Why are the morphisms in an Auslander-Reiten triangle irreducible?
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I'm working through Happel's book on triangulated categories, specifically the section on Auslander-Reiten theory. The part I'm having issues with is the proposition which states that the first two maps in an AR-triangle are irreducible.
Note: I'm reading morphism composition from right to left, not left to right as Happel does.
Let $X xrightarrow{u} Y xrightarrow{v} Z xrightarrow{w} X[1]$ be an AR-triangle, then $u$ and $v$ are irreducible.
The definition for a map $u$ to be irreducible is:
A morphism $u$ is $textbf{irreducible}$ if $u$ is neither a split epimorphism nor a split monomorphism, and for any factorization $u = u_2u_1$, either $u_1$ is a split monomorphism or $u_2$ is a split epimorphism.
Well, in the AR-triangle $X xrightarrow{u} Y xrightarrow{v} Z xrightarrow{w} X[1]$, the map $u$ is by assumption not a split monomorphism. Furthermore, I can follow Happel's proof that the factorization property for irreducibility is satisfied by $u$, but I can't show that it's not a split epimorphism. Here's my attempt:
Suppose $u$ is a split epimorphism, then there is some map $u':Yrightarrow X$ such that $uu' = 1_Y$. But then $vuu' = v1_Y = v = 0$ since $vu=0$. Shifting the triangle to $$Z[-1] xrightarrow{-w[-1]} X xrightarrow{u} Y xrightarrow{0} Z$$ one sees that it splits, so $X cong Z[-1]oplus Y$. But $X$ is indecomposable, so one of those summands must be $0$. If $Z[-1]=0$, then $-w[-1]=0$ and $w=0$, which contradicts the assumptions on AR-triangles. So $Y=0$, and we have a new triangle, isomorphic to the first:
$$X xrightarrow{0} 0 xrightarrow{0} Z xrightarrow{w} X[1]$$
This is where I get stuck. I don't see how to get a contradiction on the AR-triangle axioms with this last triangle. Any help would be appreciated.
abstract-algebra category-theory representation-theory triangulated-categories
asked Dec 10 '18 at 13:22
AuclairAuclair
772413
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add a comment |
$begingroup$
I'm working through Happel's book on triangulated categories, specifically the section on Auslander-Reiten theory. The part I'm having issues with is the proposition which states that the first two maps in an AR-triangle are irreducible.
Note: I'm reading morphism composition from right to left, not left to right as Happel does.
Let $X xrightarrow{u} Y xrightarrow{v} Z xrightarrow{w} X[1]$ be an AR-triangle, then $u$ and $v$ are irreducible.
The definition for a map $u$ to be irreducible is:
A morphism $u$ is $textbf{irreducible}$ if $u$ is neither a split epimorphism nor a split monomorphism, and for any factorization $u = u_2u_1$, either $u_1$ is a split monomorphism or $u_2$ is a split epimorphism.
Well, in the AR-triangle $X xrightarrow{u} Y xrightarrow{v} Z xrightarrow{w} X[1]$, the map $u$ is by assumption not a split monomorphism. Furthermore, I can follow Happel's proof that the factorization property for irreducibility is satisfied by $u$, but I can't show that it's not a split epimorphism. Here's my attempt:
Suppose $u$ is a split epimorphism, then there is some map $u':Yrightarrow X$ such that $uu' = 1_Y$. But then $vuu' = v1_Y = v = 0$ since $vu=0$. Shifting the triangle to $$Z[-1] xrightarrow{-w[-1]} X xrightarrow{u} Y xrightarrow{0} Z$$ one sees that it splits, so $X cong Z[-1]oplus Y$. But $X$ is indecomposable, so one of those summands must be $0$. If $Z[-1]=0$, then $-w[-1]=0$ and $w=0$, which contradicts the assumptions on AR-triangles. So $Y=0$, and we have a new triangle, isomorphic to the first:
$$X xrightarrow{0} 0 xrightarrow{0} Z xrightarrow{w} X[1]$$
This is where I get stuck. I don't see how to get a contradiction on the AR-triangle axioms with this last triangle. Any help would be appreciated.
abstract-algebra category-theory representation-theory triangulated-categories
asked Dec 10 '18 at 13:22
AuclairAuclair
772413
$endgroup$
$begingroup$
It would help to provide more of the relevant details.
$endgroup$
– Kevin Carlson
Dec 11 '18 at 4:00
$begingroup$
@KevinCarlson What details are you thinking of? Happel doesn't actually show this in his proof, so there are nothing to include there at least.
$endgroup$
– Auclair
Dec 11 '18 at 9:40
$begingroup$
I mean the definition of an AR-triangle.
