Mixing
Proof of 10.4.6 in Weibel
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Let $I$ be a bounded below complex of injectives in some Abelian category, and $Z$ any bounded below complex. Suppose $u:Ito Z$ is a quasi isomorphism. We want to proof that $u$ is split injective up to homotopy. The proof outlined in Weibel's book An Introduction to Homological Algebra (pdf) is as follows:
The complex $operatornamecone(u)$ is exact, where
$$
operatornamecone(u)^i=I^i+1oplus Z^i,
$$
and the differential is given by
$$
D=beginpmatrix-d & 0 \ -u & dendpmatrix.
$$
The map $phi:operatornamecone(u)to I[1]: (y,z)mapsto -y$ is chain map, and it can be shown, using the fact that $I$ consists of injective objects, and $operatornamecone(u)$ is exact, that $phi$ is nullhomotopic, i.e.
$$phi = fD+df$$
for some $f^i$, where
$$f^i:I^i+1oplus Z^ito I^i.$$
then we can write $f^i=(v^i,s^i)$, where $v^i:I^i+1to I^i$ and $s^i:Z^ito I^i$.
Then we get:
beginalign*
-y=phi(y,z)&=fD(y,z)+df(y,z)\
&=f(-dy,-uy+dz)+d(vy+sz)\
&=-vdy-suy+sdz+dvy+dsz\
&=-suy+(dvy-vdy)+(sdz+dsz).
endalign*
On the other hand, Weibel claims in his proof (after correcting a typo as per the errata) that
$$y=(kdy+suy+dky)+(dsz-sdz).$$
Obviously this is not the same, and my answer does lead to a proof of the lemma, so I must be doing something wrong, but I don't see what.
homological-algebra abelian-categories
edited Aug 4 at 1:10
darij grinberg
9,19132959
asked Aug 1 at 22:23
user2520938
4,02221231
add a comment |Â
up vote
4
down vote
favorite
Let $I$ be a bounded below complex of injectives in some Abelian category, and $Z$ any bounded below complex. Suppose $u:Ito Z$ is a quasi isomorphism. We want to proof that $u$ is split injective up to homotopy. The proof outlined in Weibel's book An Introduction to Homological Algebra (pdf) is as follows:
The complex $operatornamecone(u)$ is exact, where
$$
operatornamecone(u)^i=I^i+1oplus Z^i,
$$
and the differential is given by
$$
D=beginpmatrix-d & 0 \ -u & dendpmatrix.
$$
The map $phi:operatornamecone(u)to I[1]: (y,z)mapsto -y$ is chain map, and it can be shown, using the fact that $I$ consists of injective objects, and $operatornamecone(u)$ is exact, that $phi$ is nullhomotopic, i.e.
$$phi = fD+df$$
for some $f^i$, where
$$f^i:I^i+1oplus Z^ito I^i.$$
then we can write $f^i=(v^i,s^i)$, where $v^i:I^i+1to I^i$ and $s^i:Z^ito I^i$.
Then we get:
beginalign*
-y=phi(y,z)&=fD(y,z)+df(y,z)\
&=f(-dy,-uy+dz)+d(vy+sz)\
&=-vdy-suy+sdz+dvy+dsz\
&=-suy+(dvy-vdy)+(sdz+dsz).
endalign*
On the other hand, Weibel claims in his proof (after correcting a typo as per the errata) that
$$y=(kdy+suy+dky)+(dsz-sdz).$$
Obviously this is not the same, and my answer does lead to a proof of the lemma, so I must be doing something wrong, but I don't see what.
homological-algebra abelian-categories
edited Aug 4 at 1:10
darij grinberg
9,19132959
asked Aug 1 at 22:23
user2520938
4,02221231
Please allow for some generous amount of time before cross-posting to MO.
– Pedro Tamaroff♦
Aug 3 at 18:25
@PedroTamaroff If you'd prefer I'll delete it here...
– user2520938
Aug 3 at 18:28
That's alright. The question is a better fit here.
– Pedro Tamaroff♦
Aug 3 at 18:29
The difference in signs is exactly at places where you are applying the differential of $I[1]$, which is the negative of that in $I$, as noted in MO.
– Pedro Tamaroff♦
Aug 3 at 18:49
1
@user2520938 I would like to think $I[1]$ as $k[1]otimes I$ where $k$ is the ground field. When you are taking the differential, you need to follow Leibniz's rule. Because the degree of $k[1]$ is 1 and the degree of differential is also 1, when you switch those two, you will get an extra minus sign.
