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Homology of Mapping Cone is Trivial--Proof from scratch
Clash Royale CLAN TAG#URR8PPP
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Let $(A_*,partial )$ and $(B_*,partial' )$ be chain complexes, $f:A_*to B_*$ a chain map and suppose that the map $f_*$ it induces on homology is an isomorphism.
The mapping cone $(C_*,D)$ is defined by $C_n=A_n-1oplus B_n$ and $D_n(a_n-1,b_n)=(-partial_n-1a_n-1,f_n-1(a_n-1)+partial'_n b_n).$
Now, define $A_n^+=A_n-1.$ Then, using the maps $i:bmapsto (0,b)$ and $p:(a,b)mapsto a$, there arises in the usual way, the long exact sequence
$$ cdots rightarrow H_n+1(C)oversetp_*rightarrow H_n+1(A^+)oversetf_*rightarrow H_n(B)overseti_*rightarrow H_n(C)oversetp_*rightarrow H_n(A^+)rightarrow cdots $$
Then, $H_n(C)$ is trivial for all $n$. The proof is easy (but hadn't occurred to me, ugh.) I wanted to do it using the basic facts about the chain complexes involved, to get an idea of how they work.
There is a proof from scratch of essentially the same thing here but I think I found a problem with it, and made a comment there. I would appreciate feedback on my comment there as well as on my attempt here at a from scratch proof, which follows.
What I have so far (dropping the subscripts and primes on the differentiation operators):
Let $(a,b)in Z_n(C).$ Then, by exactness, we have have the following:
$1). p_*([(a,b)]=[a]Rightarrow ain Bd_nA^+$ and $2). i_*([b])=[(0,b)]Rightarrow (0,b)in Bd_nC.$
So, there is an $a'in A_n$ and a pair $(a'',b')in C_n+1$ such that
$3). partial a'=a$ and $4). (-partial a'',f(a'')+partial b')=(0,b).$
Combining these facts we get $(a,b)=(a,0)+(0,b)=(-partial (a'+a''),f(a'')+partial b').$
If I could show that either $f(a')=0$ or $partial a'=0$, I would then have $(a,b)in Bd_nC,$ which is what I want.
abstract-algebra algebraic-topology category-theory homology-cohomology homological-algebra
asked Jul 26 at 23:33
Driver 8
7,5432917
add a comment |Â
up vote
1
down vote
favorite
Let $(A_*,partial )$ and $(B_*,partial' )$ be chain complexes, $f:A_*to B_*$ a chain map and suppose that the map $f_*$ it induces on homology is an isomorphism.
The mapping cone $(C_*,D)$ is defined by $C_n=A_n-1oplus B_n$ and $D_n(a_n-1,b_n)=(-partial_n-1a_n-1,f_n-1(a_n-1)+partial'_n b_n).$
Now, define $A_n^+=A_n-1.$ Then, using the maps $i:bmapsto (0,b)$ and $p:(a,b)mapsto a$, there arises in the usual way, the long exact sequence
$$ cdots rightarrow H_n+1(C)oversetp_*rightarrow H_n+1(A^+)oversetf_*rightarrow H_n(B)overseti_*rightarrow H_n(C)oversetp_*rightarrow H_n(A^+)rightarrow cdots $$
Then, $H_n(C)$ is trivial for all $n$. The proof is easy (but hadn't occurred to me, ugh.) I wanted to do it using the basic facts about the chain complexes involved, to get an idea of how they work.
There is a proof from scratch of essentially the same thing here but I think I found a problem with it, and made a comment there. I would appreciate feedback on my comment there as well as on my attempt here at a from scratch proof, which follows.
What I have so far (dropping the subscripts and primes on the differentiation operators):
Let $(a,b)in Z_n(C).$ Then, by exactness, we have have the following:
$1). p_*([(a,b)]=[a]Rightarrow ain Bd_nA^+$ and $2). i_*([b])=[(0,b)]Rightarrow (0,b)in Bd_nC.$
So, there is an $a'in A_n$ and a pair $(a'',b')in C_n+1$ such that
$3). partial a'=a$ and $4). (-partial a'',f(a'')+partial b')=(0,b).$
Combining these facts we get $(a,b)=(a,0)+(0,b)=(-partial (a'+a''),f(a'')+partial b').$
If I could show that either $f(a')=0$ or $partial a'=0$, I would then have $(a,b)in Bd_nC,$ which is what I want.
abstract-algebra algebraic-topology category-theory homology-cohomology homological-algebra
asked Jul 26 at 23:33
Driver 8
7,5432917
It's your question why $C$ has trivial homology, or why the given sequence is well-defined and exact?
