Does the Eilenberg Moore Construction Preserve fibrations?
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Say we have a Grothendieck fibration $p : E to B$ and a monad $T$ on $B$ and a lift $T'$ of $T$ to $E$, i.e. a monad on $E$ such that $pT' = Tp$ and $p$ preserves $eta, mu$.
Then because the Eilenberg–Moore construction is functorial, we have a morphism $EM(p)$ from $T'text-Alg$ to $Ttext-Alg$. Is $EM(p)$ generally a fibration? If not, under what conditions is $EM(p)$ a fibration?
ct.category-theory monads fibration
asked Dec 29 '18 at 3:08
Max NewMax New
31018
add a comment |
Say we have a Grothendieck fibration $p : E to B$ and a monad $T$ on $B$ and a lift $T'$ of $T$ to $E$, i.e. a monad on $E$ such that $pT' = Tp$ and $p$ preserves $eta, mu$.
Then because the Eilenberg–Moore construction is functorial, we have a morphism $EM(p)$ from $T'text-Alg$ to $Ttext-Alg$. Is $EM(p)$ generally a fibration? If not, under what conditions is $EM(p)$ a fibration?
ct.category-theory monads fibration
asked Dec 29 '18 at 3:08
Max NewMax New
31018
add a comment |
Say we have a Grothendieck fibration $p : E to B$ and a monad $T$ on $B$ and a lift $T'$ of $T$ to $E$, i.e. a monad on $E$ such that $pT' = Tp$ and $p$ preserves $eta, mu$.
Then because the Eilenberg–Moore construction is functorial, we have a morphism $EM(p)$ from $T'text-Alg$ to $Ttext-Alg$. Is $EM(p)$ generally a fibration? If not, under what conditions is $EM(p)$ a fibration?
ct.category-theory monads fibration
asked Dec 29 '18 at 3:08
Max NewMax New
31018
Say we have a Grothendieck fibration $p : E to B$ and a monad $T$ on $B$ and a lift $T'$ of $T$ to $E$, i.e. a monad on $E$ such that $pT' = Tp$ and $p$ preserves $eta, mu$.
Then because the Eilenberg–Moore construction is functorial, we have a morphism $EM(p)$ from $T'text-Alg$ to $Ttext-Alg$. Is $EM(p)$ generally a fibration? If not, under what conditions is $EM(p)$ a fibration?
ct.category-theory monads fibration
ct.category-theory monads fibration
asked Dec 29 '18 at 3:08
Max NewMax New
31018
asked Dec 29 '18 at 3:08
Max NewMax New
31018
edited Dec 29 '18 at 5:46
Max New
asked Dec 29 '18 at 3:08
Max NewMax New
31018
asked Dec 29 '18 at 3:08
Max NewMax New
31018
asked Dec 29 '18 at 3:08
Max NewMax New
31018
31018
add a comment |
add a comment |
1 Answer
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Let $C$ be a 2-category and $Mnd(C)$ the 2-category of monads in $C$. As explained by Street in the formal theory of monads, the Eilenberg-Moore construction is right 2-adjoint to the inclusion 2-functor $C to Mnd(C)$ sending an object to the identity monad on it. Therefore it preserves 2-limits.
A fibration in a 2-category may be defined in terms of certain 2-limits (namely slice categories) and adjoints, and so is preserved by any 2-functor preserving 2-limits, and in particular by the Eilenberg-Moore construction.
Thus it's sufficient for $p$ to be a fibration object in the 2-category $Mnd(C)$.
answered Dec 29 '18 at 5:50
Tim CampionTim Campion
13.5k354122
1
So just need to figure out what a fibration object in Mnd is then.
– Max New
Dec 29 '18 at 16:37
1
Fibrations can be detected by homming in, so one answer will probably be a tautological one: $p: E to B$ is a fibration if $Mnd(C)(X,E) to Mnd(C)(X,B)$ is a fibration for all monads $X$, which will probably reduce in the case $C = Cat$ to the induced map of Eilenberg-Moore objects being a fibration.