– Yining Zhang
Aug 3 at 18:54
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Let $I$ be a bounded below complex of injectives in some Abelian category, and $Z$ any bounded below complex. Suppose $u:Ito Z$ is a quasi isomorphism. We want to proof that $u$ is split injective up to homotopy. The proof outlined in Weibel's book An Introduction to Homological Algebra (pdf) is as follows:
The complex $operatornamecone(u)$ is exact, where
$$
operatornamecone(u)^i=I^i+1oplus Z^i,
$$
and the differential is given by
$$
D=beginpmatrix-d & 0 \ -u & dendpmatrix.
$$
The map $phi:operatornamecone(u)to I[1]: (y,z)mapsto -y$ is chain map, and it can be shown, using the fact that $I$ consists of injective objects, and $operatornamecone(u)$ is exact, that $phi$ is nullhomotopic, i.e.
$$phi = fD+df$$
for some $f^i$, where
$$f^i:I^i+1oplus Z^ito I^i.$$
then we can write $f^i=(v^i,s^i)$, where $v^i:I^i+1to I^i$ and $s^i:Z^ito I^i$.
Then we get:
beginalign*
-y=phi(y,z)&=fD(y,z)+df(y,z)\
&=f(-dy,-uy+dz)+d(vy+sz)\
&=-vdy-suy+sdz+dvy+dsz\
&=-suy+(dvy-vdy)+(sdz+dsz).
endalign*
On the other hand, Weibel claims in his proof (after correcting a typo as per the errata) that
$$y=(kdy+suy+dky)+(dsz-sdz).$$
Obviously this is not the same, and my answer does lead to a proof of the lemma, so I must be doing something wrong, but I don't see what.
homological-algebra abelian-categories
edited Aug 4 at 1:10
darij grinberg
9,19132959
asked Aug 1 at 22:23
user2520938
4,02221231
Let $I$ be a bounded below complex of injectives in some Abelian category, and $Z$ any bounded below complex. Suppose $u:Ito Z$ is a quasi isomorphism. We want to proof that $u$ is split injective up to homotopy. The proof outlined in Weibel's book An Introduction to Homological Algebra (pdf) is as follows:
The complex $operatornamecone(u)$ is exact, where
$$
operatornamecone(u)^i=I^i+1oplus Z^i,
$$
and the differential is given by
$$
D=beginpmatrix-d & 0 \ -u & dendpmatrix.
$$
The map $phi:operatornamecone(u)to I[1]: (y,z)mapsto -y$ is chain map, and it can be shown, using the fact that $I$ consists of injective objects, and $operatornamecone(u)$ is exact, that $phi$ is nullhomotopic, i.e.
$$phi = fD+df$$
for some $f^i$, where
$$f^i:I^i+1oplus Z^ito I^i.$$
then we can write $f^i=(v^i,s^i)$, where $v^i:I^i+1to I^i$ and $s^i:Z^ito I^i$.
Then we get:
beginalign*
-y=phi(y,z)&=fD(y,z)+df(y,z)\
&=f(-dy,-uy+dz)+d(vy+sz)\
&=-vdy-suy+sdz+dvy+dsz\
&=-suy+(dvy-vdy)+(sdz+dsz).
endalign*
On the other hand, Weibel claims in his proof (after correcting a typo as per the errata) that
$$y=(kdy+suy+dky)+(dsz-sdz).$$
Obviously this is not the same, and my answer does lead to a proof of the lemma, so I must be doing something wrong, but I don't see what.
homological-algebra abelian-categories
edited Aug 4 at 1:10
darij grinberg
9,19132959
asked Aug 1 at 22:23
user2520938
4,02221231
edited Aug 4 at 1:10
darij grinberg
9,19132959
edited Aug 4 at 1:10
darij grinberg
9,19132959
edited Aug 4 at 1:10
darij grinberg
9,19132959
9,19132959
asked Aug 1 at 22:23
user2520938
4,02221231
asked Aug 1 at 22:23
user2520938
4,02221231
asked Aug 1 at 22:23
user2520938
4,02221231
4,02221231
Please allow for some generous amount of time before cross-posting to MO.
– Pedro Tamaroff♦
Aug 3 at 18:25
@PedroTamaroff If you'd prefer I'll delete it here...
– user2520938
Aug 3 at 18:28
That's alright. The question is a better fit here.
– Pedro Tamaroff♦
Aug 3 at 18:29
The difference in signs is exactly at places where you are applying the differential of $I[1]$, which is the negative of that in $I$, as noted in MO.
– Pedro Tamaroff♦
Aug 3 at 18:49
1
@user2520938 I would like to think $I[1]$ as $k[1]otimes I$ where $k$ is the ground field. When you are taking the differential, you need to follow Leibniz's rule. Because the degree of $k[1]$ is 1 and the degree of differential is also 1, when you switch those two, you will get an extra minus sign.
– Yining Zhang
Aug 3 at 18:54
add a comment |Â
Please allow for some generous amount of time before cross-posting to MO.