– Arthur
Jul 26 at 23:43
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $(A_*,partial )$ and $(B_*,partial' )$ be chain complexes, $f:A_*to B_*$ a chain map and suppose that the map $f_*$ it induces on homology is an isomorphism.
The mapping cone $(C_*,D)$ is defined by $C_n=A_n-1oplus B_n$ and $D_n(a_n-1,b_n)=(-partial_n-1a_n-1,f_n-1(a_n-1)+partial'_n b_n).$
Now, define $A_n^+=A_n-1.$ Then, using the maps $i:bmapsto (0,b)$ and $p:(a,b)mapsto a$, there arises in the usual way, the long exact sequence
$$ cdots rightarrow H_n+1(C)oversetp_*rightarrow H_n+1(A^+)oversetf_*rightarrow H_n(B)overseti_*rightarrow H_n(C)oversetp_*rightarrow H_n(A^+)rightarrow cdots $$
Then, $H_n(C)$ is trivial for all $n$. The proof is easy (but hadn't occurred to me, ugh.) I wanted to do it using the basic facts about the chain complexes involved, to get an idea of how they work.
There is a proof from scratch of essentially the same thing here but I think I found a problem with it, and made a comment there. I would appreciate feedback on my comment there as well as on my attempt here at a from scratch proof, which follows.
What I have so far (dropping the subscripts and primes on the differentiation operators):
Let $(a,b)in Z_n(C).$ Then, by exactness, we have have the following:
$1). p_*([(a,b)]=[a]Rightarrow ain Bd_nA^+$ and $2). i_*([b])=[(0,b)]Rightarrow (0,b)in Bd_nC.$
So, there is an $a'in A_n$ and a pair $(a'',b')in C_n+1$ such that
$3). partial a'=a$ and $4). (-partial a'',f(a'')+partial b')=(0,b).$
Combining these facts we get $(a,b)=(a,0)+(0,b)=(-partial (a'+a''),f(a'')+partial b').$
If I could show that either $f(a')=0$ or $partial a'=0$, I would then have $(a,b)in Bd_nC,$ which is what I want.
abstract-algebra algebraic-topology category-theory homology-cohomology homological-algebra
asked Jul 26 at 23:33
Driver 8
7,5432917
Let $(A_*,partial )$ and $(B_*,partial' )$ be chain complexes, $f:A_*to B_*$ a chain map and suppose that the map $f_*$ it induces on homology is an isomorphism.
The mapping cone $(C_*,D)$ is defined by $C_n=A_n-1oplus B_n$ and $D_n(a_n-1,b_n)=(-partial_n-1a_n-1,f_n-1(a_n-1)+partial'_n b_n).$
Now, define $A_n^+=A_n-1.$ Then, using the maps $i:bmapsto (0,b)$ and $p:(a,b)mapsto a$, there arises in the usual way, the long exact sequence
$$ cdots rightarrow H_n+1(C)oversetp_*rightarrow H_n+1(A^+)oversetf_*rightarrow H_n(B)overseti_*rightarrow H_n(C)oversetp_*rightarrow H_n(A^+)rightarrow cdots $$
Then, $H_n(C)$ is trivial for all $n$. The proof is easy (but hadn't occurred to me, ugh.) I wanted to do it using the basic facts about the chain complexes involved, to get an idea of how they work.
There is a proof from scratch of essentially the same thing here but I think I found a problem with it, and made a comment there. I would appreciate feedback on my comment there as well as on my attempt here at a from scratch proof, which follows.
What I have so far (dropping the subscripts and primes on the differentiation operators):
Let $(a,b)in Z_n(C).$ Then, by exactness, we have have the following:
$1). p_*([(a,b)]=[a]Rightarrow ain Bd_nA^+$ and $2). i_*([b])=[(0,b)]Rightarrow (0,b)in Bd_nC.$
So, there is an $a'in A_n$ and a pair $(a'',b')in C_n+1$ such that
$3). partial a'=a$ and $4). (-partial a'',f(a'')+partial b')=(0,b).$
Combining these facts we get $(a,b)=(a,0)+(0,b)=(-partial (a'+a''),f(a'')+partial b').$
If I could show that either $f(a')=0$ or $partial a'=0$, I would then have $(a,b)in Bd_nC,$ which is what I want.
abstract-algebra algebraic-topology category-theory homology-cohomology homological-algebra
asked Jul 26 at 23:33
Driver 8
7,5432917
edited Jul 27 at 2:27
asked Jul 26 at 23:33
Driver 8
7,5432917
asked Jul 26 at 23:33
Driver 8
7,5432917
asked Jul 26 at 23:33
Driver 8
7,5432917
7,5432917
It's your question why $C$ has trivial homology, or why the given sequence is well-defined and exact?
