Mistakes in Bredon's book “Topology and Geometry”?
$begingroup$
I am preparing the notes for a course in Algebraic Topology, so I decided to borrow some of the material from the classical (and wonderful) book by G. Bredon Topology and Geometry.
Looking at the part regarding the orientation of a topological $n$-manifold $M^n$, at page 341 we find the following well-known result, with its usual proof (Proposition 7.1):
So far, so good. However, after five pages we find what follows:
This makes me confused, for at least two reasons:
Point 1. The Note after the statement of Proposition 7.10 does not make any sense to me. As defined, the symbol ${}_2G$ denotes the $2$-torsion part of the abelian group $G$, so if $G$ is torsion-free (for instance, if $G=mathbf{Z}$) then ${}_2G=0$. This is clearly very different from the free-product $G ast mathbf{Z_2}$ (here $ast$ seems to denote the free-product, see pages 158-159).
Point 2. In Corollary 7.11, take $A={x}$ and $G=mathbf{Z}$. Then, when $M$ is not orientable one finds $H_n(M, , M-{x}, , mathbf{Z})=0$, and this contradicts Proposition 7.1, that yields the (correct, as far as I know) result $H_n(M, , M-{x}, , mathbf{Z})= mathbf{Z}$.
Question. Are the issues risen in Points 1, 2 above really mistakes in Bredon's book, or perhaps am I missing something trivial?
dg.differential-geometry at.algebraic-topology
asked Jan 24 at 11:12
Francesco PolizziFrancesco Polizzi
48.1k3127208
$endgroup$
|
show 9 more comments
$begingroup$
I am preparing the notes for a course in Algebraic Topology, so I decided to borrow some of the material from the classical (and wonderful) book by G. Bredon Topology and Geometry.
Looking at the part regarding the orientation of a topological $n$-manifold $M^n$, at page 341 we find the following well-known result, with its usual proof (Proposition 7.1):
So far, so good. However, after five pages we find what follows:
This makes me confused, for at least two reasons:
Point 1. The Note after the statement of Proposition 7.10 does not make any sense to me. As defined, the symbol ${}_2G$ denotes the $2$-torsion part of the abelian group $G$, so if $G$ is torsion-free (for instance, if $G=mathbf{Z}$) then ${}_2G=0$. This is clearly very different from the free-product $G ast mathbf{Z_2}$ (here $ast$ seems to denote the free-product, see pages 158-159).
Point 2. In Corollary 7.11, take $A={x}$ and $G=mathbf{Z}$. Then, when $M$ is not orientable one finds $H_n(M, , M-{x}, , mathbf{Z})=0$, and this contradicts Proposition 7.1, that yields the (correct, as far as I know) result $H_n(M, , M-{x}, , mathbf{Z})= mathbf{Z}$.
Question. Are the issues risen in Points 1, 2 above really mistakes in Bredon's book, or perhaps am I missing something trivial?
dg.differential-geometry at.algebraic-topology
asked Jan 24 at 11:12
Francesco PolizziFrancesco Polizzi
48.1k3127208
$endgroup$
2
$begingroup$
I guess $*$ might be a typo, it would rather be some sort of $otimes$.
$endgroup$
– Dima Pasechnik
Jan 24 at 11:42
4
$begingroup$
Every point has an orientable neighborhood (say, a ball), hence $M$ is always orientable along ${x}$, so corollary 7.11 says that for every manifold $M$ the formula you give holds.
$endgroup$
– Denis Nardin
Jan 24 at 11:59
7
$begingroup$
Also it seems that Bredon indicates with $ast$ what I would call $mathrm{Tor}_1$, so in particular $Aast mathbb{Z}/n$ is exactly the $n$-torsion of $A$.
$endgroup$
– Denis Nardin
Jan 24 at 12:07
3
$begingroup$
Well, at page 158 it also indicate by $*$ the free product, and in a book of 550 pages it is not easy to understand where the same notation indicates two very different things. Now it makes sense, thanks!