$endgroup$
– Kevin Carlson
Dec 11 '18 at 16:55
$begingroup$
I've just noticed that this is a duplicate, see below. There is an example at the linked question of an Auslander-Reiten triangle in which the middle object is zero, so although I haven't checked anything in Happel myself, it seems this proposition has an error. Note however that the first part of the answer there mixes up section and retraction: the second part seems to show that it is indeed possible to have $u$ be a split epi. math.stackexchange.com/questions/2372259/…
$endgroup$
– Kevin Carlson
Dec 11 '18 at 17:04
add a comment |
$begingroup$
I'm working through Happel's book on triangulated categories, specifically the section on Auslander-Reiten theory. The part I'm having issues with is the proposition which states that the first two maps in an AR-triangle are irreducible.
Note: I'm reading morphism composition from right to left, not left to right as Happel does.
Let $X xrightarrow{u} Y xrightarrow{v} Z xrightarrow{w} X[1]$ be an AR-triangle, then $u$ and $v$ are irreducible.
The definition for a map $u$ to be irreducible is:
A morphism $u$ is $textbf{irreducible}$ if $u$ is neither a split epimorphism nor a split monomorphism, and for any factorization $u = u_2u_1$, either $u_1$ is a split monomorphism or $u_2$ is a split epimorphism.
Well, in the AR-triangle $X xrightarrow{u} Y xrightarrow{v} Z xrightarrow{w} X[1]$, the map $u$ is by assumption not a split monomorphism. Furthermore, I can follow Happel's proof that the factorization property for irreducibility is satisfied by $u$, but I can't show that it's not a split epimorphism. Here's my attempt:
Suppose $u$ is a split epimorphism, then there is some map $u':Yrightarrow X$ such that $uu' = 1_Y$. But then $vuu' = v1_Y = v = 0$ since $vu=0$. Shifting the triangle to $$Z[-1] xrightarrow{-w[-1]} X xrightarrow{u} Y xrightarrow{0} Z$$ one sees that it splits, so $X cong Z[-1]oplus Y$. But $X$ is indecomposable, so one of those summands must be $0$. If $Z[-1]=0$, then $-w[-1]=0$ and $w=0$, which contradicts the assumptions on AR-triangles. So $Y=0$, and we have a new triangle, isomorphic to the first:
$$X xrightarrow{0} 0 xrightarrow{0} Z xrightarrow{w} X[1]$$
This is where I get stuck. I don't see how to get a contradiction on the AR-triangle axioms with this last triangle. Any help would be appreciated.
abstract-algebra category-theory representation-theory triangulated-categories
asked Dec 10 '18 at 13:22
AuclairAuclair
772413
$endgroup$
I'm working through Happel's book on triangulated categories, specifically the section on Auslander-Reiten theory. The part I'm having issues with is the proposition which states that the first two maps in an AR-triangle are irreducible.
Note: I'm reading morphism composition from right to left, not left to right as Happel does.
Let $X xrightarrow{u} Y xrightarrow{v} Z xrightarrow{w} X[1]$ be an AR-triangle, then $u$ and $v$ are irreducible.
The definition for a map $u$ to be irreducible is:
A morphism $u$ is $textbf{irreducible}$ if $u$ is neither a split epimorphism nor a split monomorphism, and for any factorization $u = u_2u_1$, either $u_1$ is a split monomorphism or $u_2$ is a split epimorphism.
Well, in the AR-triangle $X xrightarrow{u} Y xrightarrow{v} Z xrightarrow{w} X[1]$, the map $u$ is by assumption not a split monomorphism. Furthermore, I can follow Happel's proof that the factorization property for irreducibility is satisfied by $u$, but I can't show that it's not a split epimorphism. Here's my attempt:
Suppose $u$ is a split epimorphism, then there is some map $u':Yrightarrow X$ such that $uu' = 1_Y$. But then $vuu' = v1_Y = v = 0$ since $vu=0$. Shifting the triangle to $$Z[-1] xrightarrow{-w[-1]} X xrightarrow{u} Y xrightarrow{0} Z$$ one sees that it splits, so $X cong Z[-1]oplus Y$. But $X$ is indecomposable, so one of those summands must be $0$. If $Z[-1]=0$, then $-w[-1]=0$ and $w=0$, which contradicts the assumptions on AR-triangles. So $Y=0$, and we have a new triangle, isomorphic to the first:
$$X xrightarrow{0} 0 xrightarrow{0} Z xrightarrow{w} X[1]$$
This is where I get stuck. I don't see how to get a contradiction on the AR-triangle axioms with this last triangle. Any help would be appreciated.
abstract-algebra category-theory representation-theory triangulated-categories
abstract-algebra category-theory representation-theory triangulated-categories
asked Dec 10 '18 at 13:22
AuclairAuclair
772413
asked Dec 10 '18 at 13:22
AuclairAuclair
772413
asked Dec 10 '18 at 13:22
AuclairAuclair
772413
asked Dec 10 '18 at 13:22
AuclairAuclair
772413
asked Dec 10 '18 at 13:22
AuclairAuclair
772413
772413
$begingroup$
It would help to provide more of the relevant details.