– Pedro Tamaroff♦
Aug 3 at 18:25
@PedroTamaroff If you'd prefer I'll delete it here...
– user2520938
Aug 3 at 18:28
That's alright. The question is a better fit here.
– Pedro Tamaroff♦
Aug 3 at 18:29
The difference in signs is exactly at places where you are applying the differential of $I[1]$, which is the negative of that in $I$, as noted in MO.
– Pedro Tamaroff♦
Aug 3 at 18:49
1
@user2520938 I would like to think $I[1]$ as $k[1]otimes I$ where $k$ is the ground field. When you are taking the differential, you need to follow Leibniz's rule. Because the degree of $k[1]$ is 1 and the degree of differential is also 1, when you switch those two, you will get an extra minus sign.
– Yining Zhang
Aug 3 at 18:54
Please allow for some generous amount of time before cross-posting to MO.
– Pedro Tamaroff♦
Aug 3 at 18:25
Please allow for some generous amount of time before cross-posting to MO.
– Pedro Tamaroff♦
Aug 3 at 18:25
@PedroTamaroff If you'd prefer I'll delete it here...
– user2520938
Aug 3 at 18:28
@PedroTamaroff If you'd prefer I'll delete it here...
– user2520938
Aug 3 at 18:28
That's alright. The question is a better fit here.
– Pedro Tamaroff♦
Aug 3 at 18:29
That's alright. The question is a better fit here.
– Pedro Tamaroff♦
Aug 3 at 18:29
The difference in signs is exactly at places where you are applying the differential of $I[1]$, which is the negative of that in $I$, as noted in MO.
– Pedro Tamaroff♦
Aug 3 at 18:49
The difference in signs is exactly at places where you are applying the differential of $I[1]$, which is the negative of that in $I$, as noted in MO.
– Pedro Tamaroff♦
Aug 3 at 18:49
1
1
@user2520938 I would like to think $I[1]$ as $k[1]otimes I$ where $k$ is the ground field. When you are taking the differential, you need to follow Leibniz's rule. Because the degree of $k[1]$ is 1 and the degree of differential is also 1, when you switch those two, you will get an extra minus sign.
– Yining Zhang
Aug 3 at 18:54
@user2520938 I would like to think $I[1]$ as $k[1]otimes I$ where $k$ is the ground field. When you are taking the differential, you need to follow Leibniz's rule. Because the degree of $k[1]$ is 1 and the degree of differential is also 1, when you switch those two, you will get an extra minus sign.
– Yining Zhang
Aug 3 at 18:54
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
Notice that you are looking at $I[1]$ instead of $I$, so there is an extra minus sign involving in the differential i.e. $d_I[1]f=-df$.
answered Aug 3 at 18:47
Yining Zhang
395212
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Notice that you are looking at $I[1]$ instead of $I$, so there is an extra minus sign involving in the differential i.e. $d_I[1]f=-df$.
answered Aug 3 at 18:47
Yining Zhang
395212
add a comment |Â
up vote
3
down vote
accepted
Notice that you are looking at $I[1]$ instead of $I$, so there is an extra minus sign involving in the differential i.e. $d_I[1]f=-df$.
answered Aug 3 at 18:47
Yining Zhang
395212
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Notice that you are looking at $I[1]$ instead of $I$, so there is an extra minus sign involving in the differential i.e. $d_I[1]f=-df$.
answered Aug 3 at 18:47
Yining Zhang
395212
Notice that you are looking at $I[1]$ instead of $I$, so there is an extra minus sign involving in the differential i.e. $d_I[1]f=-df$.
answered Aug 3 at 18:47
Yining Zhang
395212
answered Aug 3 at 18:47
Yining Zhang
395212
answered Aug 3 at 18:47
Yining Zhang
395212
answered Aug 3 at 18:47
Yining Zhang
395212
395212
add a comment |Â
add a comment |Â
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Please allow for some generous amount of time before cross-posting to MO.
– Pedro Tamaroff♦
Aug 3 at 18:25
@PedroTamaroff If you'd prefer I'll delete it here...
– user2520938
Aug 3 at 18:28
That's alright. The question is a better fit here.
– Pedro Tamaroff♦
Aug 3 at 18:29
The difference in signs is exactly at places where you are applying the differential of $I[1]$, which is the negative of that in $I$, as noted in MO.
– Pedro Tamaroff♦
Aug 3 at 18:49
1
@user2520938 I would like to think $I[1]$ as $k[1]otimes I$ where $k$ is the ground field. When you are taking the differential, you need to follow Leibniz's rule. Because the degree of $k[1]$ is 1 and the degree of differential is also 1, when you switch those two, you will get an extra minus sign.
– Yining Zhang
Aug 3 at 18:54