– Arthur
Jul 26 at 23:43
add a comment |Â
It's your question why $C$ has trivial homology, or why the given sequence is well-defined and exact?
– Arthur
Jul 26 at 23:43
It's your question why $C$ has trivial homology, or why the given sequence is well-defined and exact?
– Arthur
Jul 26 at 23:43
It's your question why $C$ has trivial homology, or why the given sequence is well-defined and exact?
– Arthur
Jul 26 at 23:43
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
0
down vote
Assuming your question is why $C$ has trivial homology: $f_*$ is an isomorphism, so $p_*$ and $i_*$ must both be zero maps.
answered Jul 26 at 23:40
Arthur
98.4k793174
Yes, I know, but then how is it immediate from this that $C$ has the trivial homology? I am using the fact that $i_*$ and $p_*$ are zero maps to prove that $Z_n(C)=Bd_n(C).$ If it's trivial, could you point out why?
– Driver 8
Jul 27 at 0:06
1
If, in an exact sequence, one of the terms $X$ has both its incoming and its outgoing maps equal to zero, then $X$ has to be 0. Proof: The image of the incoming map is 0 because the incoming map is 0. But that image is also the kernel of the outgoing map, and that kernel is all of $X$ because the outgoing map is 0.
– Andreas Blass
Jul 27 at 0:17
I see. Thanks. I still would like to do a proof from scratch, to see how the "innards" of these chain complexes work. I am just starting to learn about homology theory, and am reading Rotman. If you critique my proof and/or the comment I made on a similar question (the link is cited above), I will be glad to accept your answer.
– Driver 8
Jul 27 at 2:27
add a comment |Â
up vote
0
down vote
That is probably not the answer you are looking for, but here is a proof using model category theory. Somehow, it says that this is a result not about chain complexes per se, and it is a motivation not to dive into chain complexes details.
In any model category $mathcal C$, the mapping cone $C_f$ of $f:Xto Y$ is defined as the homotopy pushout of $ast overset !gets X overset fto Y$. Computing homotopy pushout in a model category goes as follow: take a cofibrant replacement $X'overset qto X$ of $X$. Factor $fq$ as $pi$ with $p$ acyclic fibration and $i$ cofibration. Factor $!q$ as $p'i'$ with $p'$ acyclic fibration and $i'$ cofibration. Then take the ordinary pushout of $bullet overset i' gets X' overset i to bullet$. This is summed up in the following diagram:
In particular, $q,p,p'$ are weak equivalences, so if $f$ is also a weak equivalence, we get that $i$ is an acyclic fibration. It follows that its pushout $j$ also is such. In the end, we have a commutative triangle:
where $p'$ and $j$ both are weak equivalences, which yields that $C_f to ast$ is also a weak equivalence. Otherwise put, $C_f$ is contractible.
If you play that proof in your favorite model structure on chain complexes for which weak equivalences are the quasi-isomorphisms, you get the result you wanted.
answered Jul 27 at 11:11
Pece
7,92211040
I do not know much about model categories, except for the basic acyclic models theorem. I do know some category theory, though, (I have read CWM and Awodey's book) so I wonder is there a way to make your proof more understandable on that basic level? If so, I will gladly accept your answer.
– Driver 8
Jul 28 at 15:30
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Assuming your question is why $C$ has trivial homology: $f_*$ is an isomorphism, so $p_*$ and $i_*$ must both be zero maps.
answered Jul 26 at 23:40
Arthur
98.4k793174
Yes, I know, but then how is it immediate from this that $C$ has the trivial homology? I am using the fact that $i_*$ and $p_*$ are zero maps to prove that $Z_n(C)=Bd_n(C).$ If it's trivial, could you point out why?
– Driver 8
Jul 27 at 0:06
1
If, in an exact sequence, one of the terms $X$ has both its incoming and its outgoing maps equal to zero, then $X$ has to be 0. Proof: The image of the incoming map is 0 because the incoming map is 0. But that image is also the kernel of the outgoing map, and that kernel is all of $X$ because the outgoing map is 0.