$endgroup$
– Francesco Polizzi
Jan 24 at 12:09
5
$begingroup$
@GeraldEdgar: Bredon died in 2000, and there is no webpage available. On the Springer's webpage there is no errata, either. Actually, on the web I found nothing (well, maybe I did not look well enough).
$endgroup$
– Francesco Polizzi
Jan 24 at 14:51
|
show 9 more comments
$begingroup$
I am preparing the notes for a course in Algebraic Topology, so I decided to borrow some of the material from the classical (and wonderful) book by G. Bredon Topology and Geometry.
Looking at the part regarding the orientation of a topological $n$-manifold $M^n$, at page 341 we find the following well-known result, with its usual proof (Proposition 7.1):
So far, so good. However, after five pages we find what follows:
This makes me confused, for at least two reasons:
Point 1. The Note after the statement of Proposition 7.10 does not make any sense to me. As defined, the symbol ${}_2G$ denotes the $2$-torsion part of the abelian group $G$, so if $G$ is torsion-free (for instance, if $G=mathbf{Z}$) then ${}_2G=0$. This is clearly very different from the free-product $G ast mathbf{Z_2}$ (here $ast$ seems to denote the free-product, see pages 158-159).
Point 2. In Corollary 7.11, take $A={x}$ and $G=mathbf{Z}$. Then, when $M$ is not orientable one finds $H_n(M, , M-{x}, , mathbf{Z})=0$, and this contradicts Proposition 7.1, that yields the (correct, as far as I know) result $H_n(M, , M-{x}, , mathbf{Z})= mathbf{Z}$.
Question. Are the issues risen in Points 1, 2 above really mistakes in Bredon's book, or perhaps am I missing something trivial?
dg.differential-geometry at.algebraic-topology
asked Jan 24 at 11:12
Francesco PolizziFrancesco Polizzi
48.1k3127208
$endgroup$
I am preparing the notes for a course in Algebraic Topology, so I decided to borrow some of the material from the classical (and wonderful) book by G. Bredon Topology and Geometry.
Looking at the part regarding the orientation of a topological $n$-manifold $M^n$, at page 341 we find the following well-known result, with its usual proof (Proposition 7.1):
So far, so good. However, after five pages we find what follows:
This makes me confused, for at least two reasons:
Point 1. The Note after the statement of Proposition 7.10 does not make any sense to me. As defined, the symbol ${}_2G$ denotes the $2$-torsion part of the abelian group $G$, so if $G$ is torsion-free (for instance, if $G=mathbf{Z}$) then ${}_2G=0$. This is clearly very different from the free-product $G ast mathbf{Z_2}$ (here $ast$ seems to denote the free-product, see pages 158-159).
Point 2. In Corollary 7.11, take $A={x}$ and $G=mathbf{Z}$. Then, when $M$ is not orientable one finds $H_n(M, , M-{x}, , mathbf{Z})=0$, and this contradicts Proposition 7.1, that yields the (correct, as far as I know) result $H_n(M, , M-{x}, , mathbf{Z})= mathbf{Z}$.
Question. Are the issues risen in Points 1, 2 above really mistakes in Bredon's book, or perhaps am I missing something trivial?
dg.differential-geometry at.algebraic-topology
dg.differential-geometry at.algebraic-topology
asked Jan 24 at 11:12
Francesco PolizziFrancesco Polizzi
48.1k3127208
asked Jan 24 at 11:12
Francesco PolizziFrancesco Polizzi
48.1k3127208
edited Jan 24 at 21:02
Francesco Polizzi
asked Jan 24 at 11:12
Francesco PolizziFrancesco Polizzi
48.1k3127208
asked Jan 24 at 11:12
Francesco PolizziFrancesco Polizzi
48.1k3127208
asked Jan 24 at 11:12
Francesco PolizziFrancesco Polizzi
48.1k3127208
48.1k3127208
2
$begingroup$
I guess $*$ might be a typo, it would rather be some sort of $otimes$.
$endgroup$
– Dima Pasechnik
Jan 24 at 11:42
4
$begingroup$
Every point has an orientable neighborhood (say, a ball), hence $M$ is always orientable along ${x}$, so corollary 7.11 says that for every manifold $M$ the formula you give holds.
