Mistakes in Bredon's book “Topology and Geometry”?
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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 $ast$ does 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 2 hours ago
Francesco PolizziFrancesco Polizzi
47.5k3126204
$endgroup$
|
show 6 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 $ast$ does 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 2 hours ago
Francesco PolizziFrancesco Polizzi
47.5k3126204
$endgroup$
$begingroup$
I guess $*$ might be a typo, it would rather be some sort of $otimes$.
$endgroup$
– Dima Pasechnik
1 hour ago
1
$begingroup$
I suspect that the words "orientable along $A$" mean something like "a neighbourhood of $A$ is orientable", in which case corollary 7.11 would be correct
$endgroup$
– Denis Nardin
1 hour ago
1
$begingroup$
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
1 hour ago
1
$begingroup$
Yes, in fact this is true for any compact subset with a simply connected neighborhood (to see an example where this is not true, take the latitude in a torus)
$endgroup$
– Denis Nardin
1 hour ago
1
$begingroup$
These matters are also well-covered elsewhere, e.g., in Dold's "Lectures on Algebraic Topology" where one can check when in doubt.
$endgroup$
– Igor Belegradek
47 mins ago
|
show 6 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 $ast$ does 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 2 hours ago
Francesco PolizziFrancesco Polizzi
47.5k3126204
$endgroup$
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 $ast$ does 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 2 hours ago
Francesco PolizziFrancesco Polizzi
47.5k3126204
asked 2 hours ago
Francesco PolizziFrancesco Polizzi
47.5k3126204
edited 1 hour ago
Francesco Polizzi
asked 2 hours ago
Francesco PolizziFrancesco Polizzi
47.5k3126204
asked 2 hours ago
Francesco PolizziFrancesco Polizzi
47.5k3126204
asked 2 hours ago
Francesco PolizziFrancesco Polizzi
47.5k3126204
47.5k3126204
$begingroup$
I guess $*$ might be a typo, it would rather be some sort of $otimes$.
$endgroup$
– Dima Pasechnik
1 hour ago
1
$begingroup$
I suspect that the words "orientable along $A$" mean something like "a neighbourhood of $A$ is orientable", in which case corollary 7.11 would be correct
$endgroup$
– Denis Nardin
1 hour ago
1
$begingroup$
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
1 hour ago
1
$begingroup$
Yes, in fact this is true for any compact subset with a simply connected neighborhood (to see an example where this is not true, take the latitude in a torus)
$endgroup$
– Denis Nardin
1 hour ago
1
$begingroup$
These matters are also well-covered elsewhere, e.g., in Dold's "Lectures on Algebraic Topology" where one can check when in doubt.
$endgroup$
– Igor Belegradek
47 mins ago
|
show 6 more comments
$begingroup$
I guess $*$ might be a typo, it would rather be some sort of $otimes$.
$endgroup$
– Dima Pasechnik
1 hour ago
1
$begingroup$
I suspect that the words "orientable along $A$" mean something like "a neighbourhood of $A$ is orientable", in which case corollary 7.11 would be correct
$endgroup$
– Denis Nardin
1 hour ago
1
$begingroup$
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
1 hour ago
1
$begingroup$
Yes, in fact this is true for any compact subset with a simply connected neighborhood (to see an example where this is not true, take the latitude in a torus)
$endgroup$
– Denis Nardin
1 hour ago
1
$begingroup$
These matters are also well-covered elsewhere, e.g., in Dold's "Lectures on Algebraic Topology" where one can check when in doubt.
$endgroup$
– Igor Belegradek
47 mins ago
$begingroup$
I guess $*$ might be a typo, it would rather be some sort of $otimes$.
$endgroup$
– Dima Pasechnik
1 hour ago
$begingroup$
I guess $*$ might be a typo, it would rather be some sort of $otimes$.