– Andreas Blass
Jul 27 at 0:17
I see. Thanks. I still would like to do a proof from scratch, to see how the "innards" of these chain complexes work. I am just starting to learn about homology theory, and am reading Rotman. If you critique my proof and/or the comment I made on a similar question (the link is cited above), I will be glad to accept your answer.
– Driver 8
Jul 27 at 2:27
add a comment |Â
up vote
0
down vote
Assuming your question is why $C$ has trivial homology: $f_*$ is an isomorphism, so $p_*$ and $i_*$ must both be zero maps.
answered Jul 26 at 23:40
Arthur
98.4k793174
Yes, I know, but then how is it immediate from this that $C$ has the trivial homology? I am using the fact that $i_*$ and $p_*$ are zero maps to prove that $Z_n(C)=Bd_n(C).$ If it's trivial, could you point out why?
– Driver 8
Jul 27 at 0:06
1
If, in an exact sequence, one of the terms $X$ has both its incoming and its outgoing maps equal to zero, then $X$ has to be 0. Proof: The image of the incoming map is 0 because the incoming map is 0. But that image is also the kernel of the outgoing map, and that kernel is all of $X$ because the outgoing map is 0.
– Andreas Blass
Jul 27 at 0:17
I see. Thanks. I still would like to do a proof from scratch, to see how the "innards" of these chain complexes work. I am just starting to learn about homology theory, and am reading Rotman. If you critique my proof and/or the comment I made on a similar question (the link is cited above), I will be glad to accept your answer.
– Driver 8
Jul 27 at 2:27
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Assuming your question is why $C$ has trivial homology: $f_*$ is an isomorphism, so $p_*$ and $i_*$ must both be zero maps.
answered Jul 26 at 23:40
Arthur
98.4k793174
Assuming your question is why $C$ has trivial homology: $f_*$ is an isomorphism, so $p_*$ and $i_*$ must both be zero maps.
answered Jul 26 at 23:40
Arthur
98.4k793174
edited Jul 26 at 23:46
answered Jul 26 at 23:40
Arthur
98.4k793174
answered Jul 26 at 23:40
Arthur
98.4k793174
answered Jul 26 at 23:40
Arthur
98.4k793174
98.4k793174
Yes, I know, but then how is it immediate from this that $C$ has the trivial homology? I am using the fact that $i_*$ and $p_*$ are zero maps to prove that $Z_n(C)=Bd_n(C).$ If it's trivial, could you point out why?
– Driver 8
Jul 27 at 0:06
1
If, in an exact sequence, one of the terms $X$ has both its incoming and its outgoing maps equal to zero, then $X$ has to be 0. Proof: The image of the incoming map is 0 because the incoming map is 0. But that image is also the kernel of the outgoing map, and that kernel is all of $X$ because the outgoing map is 0.
– Andreas Blass
Jul 27 at 0:17
I see. Thanks. I still would like to do a proof from scratch, to see how the "innards" of these chain complexes work. I am just starting to learn about homology theory, and am reading Rotman. If you critique my proof and/or the comment I made on a similar question (the link is cited above), I will be glad to accept your answer.
– Driver 8
Jul 27 at 2:27
add a comment |Â
Yes, I know, but then how is it immediate from this that $C$ has the trivial homology? I am using the fact that $i_*$ and $p_*$ are zero maps to prove that $Z_n(C)=Bd_n(C).$ If it's trivial, could you point out why?
– Driver 8
Jul 27 at 0:06
1
If, in an exact sequence, one of the terms $X$ has both its incoming and its outgoing maps equal to zero, then $X$ has to be 0. Proof: The image of the incoming map is 0 because the incoming map is 0. But that image is also the kernel of the outgoing map, and that kernel is all of $X$ because the outgoing map is 0.
– Andreas Blass
Jul 27 at 0:17
I see. Thanks. I still would like to do a proof from scratch, to see how the "innards" of these chain complexes work. I am just starting to learn about homology theory, and am reading Rotman. If you critique my proof and/or the comment I made on a similar question (the link is cited above), I will be glad to accept your answer.
– Driver 8
Jul 27 at 2:27
Yes, I know, but then how is it immediate from this that $C$ has the trivial homology? I am using the fact that $i_*$ and $p_*$ are zero maps to prove that $Z_n(C)=Bd_n(C).$ If it's trivial, could you point out why?