$endgroup$
– Denis Nardin
Jan 24 at 11:59
7
$begingroup$
Also it seems that Bredon indicates with $ast$ what I would call $mathrm{Tor}_1$, so in particular $Aast mathbb{Z}/n$ is exactly the $n$-torsion of $A$.
$endgroup$
– Denis Nardin
Jan 24 at 12:07
3
$begingroup$
Well, at page 158 it also indicate by $*$ the free product, and in a book of 550 pages it is not easy to understand where the same notation indicates two very different things. Now it makes sense, thanks!
$endgroup$
– Francesco Polizzi
Jan 24 at 12:09
5
$begingroup$
@GeraldEdgar: Bredon died in 2000, and there is no webpage available. On the Springer's webpage there is no errata, either. Actually, on the web I found nothing (well, maybe I did not look well enough).
$endgroup$
– Francesco Polizzi
Jan 24 at 14:51
|
show 9 more comments
2
$begingroup$
I guess $*$ might be a typo, it would rather be some sort of $otimes$.
$endgroup$
– Dima Pasechnik
Jan 24 at 11:42
4
$begingroup$
Every point has an orientable neighborhood (say, a ball), hence $M$ is always orientable along ${x}$, so corollary 7.11 says that for every manifold $M$ the formula you give holds.
$endgroup$
– Denis Nardin
Jan 24 at 11:59
7
$begingroup$
Also it seems that Bredon indicates with $ast$ what I would call $mathrm{Tor}_1$, so in particular $Aast mathbb{Z}/n$ is exactly the $n$-torsion of $A$.
$endgroup$
– Denis Nardin
Jan 24 at 12:07
3
$begingroup$
Well, at page 158 it also indicate by $*$ the free product, and in a book of 550 pages it is not easy to understand where the same notation indicates two very different things. Now it makes sense, thanks!
$endgroup$
– Francesco Polizzi
Jan 24 at 12:09
5
$begingroup$
@GeraldEdgar: Bredon died in 2000, and there is no webpage available. On the Springer's webpage there is no errata, either. Actually, on the web I found nothing (well, maybe I did not look well enough).
$endgroup$
– Francesco Polizzi
Jan 24 at 14:51
2
2
$begingroup$
I guess $*$ might be a typo, it would rather be some sort of $otimes$.
$endgroup$
– Dima Pasechnik
Jan 24 at 11:42
$begingroup$
I guess $*$ might be a typo, it would rather be some sort of $otimes$.
$endgroup$
– Dima Pasechnik
Jan 24 at 11:42
4
4
$begingroup$
Every point has an orientable neighborhood (say, a ball), hence $M$ is always orientable along ${x}$, so corollary 7.11 says that for every manifold $M$ the formula you give holds.
$endgroup$
– Denis Nardin
Jan 24 at 11:59
$begingroup$
Every point has an orientable neighborhood (say, a ball), hence $M$ is always orientable along ${x}$, so corollary 7.11 says that for every manifold $M$ the formula you give holds.
$endgroup$
– Denis Nardin
Jan 24 at 11:59
7
7
$begingroup$
Also it seems that Bredon indicates with $ast$ what I would call $mathrm{Tor}_1$, so in particular $Aast mathbb{Z}/n$ is exactly the $n$-torsion of $A$.
$endgroup$
– Denis Nardin
Jan 24 at 12:07
$begingroup$
Also it seems that Bredon indicates with $ast$ what I would call $mathrm{Tor}_1$, so in particular $Aast mathbb{Z}/n$ is exactly the $n$-torsion of $A$.
$endgroup$
– Denis Nardin
Jan 24 at 12:07
3
3
$begingroup$
Well, at page 158 it also indicate by $*$ the free product, and in a book of 550 pages it is not easy to understand where the same notation indicates two very different things. Now it makes sense, thanks!