$endgroup$
– Dima Pasechnik
1 hour ago
1
1
$begingroup$
I suspect that the words "orientable along $A$" mean something like "a neighbourhood of $A$ is orientable", in which case corollary 7.11 would be correct
$endgroup$
– Denis Nardin
1 hour ago
$begingroup$
I suspect that the words "orientable along $A$" mean something like "a neighbourhood of $A$ is orientable", in which case corollary 7.11 would be correct
$endgroup$
– Denis Nardin
1 hour ago
1
1
$begingroup$
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
1 hour ago
$begingroup$
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
1 hour ago
1
1
$begingroup$
Yes, in fact this is true for any compact subset with a simply connected neighborhood (to see an example where this is not true, take the latitude in a torus)
$endgroup$
– Denis Nardin
1 hour ago
$begingroup$
Yes, in fact this is true for any compact subset with a simply connected neighborhood (to see an example where this is not true, take the latitude in a torus)
$endgroup$
– Denis Nardin
1 hour ago
1
1
$begingroup$
These matters are also well-covered elsewhere, e.g., in Dold's "Lectures on Algebraic Topology" where one can check when in doubt.
$endgroup$
– Igor Belegradek
47 mins ago
$begingroup$
These matters are also well-covered elsewhere, e.g., in Dold's "Lectures on Algebraic Topology" where one can check when in doubt.
$endgroup$
– Igor Belegradek
47 mins ago
|
show 6 more comments
1 Answer
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$begingroup$
Star (in older topology texts) often indicate torsion product of abelian groups, that is, $A * B := Tor_{Bbb Z}(A, B)$. Usually it is clear from the context whether free product or torsion product is meant.
answered 31 mins ago
Denis T.Denis T.
944614
$endgroup$
$begingroup$
Thanks. I was not aware of this (old) notation.
$endgroup$
– Francesco Polizzi
1 min ago
add a comment |
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$begingroup$
Star (in older topology texts) often indicate torsion product of abelian groups, that is, $A * B := Tor_{Bbb Z}(A, B)$. Usually it is clear from the context whether free product or torsion product is meant.
answered 31 mins ago
Denis T.Denis T.
944614
$endgroup$
$begingroup$
Thanks. I was not aware of this (old) notation.
$endgroup$
– Francesco Polizzi
1 min ago
add a comment |
$begingroup$
Star (in older topology texts) often indicate torsion product of abelian groups, that is, $A * B := Tor_{Bbb Z}(A, B)$. Usually it is clear from the context whether free product or torsion product is meant.
answered 31 mins ago
Denis T.Denis T.
944614
$endgroup$
$begingroup$
Thanks. I was not aware of this (old) notation.
$endgroup$
– Francesco Polizzi
1 min ago
add a comment |
$begingroup$
Star (in older topology texts) often indicate torsion product of abelian groups, that is, $A * B := Tor_{Bbb Z}(A, B)$. Usually it is clear from the context whether free product or torsion product is meant.
answered 31 mins ago
Denis T.Denis T.
944614
$endgroup$
Star (in older topology texts) often indicate torsion product of abelian groups, that is, $A * B := Tor_{Bbb Z}(A, B)$. Usually it is clear from the context whether free product or torsion product is meant.
answered 31 mins ago
Denis T.Denis T.
944614
answered 31 mins ago
Denis T.Denis T.
944614
answered 31 mins ago
Denis T.Denis T.
944614
answered 31 mins ago
Denis T.Denis T.
944614
944614
$begingroup$
Thanks. I was not aware of this (old) notation.
$endgroup$
– Francesco Polizzi
1 min ago
add a comment |
$begingroup$
Thanks. I was not aware of this (old) notation.
$endgroup$
– Francesco Polizzi
1 min ago
$begingroup$
Thanks. I was not aware of this (old) notation.
$endgroup$
– Francesco Polizzi
1 min ago
$begingroup$
Thanks. I was not aware of this (old) notation.
$endgroup$
– Francesco Polizzi
1 min ago
add a comment |
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$begingroup$
I guess $*$ might be a typo, it would rather be some sort of $otimes$.
$endgroup$
– Dima Pasechnik
1 hour ago
1
$begingroup$
I suspect that the words "orientable along $A$" mean something like "a neighbourhood of $A$ is orientable", in which case corollary 7.11 would be correct
$endgroup$
– Denis Nardin
1 hour ago
1
$begingroup$
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
1 hour ago
1
$begingroup$
Yes, in fact this is true for any compact subset with a simply connected neighborhood (to see an example where this is not true, take the latitude in a torus)
$endgroup$
– Denis Nardin
1 hour ago
1
$begingroup$
These matters are also well-covered elsewhere, e.g., in Dold's "Lectures on Algebraic Topology" where one can check when in doubt.
$endgroup$
– Igor Belegradek
47 mins ago