$endgroup$
– Kevin Carlson
Dec 11 '18 at 4:00
$begingroup$
@KevinCarlson What details are you thinking of? Happel doesn't actually show this in his proof, so there are nothing to include there at least.
$endgroup$
– Auclair
Dec 11 '18 at 9:40
$begingroup$
I mean the definition of an AR-triangle.
$endgroup$
– Kevin Carlson
Dec 11 '18 at 16:55
$begingroup$
I've just noticed that this is a duplicate, see below. There is an example at the linked question of an Auslander-Reiten triangle in which the middle object is zero, so although I haven't checked anything in Happel myself, it seems this proposition has an error. Note however that the first part of the answer there mixes up section and retraction: the second part seems to show that it is indeed possible to have $u$ be a split epi. math.stackexchange.com/questions/2372259/…
$endgroup$
– Kevin Carlson
Dec 11 '18 at 17:04
add a comment |
$begingroup$
It would help to provide more of the relevant details.
$endgroup$
– Kevin Carlson
Dec 11 '18 at 4:00
$begingroup$
@KevinCarlson What details are you thinking of? Happel doesn't actually show this in his proof, so there are nothing to include there at least.
$endgroup$
– Auclair
Dec 11 '18 at 9:40
$begingroup$
I mean the definition of an AR-triangle.
$endgroup$
– Kevin Carlson
Dec 11 '18 at 16:55
$begingroup$
I've just noticed that this is a duplicate, see below. There is an example at the linked question of an Auslander-Reiten triangle in which the middle object is zero, so although I haven't checked anything in Happel myself, it seems this proposition has an error. Note however that the first part of the answer there mixes up section and retraction: the second part seems to show that it is indeed possible to have $u$ be a split epi. math.stackexchange.com/questions/2372259/…
$endgroup$
– Kevin Carlson
Dec 11 '18 at 17:04
$begingroup$
It would help to provide more of the relevant details.
$endgroup$
– Kevin Carlson
Dec 11 '18 at 4:00
$begingroup$
It would help to provide more of the relevant details.
$endgroup$
– Kevin Carlson
Dec 11 '18 at 4:00
$begingroup$
@KevinCarlson What details are you thinking of? Happel doesn't actually show this in his proof, so there are nothing to include there at least.
$endgroup$
– Auclair
Dec 11 '18 at 9:40
$begingroup$
@KevinCarlson What details are you thinking of? Happel doesn't actually show this in his proof, so there are nothing to include there at least.
$endgroup$
– Auclair
Dec 11 '18 at 9:40
$begingroup$
I mean the definition of an AR-triangle.
$endgroup$
– Kevin Carlson
Dec 11 '18 at 16:55
$begingroup$
I mean the definition of an AR-triangle.
$endgroup$
– Kevin Carlson
Dec 11 '18 at 16:55
$begingroup$
I've just noticed that this is a duplicate, see below. There is an example at the linked question of an Auslander-Reiten triangle in which the middle object is zero, so although I haven't checked anything in Happel myself, it seems this proposition has an error. Note however that the first part of the answer there mixes up section and retraction: the second part seems to show that it is indeed possible to have $u$ be a split epi. math.stackexchange.com/questions/2372259/…
$endgroup$
– Kevin Carlson
Dec 11 '18 at 17:04
$begingroup$
I've just noticed that this is a duplicate, see below. There is an example at the linked question of an Auslander-Reiten triangle in which the middle object is zero, so although I haven't checked anything in Happel myself, it seems this proposition has an error. Note however that the first part of the answer there mixes up section and retraction: the second part seems to show that it is indeed possible to have $u$ be a split epi. math.stackexchange.com/questions/2372259/…
$endgroup$
– Kevin Carlson
Dec 11 '18 at 17:04
add a comment |
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$begingroup$
It would help to provide more of the relevant details.
$endgroup$
– Kevin Carlson
Dec 11 '18 at 4:00
$begingroup$
@KevinCarlson What details are you thinking of? Happel doesn't actually show this in his proof, so there are nothing to include there at least.
$endgroup$
– Auclair
Dec 11 '18 at 9:40
$begingroup$
I mean the definition of an AR-triangle.
$endgroup$
– Kevin Carlson
Dec 11 '18 at 16:55
$begingroup$
I've just noticed that this is a duplicate, see below. There is an example at the linked question of an Auslander-Reiten triangle in which the middle object is zero, so although I haven't checked anything in Happel myself, it seems this proposition has an error. Note however that the first part of the answer there mixes up section and retraction: the second part seems to show that it is indeed possible to have $u$ be a split epi. math.stackexchange.com/questions/2372259/…
$endgroup$
– Kevin Carlson
Dec 11 '18 at 17:04