M$, we have a map $widetildevarphi_alpha : pi^-1(U_alpha) to BbbR^2n$ defined as
$$
widetildevarphi_alpha (v_p) = (x^1(p),dots,x^n(p),v^1,dots,v^n) .
$$
This map is a bijection onto its image $varphi_alpha(U) times BbbR^n$.
The topology defined on $TM$ is the one that generated by $widetildevarphi_alpha^-1(V) : forall alpha in A, Vsubseteq BbbR^2n text is open$. You can check directly that the collection above indeed generate a topology.
To show the Hausdorff property, let $p in U subseteq M$, where $U$ is the domain of some smooth chart $(U,varphi)$ of $M$. And $v_p,w_p in pi^-1(p) subseteq pi^-1(U)subseteq TM$, be a pair of points that lie in the same fiber.
With this topology, the map $widetildevarphi : pi^-1(U) to BbbR^2n$ is a homeomorphism onto its image. We can write $widetildevarphi(v_p) = (hatp,v)$, with $hatp = varphi(p)$. So let $widetildevarphi(v_p)=(hatp,v)$ and $widetildevarphi(w_p) = (hatp,w)$ be their images and $v,w in BbbR^n$ be two distinct vectors. To obtain disjoint neighbourhoods of $v_p$ and $w_p$, just take disjoint neighbourhoods $B_1,B_2 subseteq BbbR^n$ of $v$ and $w$ resp. and maped back the disjoint open sets $varphi(U)times B_1$ and $varphi(U)times B_2$ to $pi^-1(U)$. So $widetildevarphi^-1(varphi(U) times B_1)$ and $widetildevarphi^-1(varphi(U) times B_2)$ are the desired neighbourhoods for $v_p$ and $w_p$.
answered Jul 24 at 19:19
Sou
2,7062820
edited Jul 25 at 20:28
answered Jul 24 at 19:19
Sou
2,7062820
answered Jul 24 at 19:19
Sou
2,7062820
answered Jul 24 at 19:19
Sou
2,7062820
2,7062820
Thanks, in my book is Lemma 4.1 (There is also, no prop 3.18), there is a little subtletly here since you are asserting an homeomorphism, that might not be true with your defined topology. Maybe define a subbasis with those pre-images, taking a countable basis of $mathbbR^n$. I mixed your answer with Frost's and I think it works, thanks.
– Bajo Fondo
Jul 25 at 2:20
@BajoFondo : Maybe you use different edition (mine was 2ed). Why you say that $widetildevarphi$ might not be a homeomorphism ? I think its clearly stated in the proposition (by exhibit its continous inverse).
– Sou
Jul 25 at 2:35
Take $U times V subset mathbbR^n times mathbbR^n$ open set. Then $widetildevarphi^-1(U times V)$ is not $pi^-1(V_alpha)$ for any $V_alpha$. Fix $p in U$, then $p times T_pM subset pi^-1(U)$ but take a $v_p$ that is not in $V$, then $(p,v_p) in p times T_pM subset pi^-1(U)$, but it is not in $widetildevarphi^-1(U times V)$, hence $widetildevarphi^-1(U times V)$ cannot be of the form $pi^-1(U)$. The idea is that with your topology your open sets are the union of the elements of a class $p times T_pM; p in U_alpha$.
– Bajo Fondo
Jul 25 at 19:30
@BajoFondo You're right i made a mistake about the topology. Let me amend it.
– Sou
Jul 25 at 20:06
add a comment |Â
Thanks, in my book is Lemma 4.1 (There is also, no prop 3.18), there is a little subtletly here since you are asserting an homeomorphism, that might not be true with your defined topology. Maybe define a subbasis with those pre-images, taking a countable basis of $mathbbR^n$. I mixed your answer with Frost's and I think it works, thanks.
– Bajo Fondo
Jul 25 at 2:20
@BajoFondo : Maybe you use different edition (mine was 2ed). Why you say that $widetildevarphi$ might not be a homeomorphism ? I think its clearly stated in the proposition (by exhibit its continous inverse).
– Sou
Jul 25 at 2:35
Take $U times V subset mathbbR^n times mathbbR^n$ open set. Then $widetildevarphi^-1(U times V)$ is not $pi^-1(V_alpha)$ for any $V_alpha$. Fix $p in U$, then $p times T_pM subset pi^-1(U)$ but take a $v_p$ that is not in $V$, then $(p,v_p) in p times T_pM subset pi^-1(U)$, but it is not in $widetildevarphi^-1(U times V)$, hence $widetildevarphi^-1(U times V)$ cannot be of the form $pi^-1(U)$. The idea is that with your topology your open sets are the union of the elements of a class $p times T_pM; p in U_alpha$.
– Bajo Fondo
Jul 25 at 19:30
@BajoFondo You're right i made a mistake about the topology. Let me amend it.
– Sou
Jul 25 at 20:06
Thanks, in my book is Lemma 4.1 (There is also, no prop 3.18), there is a little subtletly here since you are asserting an homeomorphism, that might not be true with your defined topology. Maybe define a subbasis with those pre-images, taking a countable basis of $mathbbR^n$. I mixed your answer with Frost's and I think it works, thanks.
– Bajo Fondo
Jul 25 at 2:20
Thanks, in my book is Lemma 4.1 (There is also, no prop 3.18), there is a little subtletly here since you are asserting an homeomorphism, that might not be true with your defined topology. Maybe define a subbasis with those pre-images, taking a countable basis of $mathbbR^n$. I mixed your answer with Frost's and I think it works, thanks.
– Bajo Fondo
Jul 25 at 2:20
@BajoFondo : Maybe you use different edition (mine was 2ed). Why you say that $widetildevarphi$ might not be a homeomorphism ? I think its clearly stated in the proposition (by exhibit its continous inverse).
– Sou
Jul 25 at 2:35
@BajoFondo : Maybe you use different edition (mine was 2ed). Why you say that $widetildevarphi$ might not be a homeomorphism ? I think its clearly stated in the proposition (by exhibit its continous inverse).
– Sou
Jul 25 at 2:35
Take $U times V subset mathbbR^n times mathbbR^n$ open set. Then $widetildevarphi^-1(U times V)$ is not $pi^-1(V_alpha)$ for any $V_alpha$. Fix $p in U$, then $p times T_pM subset pi^-1(U)$ but take a $v_p$ that is not in $V$, then $(p,v_p) in p times T_pM subset pi^-1(U)$, but it is not in $widetildevarphi^-1(U times V)$, hence $widetildevarphi^-1(U times V)$ cannot be of the form $pi^-1(U)$. The idea is that with your topology your open sets are the union of the elements of a class $p times T_pM; p in U_alpha$.
– Bajo Fondo
Jul 25 at 19:30
Take $U times V subset mathbbR^n times mathbbR^n$ open set. Then $widetildevarphi^-1(U times V)$ is not $pi^-1(V_alpha)$ for any $V_alpha$. Fix $p in U$, then $p times T_pM subset pi^-1(U)$ but take a $v_p$ that is not in $V$, then $(p,v_p) in p times T_pM subset pi^-1(U)$, but it is not in $widetildevarphi^-1(U times V)$, hence $widetildevarphi^-1(U times V)$ cannot be of the form $pi^-1(U)$. The idea is that with your topology your open sets are the union of the elements of a class $p times T_pM; p in U_alpha$.
– Bajo Fondo
Jul 25 at 19:30
@BajoFondo You're right i made a mistake about the topology. Let me amend it.
– Sou
Jul 25 at 20:06
@BajoFondo You're right i made a mistake about the topology. Let me amend it.
– Sou
Jul 25 at 20:06
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