– Driver 8
Jul 27 at 0:06
Yes, I know, but then how is it immediate from this that $C$ has the trivial homology? I am using the fact that $i_*$ and $p_*$ are zero maps to prove that $Z_n(C)=Bd_n(C).$ If it's trivial, could you point out why?
– Driver 8
Jul 27 at 0:06
1
1
If, in an exact sequence, one of the terms $X$ has both its incoming and its outgoing maps equal to zero, then $X$ has to be 0. Proof: The image of the incoming map is 0 because the incoming map is 0. But that image is also the kernel of the outgoing map, and that kernel is all of $X$ because the outgoing map is 0.
– Andreas Blass
Jul 27 at 0:17
If, in an exact sequence, one of the terms $X$ has both its incoming and its outgoing maps equal to zero, then $X$ has to be 0. Proof: The image of the incoming map is 0 because the incoming map is 0. But that image is also the kernel of the outgoing map, and that kernel is all of $X$ because the outgoing map is 0.
– Andreas Blass
Jul 27 at 0:17
I see. Thanks. I still would like to do a proof from scratch, to see how the "innards" of these chain complexes work. I am just starting to learn about homology theory, and am reading Rotman. If you critique my proof and/or the comment I made on a similar question (the link is cited above), I will be glad to accept your answer.
– Driver 8
Jul 27 at 2:27
I see. Thanks. I still would like to do a proof from scratch, to see how the "innards" of these chain complexes work. I am just starting to learn about homology theory, and am reading Rotman. If you critique my proof and/or the comment I made on a similar question (the link is cited above), I will be glad to accept your answer.
– Driver 8
Jul 27 at 2:27
add a comment |Â
up vote
0
down vote
That is probably not the answer you are looking for, but here is a proof using model category theory. Somehow, it says that this is a result not about chain complexes per se, and it is a motivation not to dive into chain complexes details.
In any model category $mathcal C$, the mapping cone $C_f$ of $f:Xto Y$ is defined as the homotopy pushout of $ast overset !gets X overset fto Y$. Computing homotopy pushout in a model category goes as follow: take a cofibrant replacement $X'overset qto X$ of $X$. Factor $fq$ as $pi$ with $p$ acyclic fibration and $i$ cofibration. Factor $!q$ as $p'i'$ with $p'$ acyclic fibration and $i'$ cofibration. Then take the ordinary pushout of $bullet overset i' gets X' overset i to bullet$. This is summed up in the following diagram:
In particular, $q,p,p'$ are weak equivalences, so if $f$ is also a weak equivalence, we get that $i$ is an acyclic fibration. It follows that its pushout $j$ also is such. In the end, we have a commutative triangle:
where $p'$ and $j$ both are weak equivalences, which yields that $C_f to ast$ is also a weak equivalence. Otherwise put, $C_f$ is contractible.
If you play that proof in your favorite model structure on chain complexes for which weak equivalences are the quasi-isomorphisms, you get the result you wanted.
answered Jul 27 at 11:11
Pece
7,92211040
I do not know much about model categories, except for the basic acyclic models theorem. I do know some category theory, though, (I have read CWM and Awodey's book) so I wonder is there a way to make your proof more understandable on that basic level? If so, I will gladly accept your answer.
– Driver 8
Jul 28 at 15:30
add a comment |Â
up vote
0
down vote
That is probably not the answer you are looking for, but here is a proof using model category theory. Somehow, it says that this is a result not about chain complexes per se, and it is a motivation not to dive into chain complexes details.
In any model category $mathcal C$, the mapping cone $C_f$ of $f:Xto Y$ is defined as the homotopy pushout of $ast overset !gets X overset fto Y$. Computing homotopy pushout in a model category goes as follow: take a cofibrant replacement $X'overset qto X$ of $X$. Factor $fq$ as $pi$ with $p$ acyclic fibration and $i$ cofibration. Factor $!q$ as $p'i'$ with $p'$ acyclic fibration and $i'$ cofibration. Then take the ordinary pushout of $bullet overset i' gets X' overset i to bullet$. This is summed up in the following diagram:
In particular, $q,p,p'$ are weak equivalences, so if $f$ is also a weak equivalence, we get that $i$ is an acyclic fibration. It follows that its pushout $j$ also is such. In the end, we have a commutative triangle:
where $p'$ and $j$ both are weak equivalences, which yields that $C_f to ast$ is also a weak equivalence. Otherwise put, $C_f$ is contractible.