$endgroup$
– Francesco Polizzi
Jan 24 at 12:09
$begingroup$
Well, at page 158 it also indicate by $*$ the free product, and in a book of 550 pages it is not easy to understand where the same notation indicates two very different things. Now it makes sense, thanks!
$endgroup$
– Francesco Polizzi
Jan 24 at 12:09
5
5
$begingroup$
@GeraldEdgar: Bredon died in 2000, and there is no webpage available. On the Springer's webpage there is no errata, either. Actually, on the web I found nothing (well, maybe I did not look well enough).
$endgroup$
– Francesco Polizzi
Jan 24 at 14:51
$begingroup$
@GeraldEdgar: Bredon died in 2000, and there is no webpage available. On the Springer's webpage there is no errata, either. Actually, on the web I found nothing (well, maybe I did not look well enough).
$endgroup$
– Francesco Polizzi
Jan 24 at 14:51
|
show 9 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Star (in older topology texts) often indicate torsion product of abelian groups, that is, $A * B := operatorname{Tor}_{Bbb Z}(A, B)$. Usually it is clear from the context whether free product or torsion product is meant.
edited Jan 26 at 8:40
Community♦
123
answered Jan 24 at 13:04
Denis T.Denis T.
1,263816
$endgroup$
8
$begingroup$
Thanks. I was not aware of this (old) notation.
$endgroup$
– Francesco Polizzi
Jan 24 at 13:34
5
$begingroup$
(This notation is also used in Spanier's text, for example.)
$endgroup$
– Pedro Tamaroff
Jan 24 at 13:51
1
$begingroup$
And Munkres!...
$endgroup$
– Greg Friedman
Jan 25 at 5:00
1
$begingroup$
I'm not a specialist in Algebraic Topology, and my background on these basic topics is mainly from Massey's and Hatcher's books, where I never found this notation for $mathrm{Tor}_1$ (at least, as far as I can remember). I am actually quite surprised that it seems to be rather common in older textbooks.
$endgroup$
– Francesco Polizzi
Jan 25 at 8:29
6
$begingroup$
@FrancescoPolizzi I'm allegedly a specialist in Algebraic Topology, and I didn't know either, so don't feel too bad :)
$endgroup$
– Denis Nardin
Jan 25 at 12:22
|
show 2 more comments
$begingroup$
I think that you are missing the definition of 'orientable along $A$'. I haven't got that book of Bredon to hand, but presumably 'orientable along $A$' means that if you move a local orientation of $M$ around a closed path that stays in $A$ then it will come back to the same local orientation. In particular, in the case when $A$ is a single point, then $M$ will always be orientable along $A$, regardless of whether $M$ is orientable or not, so the case that you view as wrong doesn't arise.
I agree with Denis T's interpretation of the notation $A*B$.
answered Jan 25 at 11:59
IJLIJL
701311
$endgroup$
$begingroup$
Yes, definitely I was confused about the definition of "orientable along $A$". And I was unaware of the old notation $A*B$ for $mathrm{Tor}_1(A, , B)$.
$endgroup$
– Francesco Polizzi
Jan 25 at 17:27
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Star (in older topology texts) often indicate torsion product of abelian groups, that is, $A * B := operatorname{Tor}_{Bbb Z}(A, B)$. Usually it is clear from the context whether free product or torsion product is meant.
edited Jan 26 at 8:40
Community♦
123
answered Jan 24 at 13:04
Denis T.Denis T.
1,263816
$endgroup$
8
$begingroup$
Thanks. I was not aware of this (old) notation.
$endgroup$
– Francesco Polizzi
Jan 24 at 13:34
5
$begingroup$
(This notation is also used in Spanier's text, for example.)
$endgroup$
– Pedro Tamaroff
Jan 24 at 13:51
1
$begingroup$
And Munkres!...
$endgroup$
– Greg Friedman
Jan 25 at 5:00
1
$begingroup$
I'm not a specialist in Algebraic Topology, and my background on these basic topics is mainly from Massey's and Hatcher's books, where I never found this notation for $mathrm{Tor}_1$ (at least, as far as I can remember). I am actually quite surprised that it seems to be rather common in older textbooks.
