Are all (unbounded) chain complexes over a field bifibrant?
$begingroup$
In his paper Homological algebra of homotopy algebras, Hinich defines a model structure on unbounded chain complexes where
- weak equivalences are the quasi-isomorphisms,
- fibrations are degree-wise surjections, and
- cofibrations are defined by the left lifting property.
Obviously, every object is fibrant. Now assume that we are working over a field $k$ instead of a general ring.
Question: Is it true that in this case every chain complex is also cofibrant?
I believe the answer to be yes, and I have an ugly argument that should prove it (even though I still have to check the details). Is there a sly/elegant way to show it?
homological-algebra model-categories
asked Oct 4 '17 at 5:34
Daniel Robert-NicoudDaniel Robert-Nicoud
20.5k33797
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add a comment |
$begingroup$
In his paper Homological algebra of homotopy algebras, Hinich defines a model structure on unbounded chain complexes where
- weak equivalences are the quasi-isomorphisms,
- fibrations are degree-wise surjections, and
- cofibrations are defined by the left lifting property.
Obviously, every object is fibrant. Now assume that we are working over a field $k$ instead of a general ring.
Question: Is it true that in this case every chain complex is also cofibrant?
I believe the answer to be yes, and I have an ugly argument that should prove it (even though I still have to check the details). Is there a sly/elegant way to show it?
homological-algebra model-categories
asked Oct 4 '17 at 5:34
Daniel Robert-NicoudDaniel Robert-Nicoud
20.5k33797
$endgroup$
add a comment |
$begingroup$
In his paper Homological algebra of homotopy algebras, Hinich defines a model structure on unbounded chain complexes where
- weak equivalences are the quasi-isomorphisms,
- fibrations are degree-wise surjections, and
- cofibrations are defined by the left lifting property.
Obviously, every object is fibrant. Now assume that we are working over a field $k$ instead of a general ring.
Question: Is it true that in this case every chain complex is also cofibrant?
I believe the answer to be yes, and I have an ugly argument that should prove it (even though I still have to check the details). Is there a sly/elegant way to show it?
homological-algebra model-categories
asked Oct 4 '17 at 5:34
Daniel Robert-NicoudDaniel Robert-Nicoud
20.5k33797
$endgroup$
In his paper Homological algebra of homotopy algebras, Hinich defines a model structure on unbounded chain complexes where
- weak equivalences are the quasi-isomorphisms,
- fibrations are degree-wise surjections, and
- cofibrations are defined by the left lifting property.
Obviously, every object is fibrant. Now assume that we are working over a field $k$ instead of a general ring.
Question: Is it true that in this case every chain complex is also cofibrant?
I believe the answer to be yes, and I have an ugly argument that should prove it (even though I still have to check the details). Is there a sly/elegant way to show it?
homological-algebra model-categories
homological-algebra model-categories
asked Oct 4 '17 at 5:34
Daniel Robert-NicoudDaniel Robert-Nicoud
20.5k33797
asked Oct 4 '17 at 5:34
Daniel Robert-NicoudDaniel Robert-Nicoud
20.5k33797
asked Oct 4 '17 at 5:34
Daniel Robert-NicoudDaniel Robert-Nicoud
20.5k33797
asked Oct 4 '17 at 5:34
Daniel Robert-NicoudDaniel Robert-Nicoud
20.5k33797
asked Oct 4 '17 at 5:34
Daniel Robert-NicoudDaniel Robert-Nicoud
20.5k33797
20.5k33797
add a comment |
add a comment |
2 Answers
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Yes.
Every (possibly unbounded) chain complex of $k$-vector spaces is a (possibly infinite) direct sum of complexes of the form
$$dotsto0to kto0todots$$
and
$$dotsto0to kstackrel{sim}{to}kto0todots$$
and their shifts.
It is easy to show that each of these complexes is cofibrant, and so every complex is cofibrant.
answered Oct 4 '17 at 12:19
Jeremy RickardJeremy Rickard
16.7k11745
$endgroup$
$begingroup$
Good point. Thanks.
$endgroup$
– Daniel Robert-Nicoud
Oct 4 '17 at 12:50
add a comment |
$begingroup$
it is not difficult to see that when $k$ is a field, then a cofibration is exactly a degree-wise injective map.
The key lemma is the following: a chain complex whose homology is trivial is homotopic to zero.
answered Jan 11 at 6:21
Damien LDamien L
5,047933
$endgroup$
$begingroup$
Merci Damien ;)
$endgroup$
– Daniel Robert-Nicoud
Jan 12 at 17:13
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
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$begingroup$
Yes.
Every (possibly unbounded) chain complex of $k$-vector spaces is a (possibly infinite) direct sum of complexes of the form
$$dotsto0to kto0todots$$
and
$$dotsto0to kstackrel{sim}{to}kto0todots$$
and their shifts.