If you play that proof in your favorite model structure on chain complexes for which weak equivalences are the quasi-isomorphisms, you get the result you wanted.
answered Jul 27 at 11:11
Pece
7,92211040
I do not know much about model categories, except for the basic acyclic models theorem. I do know some category theory, though, (I have read CWM and Awodey's book) so I wonder is there a way to make your proof more understandable on that basic level? If so, I will gladly accept your answer.
– Driver 8
Jul 28 at 15:30
add a comment |Â
up vote
0
down vote
up vote
0
down vote
That is probably not the answer you are looking for, but here is a proof using model category theory. Somehow, it says that this is a result not about chain complexes per se, and it is a motivation not to dive into chain complexes details.
In any model category $mathcal C$, the mapping cone $C_f$ of $f:Xto Y$ is defined as the homotopy pushout of $ast overset !gets X overset fto Y$. Computing homotopy pushout in a model category goes as follow: take a cofibrant replacement $X'overset qto X$ of $X$. Factor $fq$ as $pi$ with $p$ acyclic fibration and $i$ cofibration. Factor $!q$ as $p'i'$ with $p'$ acyclic fibration and $i'$ cofibration. Then take the ordinary pushout of $bullet overset i' gets X' overset i to bullet$. This is summed up in the following diagram:
In particular, $q,p,p'$ are weak equivalences, so if $f$ is also a weak equivalence, we get that $i$ is an acyclic fibration. It follows that its pushout $j$ also is such. In the end, we have a commutative triangle:
where $p'$ and $j$ both are weak equivalences, which yields that $C_f to ast$ is also a weak equivalence. Otherwise put, $C_f$ is contractible.
If you play that proof in your favorite model structure on chain complexes for which weak equivalences are the quasi-isomorphisms, you get the result you wanted.
answered Jul 27 at 11:11
Pece
7,92211040
That is probably not the answer you are looking for, but here is a proof using model category theory. Somehow, it says that this is a result not about chain complexes per se, and it is a motivation not to dive into chain complexes details.
In any model category $mathcal C$, the mapping cone $C_f$ of $f:Xto Y$ is defined as the homotopy pushout of $ast overset !gets X overset fto Y$. Computing homotopy pushout in a model category goes as follow: take a cofibrant replacement $X'overset qto X$ of $X$. Factor $fq$ as $pi$ with $p$ acyclic fibration and $i$ cofibration. Factor $!q$ as $p'i'$ with $p'$ acyclic fibration and $i'$ cofibration. Then take the ordinary pushout of $bullet overset i' gets X' overset i to bullet$. This is summed up in the following diagram:
In particular, $q,p,p'$ are weak equivalences, so if $f$ is also a weak equivalence, we get that $i$ is an acyclic fibration. It follows that its pushout $j$ also is such. In the end, we have a commutative triangle:
where $p'$ and $j$ both are weak equivalences, which yields that $C_f to ast$ is also a weak equivalence. Otherwise put, $C_f$ is contractible.
If you play that proof in your favorite model structure on chain complexes for which weak equivalences are the quasi-isomorphisms, you get the result you wanted.
answered Jul 27 at 11:11
Pece
7,92211040
answered Jul 27 at 11:11
Pece
7,92211040
answered Jul 27 at 11:11
Pece
7,92211040
answered Jul 27 at 11:11
Pece
7,92211040
7,92211040
I do not know much about model categories, except for the basic acyclic models theorem. I do know some category theory, though, (I have read CWM and Awodey's book) so I wonder is there a way to make your proof more understandable on that basic level? If so, I will gladly accept your answer.
– Driver 8
Jul 28 at 15:30
add a comment |Â
I do not know much about model categories, except for the basic acyclic models theorem. I do know some category theory, though, (I have read CWM and Awodey's book) so I wonder is there a way to make your proof more understandable on that basic level? If so, I will gladly accept your answer.
– Driver 8
Jul 28 at 15:30
I do not know much about model categories, except for the basic acyclic models theorem. I do know some category theory, though, (I have read CWM and Awodey's book) so I wonder is there a way to make your proof more understandable on that basic level? If so, I will gladly accept your answer.
– Driver 8
Jul 28 at 15:30
I do not know much about model categories, except for the basic acyclic models theorem. I do know some category theory, though, (I have read CWM and Awodey's book) so I wonder is there a way to make your proof more understandable on that basic level? If so, I will gladly accept your answer.
– Driver 8
Jul 28 at 15:30
add a comment |Â
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It's your question why $C$ has trivial homology, or why the given sequence is well-defined and exact?
– Arthur
Jul 26 at 23:43