– Tim Campion
Dec 30 '18 at 15:00
2
I think it would probably be easier to understand fibrations in Mnd(C) using the characterization of fibrations in terms of limits. Since the forgetful functor $mathitMnd(C)to C$ also preserves limits (when C has Kleisli objects it has a left adjoint), being a fibration in Mnd(C) just means being a fibration in C together with the extra structure that the universal lift-assigning functor and transformation lift to Mnd(C). Then just work out what that means explicitly.
– Mike Shulman
Dec 30 '18 at 15:42
When you say the forgetful functor $Mnd(C) to C$ do you mean the inclusion $C to Mnd(C)$ that picks the identity monad instead? That's what it looks like it says in Street.
– Max New
Jan 2 at 20:46
Oh wow! Yes, thanks for catching that!
– Tim Campion
Jan 4 at 15:16
add a comment |
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Let $C$ be a 2-category and $Mnd(C)$ the 2-category of monads in $C$. As explained by Street in the formal theory of monads, the Eilenberg-Moore construction is right 2-adjoint to the inclusion 2-functor $C to Mnd(C)$ sending an object to the identity monad on it. Therefore it preserves 2-limits.
A fibration in a 2-category may be defined in terms of certain 2-limits (namely slice categories) and adjoints, and so is preserved by any 2-functor preserving 2-limits, and in particular by the Eilenberg-Moore construction.
Thus it's sufficient for $p$ to be a fibration object in the 2-category $Mnd(C)$.
answered Dec 29 '18 at 5:50
Tim CampionTim Campion
13.5k354122
1
So just need to figure out what a fibration object in Mnd is then.
– Max New
Dec 29 '18 at 16:37
1
Fibrations can be detected by homming in, so one answer will probably be a tautological one: $p: E to B$ is a fibration if $Mnd(C)(X,E) to Mnd(C)(X,B)$ is a fibration for all monads $X$, which will probably reduce in the case $C = Cat$ to the induced map of Eilenberg-Moore objects being a fibration.
– Tim Campion
Dec 30 '18 at 15:00
2
I think it would probably be easier to understand fibrations in Mnd(C) using the characterization of fibrations in terms of limits. Since the forgetful functor $mathitMnd(C)to C$ also preserves limits (when C has Kleisli objects it has a left adjoint), being a fibration in Mnd(C) just means being a fibration in C together with the extra structure that the universal lift-assigning functor and transformation lift to Mnd(C). Then just work out what that means explicitly.
– Mike Shulman
Dec 30 '18 at 15:42
When you say the forgetful functor $Mnd(C) to C$ do you mean the inclusion $C to Mnd(C)$ that picks the identity monad instead? That's what it looks like it says in Street.
– Max New
Jan 2 at 20:46
Oh wow! Yes, thanks for catching that!
– Tim Campion
Jan 4 at 15:16
add a comment |
Let $C$ be a 2-category and $Mnd(C)$ the 2-category of monads in $C$. As explained by Street in the formal theory of monads, the Eilenberg-Moore construction is right 2-adjoint to the inclusion 2-functor $C to Mnd(C)$ sending an object to the identity monad on it. Therefore it preserves 2-limits.
A fibration in a 2-category may be defined in terms of certain 2-limits (namely slice categories) and adjoints, and so is preserved by any 2-functor preserving 2-limits, and in particular by the Eilenberg-Moore construction.
Thus it's sufficient for $p$ to be a fibration object in the 2-category $Mnd(C)$.
answered Dec 29 '18 at 5:50
Tim CampionTim Campion
13.5k354122
1
So just need to figure out what a fibration object in Mnd is then.
– Max New
Dec 29 '18 at 16:37
1
Fibrations can be detected by homming in, so one answer will probably be a tautological one: $p: E to B$ is a fibration if $Mnd(C)(X,E) to Mnd(C)(X,B)$ is a fibration for all monads $X$, which will probably reduce in the case $C = Cat$ to the induced map of Eilenberg-Moore objects being a fibration.