$endgroup$
– Francesco Polizzi
Jan 25 at 8:29
6
$begingroup$
@FrancescoPolizzi I'm allegedly a specialist in Algebraic Topology, and I didn't know either, so don't feel too bad :)
$endgroup$
– Denis Nardin
Jan 25 at 12:22
|
show 2 more comments
$begingroup$
Star (in older topology texts) often indicate torsion product of abelian groups, that is, $A * B := operatorname{Tor}_{Bbb Z}(A, B)$. Usually it is clear from the context whether free product or torsion product is meant.
edited Jan 26 at 8:40
Community♦
123
answered Jan 24 at 13:04
Denis T.Denis T.
1,263816
$endgroup$
8
$begingroup$
Thanks. I was not aware of this (old) notation.
$endgroup$
– Francesco Polizzi
Jan 24 at 13:34
5
$begingroup$
(This notation is also used in Spanier's text, for example.)
$endgroup$
– Pedro Tamaroff
Jan 24 at 13:51
1
$begingroup$
And Munkres!...
$endgroup$
– Greg Friedman
Jan 25 at 5:00
1
$begingroup$
I'm not a specialist in Algebraic Topology, and my background on these basic topics is mainly from Massey's and Hatcher's books, where I never found this notation for $mathrm{Tor}_1$ (at least, as far as I can remember). I am actually quite surprised that it seems to be rather common in older textbooks.
$endgroup$
– Francesco Polizzi
Jan 25 at 8:29
6
$begingroup$
@FrancescoPolizzi I'm allegedly a specialist in Algebraic Topology, and I didn't know either, so don't feel too bad :)
$endgroup$
– Denis Nardin
Jan 25 at 12:22
|
show 2 more comments
$begingroup$
Star (in older topology texts) often indicate torsion product of abelian groups, that is, $A * B := operatorname{Tor}_{Bbb Z}(A, B)$. Usually it is clear from the context whether free product or torsion product is meant.
edited Jan 26 at 8:40
Community♦
123
answered Jan 24 at 13:04
Denis T.Denis T.
1,263816
$endgroup$
Star (in older topology texts) often indicate torsion product of abelian groups, that is, $A * B := operatorname{Tor}_{Bbb Z}(A, B)$. Usually it is clear from the context whether free product or torsion product is meant.
edited Jan 26 at 8:40
Community♦
123
answered Jan 24 at 13:04
Denis T.Denis T.
1,263816
edited Jan 26 at 8:40
Community♦
123
edited Jan 26 at 8:40
Community♦
123
edited Jan 26 at 8:40
Community♦
123
123
answered Jan 24 at 13:04
Denis T.Denis T.
1,263816
answered Jan 24 at 13:04
Denis T.Denis T.
1,263816
answered Jan 24 at 13:04
Denis T.Denis T.
1,263816
1,263816
8
$begingroup$
Thanks. I was not aware of this (old) notation.
$endgroup$
– Francesco Polizzi
Jan 24 at 13:34
5
$begingroup$
(This notation is also used in Spanier's text, for example.)
$endgroup$
– Pedro Tamaroff
Jan 24 at 13:51
1
$begingroup$
And Munkres!...
$endgroup$
– Greg Friedman
Jan 25 at 5:00
1
$begingroup$
I'm not a specialist in Algebraic Topology, and my background on these basic topics is mainly from Massey's and Hatcher's books, where I never found this notation for $mathrm{Tor}_1$ (at least, as far as I can remember). I am actually quite surprised that it seems to be rather common in older textbooks.
$endgroup$
– Francesco Polizzi
Jan 25 at 8:29
6
$begingroup$
@FrancescoPolizzi I'm allegedly a specialist in Algebraic Topology, and I didn't know either, so don't feel too bad :)
$endgroup$
– Denis Nardin
Jan 25 at 12:22
|
show 2 more comments
8
$begingroup$
Thanks. I was not aware of this (old) notation.
