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Question about free resolution of an abelian group $H$
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In Hatcher's Algebraic Topology book on page 193 he defines a free resolution of an abelian group $H$ to be an exact sequence of the from
$... rightarrow F_2 rightarrow F_1 rightarrow F_0 rightarrow H rightarrow 0$
where each $F_n$ is a free.
Later, in the second paragraph of page 195 he allows $F_0$ to be a free abelian group. My problem here is that free abelian groups are not necessarily free groups.
What is going on here?
abstract-algebra algebraic-topology
edited 1 hour ago
Paul Frost
3,298320
asked 4 hours ago
TuoTuo
1,477512
add a comment |Â
up vote
1
down vote
favorite
In Hatcher's Algebraic Topology book on page 193 he defines a free resolution of an abelian group $H$ to be an exact sequence of the from
$... rightarrow F_2 rightarrow F_1 rightarrow F_0 rightarrow H rightarrow 0$
where each $F_n$ is a free.
Later, in the second paragraph of page 195 he allows $F_0$ to be a free abelian group. My problem here is that free abelian groups are not necessarily free groups.
What is going on here?
abstract-algebra algebraic-topology
edited 1 hour ago
Paul Frost
3,298320
asked 4 hours ago
TuoTuo
1,477512
Are you sure that “abelian†is not implied on page 193?
– egreg
4 hours ago
4
In my edition he doesn't say "free group" he just says "free". As the context is Abelian groups, he means free Abelian group.
– Lord Shark the Unknown
4 hours ago
@LordSharktheUnknown I see, this makes sense.
– TuoTuo
3 hours ago
A more general context resolutions make sense is $R$ modules for $R$ a commutative ring and abelian groups are used here precisely because they are modules for $mathbb Z$.
– ArithmeticGeometer
2 hours ago
Lord Shark the Unknown is right. I therefore edited your question to make clear what Hatcher really said: I replaced "each $F_n$ is a free group" by the original "each $F_n$ free".
– Paul Frost
1 hour ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
In Hatcher's Algebraic Topology book on page 193 he defines a free resolution of an abelian group $H$ to be an exact sequence of the from
$... rightarrow F_2 rightarrow F_1 rightarrow F_0 rightarrow H rightarrow 0$
where each $F_n$ is a free.
Later, in the second paragraph of page 195 he allows $F_0$ to be a free abelian group. My problem here is that free abelian groups are not necessarily free groups.
What is going on here?
abstract-algebra algebraic-topology
edited 1 hour ago
Paul Frost
3,298320
asked 4 hours ago
TuoTuo
1,477512
In Hatcher's Algebraic Topology book on page 193 he defines a free resolution of an abelian group $H$ to be an exact sequence of the from
$... rightarrow F_2 rightarrow F_1 rightarrow F_0 rightarrow H rightarrow 0$
where each $F_n$ is a free.
Later, in the second paragraph of page 195 he allows $F_0$ to be a free abelian group. My problem here is that free abelian groups are not necessarily free groups.
What is going on here?
abstract-algebra algebraic-topology
edited 1 hour ago
Paul Frost
3,298320
asked 4 hours ago
TuoTuo
1,477512
edited 1 hour ago
Paul Frost
3,298320
edited 1 hour ago
Paul Frost
3,298320
edited 1 hour ago
Paul Frost
3,298320
3,298320
asked 4 hours ago
TuoTuo
1,477512
asked 4 hours ago
TuoTuo
1,477512
asked 4 hours ago
TuoTuo
1,477512
1,477512
Are you sure that “abelian†is not implied on page 193?
– egreg
4 hours ago
4
In my edition he doesn't say "free group" he just says "free". As the context is Abelian groups, he means free Abelian group.
– Lord Shark the Unknown
4 hours ago
@LordSharktheUnknown I see, this makes sense.
– TuoTuo
3 hours ago
A more general context resolutions make sense is $R$ modules for $R$ a commutative ring and abelian groups are used here precisely because they are modules for $mathbb Z$.
– ArithmeticGeometer
2 hours ago
Lord Shark the Unknown is right. I therefore edited your question to make clear what Hatcher really said: I replaced "each $F_n$ is a free group" by the original "each $F_n$ free".
– Paul Frost
1 hour ago
add a comment |Â
Are you sure that “abelian†is not implied on page 193?
– egreg
4 hours ago
4
In my edition he doesn't say "free group" he just says "free". As the context is Abelian groups, he means free Abelian group.
– Lord Shark the Unknown
4 hours ago
@LordSharktheUnknown I see, this makes sense.
– TuoTuo
3 hours ago
A more general context resolutions make sense is $R$ modules for $R$ a commutative ring and abelian groups are used here precisely because they are modules for $mathbb Z$.
– ArithmeticGeometer
2 hours ago
Lord Shark the Unknown is right. I therefore edited your question to make clear what Hatcher really said: I replaced "each $F_n$ is a free group" by the original "each $F_n$ free".
– Paul Frost
1 hour ago
Are you sure that “abelian†is not implied on page 193?
– egreg
4 hours ago
Are you sure that “abelian†is not implied on page 193?
– egreg
4 hours ago
4
4
In my edition he doesn't say "free group" he just says "free". As the context is Abelian groups, he means free Abelian group.
– Lord Shark the Unknown
4 hours ago
In my edition he doesn't say "free group" he just says "free". As the context is Abelian groups, he means free Abelian group.
– Lord Shark the Unknown
4 hours ago
@LordSharktheUnknown I see, this makes sense.
– TuoTuo
3 hours ago
@LordSharktheUnknown I see, this makes sense.
– TuoTuo
3 hours ago
A more general context resolutions make sense is $R$ modules for $R$ a commutative ring and abelian groups are used here precisely because they are modules for $mathbb Z$.
– ArithmeticGeometer
2 hours ago
A more general context resolutions make sense is $R$ modules for $R$ a commutative ring and abelian groups are used here precisely because they are modules for $mathbb Z$.
– ArithmeticGeometer
2 hours ago
Lord Shark the Unknown is right. I therefore edited your question to make clear what Hatcher really said: I replaced "each $F_n$ is a free group" by the original "each $F_n$ free".
– Paul Frost
1 hour ago
Lord Shark the Unknown is right. I therefore edited your question to make clear what Hatcher really said: I replaced "each $F_n$ is a free group" by the original "each $F_n$ free".
– Paul Frost
1 hour ago
add a comment |Â
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Are you sure that “abelian†is not implied on page 193?
– egreg
4 hours ago
4
In my edition he doesn't say "free group" he just says "free". As the context is Abelian groups, he means free Abelian group.
– Lord Shark the Unknown
4 hours ago
@LordSharktheUnknown I see, this makes sense.
– TuoTuo
3 hours ago
A more general context resolutions make sense is $R$ modules for $R$ a commutative ring and abelian groups are used here precisely because they are modules for $mathbb Z$.
– ArithmeticGeometer
2 hours ago
Lord Shark the Unknown is right. I therefore edited your question to make clear what Hatcher really said: I replaced "each $F_n$ is a free group" by the original "each $F_n$ free".
– Paul Frost
1 hour ago