– Tim Campion
Dec 30 '18 at 15:00
2
I think it would probably be easier to understand fibrations in Mnd(C) using the characterization of fibrations in terms of limits. Since the forgetful functor $mathitMnd(C)to C$ also preserves limits (when C has Kleisli objects it has a left adjoint), being a fibration in Mnd(C) just means being a fibration in C together with the extra structure that the universal lift-assigning functor and transformation lift to Mnd(C). Then just work out what that means explicitly.
– Mike Shulman
Dec 30 '18 at 15:42
When you say the forgetful functor $Mnd(C) to C$ do you mean the inclusion $C to Mnd(C)$ that picks the identity monad instead? That's what it looks like it says in Street.
– Max New
Jan 2 at 20:46
Oh wow! Yes, thanks for catching that!
– Tim Campion
Jan 4 at 15:16
add a comment |
Let $C$ be a 2-category and $Mnd(C)$ the 2-category of monads in $C$. As explained by Street in the formal theory of monads, the Eilenberg-Moore construction is right 2-adjoint to the inclusion 2-functor $C to Mnd(C)$ sending an object to the identity monad on it. Therefore it preserves 2-limits.
A fibration in a 2-category may be defined in terms of certain 2-limits (namely slice categories) and adjoints, and so is preserved by any 2-functor preserving 2-limits, and in particular by the Eilenberg-Moore construction.
Thus it's sufficient for $p$ to be a fibration object in the 2-category $Mnd(C)$.
answered Dec 29 '18 at 5:50
Tim CampionTim Campion
13.5k354122
Let $C$ be a 2-category and $Mnd(C)$ the 2-category of monads in $C$. As explained by Street in the formal theory of monads, the Eilenberg-Moore construction is right 2-adjoint to the inclusion 2-functor $C to Mnd(C)$ sending an object to the identity monad on it. Therefore it preserves 2-limits.
A fibration in a 2-category may be defined in terms of certain 2-limits (namely slice categories) and adjoints, and so is preserved by any 2-functor preserving 2-limits, and in particular by the Eilenberg-Moore construction.
Thus it's sufficient for $p$ to be a fibration object in the 2-category $Mnd(C)$.
answered Dec 29 '18 at 5:50
Tim CampionTim Campion
13.5k354122
edited Jan 4 at 15:17
answered Dec 29 '18 at 5:50
Tim CampionTim Campion
13.5k354122
answered Dec 29 '18 at 5:50
Tim CampionTim Campion
13.5k354122
answered Dec 29 '18 at 5:50
Tim CampionTim Campion
13.5k354122
13.5k354122
1
So just need to figure out what a fibration object in Mnd is then.
– Max New
Dec 29 '18 at 16:37
1
Fibrations can be detected by homming in, so one answer will probably be a tautological one: $p: E to B$ is a fibration if $Mnd(C)(X,E) to Mnd(C)(X,B)$ is a fibration for all monads $X$, which will probably reduce in the case $C = Cat$ to the induced map of Eilenberg-Moore objects being a fibration.
– Tim Campion
Dec 30 '18 at 15:00
2
I think it would probably be easier to understand fibrations in Mnd(C) using the characterization of fibrations in terms of limits. Since the forgetful functor $mathitMnd(C)to C$ also preserves limits (when C has Kleisli objects it has a left adjoint), being a fibration in Mnd(C) just means being a fibration in C together with the extra structure that the universal lift-assigning functor and transformation lift to Mnd(C). Then just work out what that means explicitly.
– Mike Shulman
Dec 30 '18 at 15:42
When you say the forgetful functor $Mnd(C) to C$ do you mean the inclusion $C to Mnd(C)$ that picks the identity monad instead? That's what it looks like it says in Street.
– Max New
Jan 2 at 20:46
Oh wow! Yes, thanks for catching that!