$endgroup$
– Francesco Polizzi
Jan 24 at 13:34
5
$begingroup$
(This notation is also used in Spanier's text, for example.)
$endgroup$
– Pedro Tamaroff
Jan 24 at 13:51
1
$begingroup$
And Munkres!...
$endgroup$
– Greg Friedman
Jan 25 at 5:00
1
$begingroup$
I'm not a specialist in Algebraic Topology, and my background on these basic topics is mainly from Massey's and Hatcher's books, where I never found this notation for $mathrm{Tor}_1$ (at least, as far as I can remember). I am actually quite surprised that it seems to be rather common in older textbooks.
$endgroup$
– Francesco Polizzi
Jan 25 at 8:29
6
$begingroup$
@FrancescoPolizzi I'm allegedly a specialist in Algebraic Topology, and I didn't know either, so don't feel too bad :)
$endgroup$
– Denis Nardin
Jan 25 at 12:22
8
8
$begingroup$
Thanks. I was not aware of this (old) notation.
$endgroup$
– Francesco Polizzi
Jan 24 at 13:34
$begingroup$
Thanks. I was not aware of this (old) notation.
$endgroup$
– Francesco Polizzi
Jan 24 at 13:34
5
5
$begingroup$
(This notation is also used in Spanier's text, for example.)
$endgroup$
– Pedro Tamaroff
Jan 24 at 13:51
$begingroup$
(This notation is also used in Spanier's text, for example.)
$endgroup$
– Pedro Tamaroff
Jan 24 at 13:51
1
1
$begingroup$
And Munkres!...
$endgroup$
– Greg Friedman
Jan 25 at 5:00
$begingroup$
And Munkres!...
$endgroup$
– Greg Friedman
Jan 25 at 5:00
1
1
$begingroup$
I'm not a specialist in Algebraic Topology, and my background on these basic topics is mainly from Massey's and Hatcher's books, where I never found this notation for $mathrm{Tor}_1$ (at least, as far as I can remember). I am actually quite surprised that it seems to be rather common in older textbooks.
$endgroup$
– Francesco Polizzi
Jan 25 at 8:29
$begingroup$
I'm not a specialist in Algebraic Topology, and my background on these basic topics is mainly from Massey's and Hatcher's books, where I never found this notation for $mathrm{Tor}_1$ (at least, as far as I can remember). I am actually quite surprised that it seems to be rather common in older textbooks.
$endgroup$
– Francesco Polizzi
Jan 25 at 8:29
6
6
$begingroup$
@FrancescoPolizzi I'm allegedly a specialist in Algebraic Topology, and I didn't know either, so don't feel too bad :)
$endgroup$
– Denis Nardin
Jan 25 at 12:22
$begingroup$
@FrancescoPolizzi I'm allegedly a specialist in Algebraic Topology, and I didn't know either, so don't feel too bad :)
$endgroup$
– Denis Nardin
Jan 25 at 12:22
|
show 2 more comments
$begingroup$
I think that you are missing the definition of 'orientable along $A$'. I haven't got that book of Bredon to hand, but presumably 'orientable along $A$' means that if you move a local orientation of $M$ around a closed path that stays in $A$ then it will come back to the same local orientation. In particular, in the case when $A$ is a single point, then $M$ will always be orientable along $A$, regardless of whether $M$ is orientable or not, so the case that you view as wrong doesn't arise.
I agree with Denis T's interpretation of the notation $A*B$.
answered Jan 25 at 11:59
IJLIJL
701311
$endgroup$
$begingroup$
Yes, definitely I was confused about the definition of "orientable along $A$". And I was unaware of the old notation $A*B$ for $mathrm{Tor}_1(A, , B)$.
$endgroup$
– Francesco Polizzi
Jan 25 at 17:27
add a comment |
$begingroup$
I think that you are missing the definition of 'orientable along $A$'. I haven't got that book of Bredon to hand, but presumably 'orientable along $A$' means that if you move a local orientation of $M$ around a closed path that stays in $A$ then it will come back to the same local orientation. In particular, in the case when $A$ is a single point, then $M$ will always be orientable along $A$, regardless of whether $M$ is orientable or not, so the case that you view as wrong doesn't arise.