It is easy to show that each of these complexes is cofibrant, and so every complex is cofibrant.
answered Oct 4 '17 at 12:19
Jeremy RickardJeremy Rickard
16.7k11745
$endgroup$
$begingroup$
Good point. Thanks.
$endgroup$
– Daniel Robert-Nicoud
Oct 4 '17 at 12:50
add a comment |
$begingroup$
Yes.
Every (possibly unbounded) chain complex of $k$-vector spaces is a (possibly infinite) direct sum of complexes of the form
$$dotsto0to kto0todots$$
and
$$dotsto0to kstackrel{sim}{to}kto0todots$$
and their shifts.
It is easy to show that each of these complexes is cofibrant, and so every complex is cofibrant.
answered Oct 4 '17 at 12:19
Jeremy RickardJeremy Rickard
16.7k11745
$endgroup$
$begingroup$
Good point. Thanks.
$endgroup$
– Daniel Robert-Nicoud
Oct 4 '17 at 12:50
add a comment |
$begingroup$
Yes.
Every (possibly unbounded) chain complex of $k$-vector spaces is a (possibly infinite) direct sum of complexes of the form
$$dotsto0to kto0todots$$
and
$$dotsto0to kstackrel{sim}{to}kto0todots$$
and their shifts.
It is easy to show that each of these complexes is cofibrant, and so every complex is cofibrant.
answered Oct 4 '17 at 12:19
Jeremy RickardJeremy Rickard
16.7k11745
$endgroup$
Yes.
Every (possibly unbounded) chain complex of $k$-vector spaces is a (possibly infinite) direct sum of complexes of the form
$$dotsto0to kto0todots$$
and
$$dotsto0to kstackrel{sim}{to}kto0todots$$
and their shifts.
It is easy to show that each of these complexes is cofibrant, and so every complex is cofibrant.
answered Oct 4 '17 at 12:19
Jeremy RickardJeremy Rickard
16.7k11745
answered Oct 4 '17 at 12:19
Jeremy RickardJeremy Rickard
16.7k11745
answered Oct 4 '17 at 12:19
Jeremy RickardJeremy Rickard
16.7k11745
answered Oct 4 '17 at 12:19
Jeremy RickardJeremy Rickard
16.7k11745
16.7k11745
$begingroup$
Good point. Thanks.
$endgroup$
– Daniel Robert-Nicoud
Oct 4 '17 at 12:50
add a comment |
$begingroup$
Good point. Thanks.
$endgroup$
– Daniel Robert-Nicoud
Oct 4 '17 at 12:50
$begingroup$
Good point. Thanks.
$endgroup$
– Daniel Robert-Nicoud
Oct 4 '17 at 12:50
$begingroup$
Good point. Thanks.
$endgroup$
– Daniel Robert-Nicoud
Oct 4 '17 at 12:50
add a comment |
$begingroup$
it is not difficult to see that when $k$ is a field, then a cofibration is exactly a degree-wise injective map.
The key lemma is the following: a chain complex whose homology is trivial is homotopic to zero.
answered Jan 11 at 6:21
Damien LDamien L
5,047933
$endgroup$
$begingroup$
Merci Damien ;)
$endgroup$
– Daniel Robert-Nicoud
Jan 12 at 17:13
add a comment |
$begingroup$
it is not difficult to see that when $k$ is a field, then a cofibration is exactly a degree-wise injective map.
The key lemma is the following: a chain complex whose homology is trivial is homotopic to zero.
answered Jan 11 at 6:21
Damien LDamien L
5,047933
$endgroup$
$begingroup$
Merci Damien ;)
$endgroup$
– Daniel Robert-Nicoud
Jan 12 at 17:13
add a comment |
$begingroup$
it is not difficult to see that when $k$ is a field, then a cofibration is exactly a degree-wise injective map.
The key lemma is the following: a chain complex whose homology is trivial is homotopic to zero.
answered Jan 11 at 6:21
Damien LDamien L
5,047933
$endgroup$
it is not difficult to see that when $k$ is a field, then a cofibration is exactly a degree-wise injective map.
The key lemma is the following: a chain complex whose homology is trivial is homotopic to zero.
answered Jan 11 at 6:21
Damien LDamien L
5,047933
answered Jan 11 at 6:21
Damien LDamien L
5,047933
answered Jan 11 at 6:21
Damien LDamien L
5,047933
answered Jan 11 at 6:21
Damien LDamien L
5,047933
5,047933
$begingroup$
Merci Damien ;)
$endgroup$
– Daniel Robert-Nicoud
Jan 12 at 17:13
add a comment |
$begingroup$
Merci Damien ;)
$endgroup$
– Daniel Robert-Nicoud
Jan 12 at 17:13
$begingroup$
Merci Damien ;)
$endgroup$
– Daniel Robert-Nicoud
Jan 12 at 17:13
$begingroup$
Merci Damien ;)
$endgroup$
– Daniel Robert-Nicoud
Jan 12 at 17:13
add a comment |
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