– Tim Campion
Jan 4 at 15:16
add a comment |
1
So just need to figure out what a fibration object in Mnd is then.
– Max New
Dec 29 '18 at 16:37
1
Fibrations can be detected by homming in, so one answer will probably be a tautological one: $p: E to B$ is a fibration if $Mnd(C)(X,E) to Mnd(C)(X,B)$ is a fibration for all monads $X$, which will probably reduce in the case $C = Cat$ to the induced map of Eilenberg-Moore objects being a fibration.
– Tim Campion
Dec 30 '18 at 15:00
2
I think it would probably be easier to understand fibrations in Mnd(C) using the characterization of fibrations in terms of limits. Since the forgetful functor $mathitMnd(C)to C$ also preserves limits (when C has Kleisli objects it has a left adjoint), being a fibration in Mnd(C) just means being a fibration in C together with the extra structure that the universal lift-assigning functor and transformation lift to Mnd(C). Then just work out what that means explicitly.
– Mike Shulman
Dec 30 '18 at 15:42
When you say the forgetful functor $Mnd(C) to C$ do you mean the inclusion $C to Mnd(C)$ that picks the identity monad instead? That's what it looks like it says in Street.
– Max New
Jan 2 at 20:46
Oh wow! Yes, thanks for catching that!
– Tim Campion
Jan 4 at 15:16
1
1
So just need to figure out what a fibration object in Mnd is then.
– Max New
Dec 29 '18 at 16:37
So just need to figure out what a fibration object in Mnd is then.
– Max New
Dec 29 '18 at 16:37
1
1
Fibrations can be detected by homming in, so one answer will probably be a tautological one: $p: E to B$ is a fibration if $Mnd(C)(X,E) to Mnd(C)(X,B)$ is a fibration for all monads $X$, which will probably reduce in the case $C = Cat$ to the induced map of Eilenberg-Moore objects being a fibration.
– Tim Campion
Dec 30 '18 at 15:00
Fibrations can be detected by homming in, so one answer will probably be a tautological one: $p: E to B$ is a fibration if $Mnd(C)(X,E) to Mnd(C)(X,B)$ is a fibration for all monads $X$, which will probably reduce in the case $C = Cat$ to the induced map of Eilenberg-Moore objects being a fibration.
– Tim Campion
Dec 30 '18 at 15:00
2
2
I think it would probably be easier to understand fibrations in Mnd(C) using the characterization of fibrations in terms of limits. Since the forgetful functor $mathitMnd(C)to C$ also preserves limits (when C has Kleisli objects it has a left adjoint), being a fibration in Mnd(C) just means being a fibration in C together with the extra structure that the universal lift-assigning functor and transformation lift to Mnd(C). Then just work out what that means explicitly.
– Mike Shulman
Dec 30 '18 at 15:42
I think it would probably be easier to understand fibrations in Mnd(C) using the characterization of fibrations in terms of limits. Since the forgetful functor $mathitMnd(C)to C$ also preserves limits (when C has Kleisli objects it has a left adjoint), being a fibration in Mnd(C) just means being a fibration in C together with the extra structure that the universal lift-assigning functor and transformation lift to Mnd(C). Then just work out what that means explicitly.
– Mike Shulman
Dec 30 '18 at 15:42
When you say the forgetful functor $Mnd(C) to C$ do you mean the inclusion $C to Mnd(C)$ that picks the identity monad instead? That's what it looks like it says in Street.
– Max New
Jan 2 at 20:46
When you say the forgetful functor $Mnd(C) to C$ do you mean the inclusion $C to Mnd(C)$ that picks the identity monad instead? That's what it looks like it says in Street.
– Max New
Jan 2 at 20:46
Oh wow! Yes, thanks for catching that!
– Tim Campion
Jan 4 at 15:16
Oh wow! Yes, thanks for catching that!
– Tim Campion
Jan 4 at 15:16
add a comment |
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