I agree with Denis T's interpretation of the notation $A*B$.
answered Jan 25 at 11:59
IJLIJL
701311
$endgroup$
$begingroup$
Yes, definitely I was confused about the definition of "orientable along $A$". And I was unaware of the old notation $A*B$ for $mathrm{Tor}_1(A, , B)$.
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– Francesco Polizzi
Jan 25 at 17:27
add a comment |
$begingroup$
I think that you are missing the definition of 'orientable along $A$'. I haven't got that book of Bredon to hand, but presumably 'orientable along $A$' means that if you move a local orientation of $M$ around a closed path that stays in $A$ then it will come back to the same local orientation. In particular, in the case when $A$ is a single point, then $M$ will always be orientable along $A$, regardless of whether $M$ is orientable or not, so the case that you view as wrong doesn't arise.
I agree with Denis T's interpretation of the notation $A*B$.
answered Jan 25 at 11:59
IJLIJL
701311
$endgroup$
I think that you are missing the definition of 'orientable along $A$'. I haven't got that book of Bredon to hand, but presumably 'orientable along $A$' means that if you move a local orientation of $M$ around a closed path that stays in $A$ then it will come back to the same local orientation. In particular, in the case when $A$ is a single point, then $M$ will always be orientable along $A$, regardless of whether $M$ is orientable or not, so the case that you view as wrong doesn't arise.
I agree with Denis T's interpretation of the notation $A*B$.
answered Jan 25 at 11:59
IJLIJL
701311
edited Jan 25 at 14:43
answered Jan 25 at 11:59
IJLIJL
701311
answered Jan 25 at 11:59
IJLIJL
701311
answered Jan 25 at 11:59
IJLIJL
701311
701311
$begingroup$
Yes, definitely I was confused about the definition of "orientable along $A$". And I was unaware of the old notation $A*B$ for $mathrm{Tor}_1(A, , B)$.
$endgroup$
– Francesco Polizzi
Jan 25 at 17:27
add a comment |
$begingroup$
Yes, definitely I was confused about the definition of "orientable along $A$". And I was unaware of the old notation $A*B$ for $mathrm{Tor}_1(A, , B)$.
$endgroup$
– Francesco Polizzi
Jan 25 at 17:27
$begingroup$
Yes, definitely I was confused about the definition of "orientable along $A$". And I was unaware of the old notation $A*B$ for $mathrm{Tor}_1(A, , B)$.
$endgroup$
– Francesco Polizzi
Jan 25 at 17:27
$begingroup$
Yes, definitely I was confused about the definition of "orientable along $A$". And I was unaware of the old notation $A*B$ for $mathrm{Tor}_1(A, , B)$.
$endgroup$
– Francesco Polizzi
Jan 25 at 17:27
add a comment |
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2
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I guess $*$ might be a typo, it would rather be some sort of $otimes$.
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– Dima Pasechnik
Jan 24 at 11:42
4
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Every point has an orientable neighborhood (say, a ball), hence $M$ is always orientable along ${x}$, so corollary 7.11 says that for every manifold $M$ the formula you give holds.
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– Denis Nardin
Jan 24 at 11:59
7
$begingroup$
Also it seems that Bredon indicates with $ast$ what I would call $mathrm{Tor}_1$, so in particular $Aast mathbb{Z}/n$ is exactly the $n$-torsion of $A$.
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– Denis Nardin
Jan 24 at 12:07
3
$begingroup$
Well, at page 158 it also indicate by $*$ the free product, and in a book of 550 pages it is not easy to understand where the same notation indicates two very different things. Now it makes sense, thanks!
$endgroup$
– Francesco Polizzi
Jan 24 at 12:09
5
$begingroup$
@GeraldEdgar: Bredon died in 2000, and there is no webpage available. On the Springer's webpage there is no errata, either. Actually, on the web I found nothing (well, maybe I did not look well enough).
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– Francesco Polizzi
Jan 24 at 14:51