Counit of the Kleisli adjunction
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In the Kleisli adjunction
$Gvarepsilon F = mu$ where $varepsilon$ is a natural transformation called the counit.
By definition $F(X) = X$ so isn't $F$ superfluous in the definition of the counit? I.e.:
$Gvarepsilon F = Gvarepsilon = mu$
category-theory adjoint-functors functors natural-transformations
asked Nov 24 at 9:38
Roland
19311
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In the Kleisli adjunction
$Gvarepsilon F = mu$ where $varepsilon$ is a natural transformation called the counit.
By definition $F(X) = X$ so isn't $F$ superfluous in the definition of the counit? I.e.:
$Gvarepsilon F = Gvarepsilon = mu$
category-theory adjoint-functors functors natural-transformations
asked Nov 24 at 9:38
Roland
19311
1
No, because $F(X) = X$ is only half the definition of functor. Look at how it acts on morphisms, it's not identity.
– Roll up and smoke Adjoint
Nov 24 at 9:56
@RollupandsmokeAdjoint Understood. But what concerns natural transformations does it matter what $F$ does to morphisms?
– Roland
Nov 24 at 9:59
1
@Roland It matters because $X$ is an object of the base category, which is not equal to the object named $X$ in the Kleisli category, just as happens in the case of opposite categories.
– Kevin Carlson
Nov 24 at 21:47
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
In the Kleisli adjunction
$Gvarepsilon F = mu$ where $varepsilon$ is a natural transformation called the counit.
By definition $F(X) = X$ so isn't $F$ superfluous in the definition of the counit? I.e.:
$Gvarepsilon F = Gvarepsilon = mu$
category-theory adjoint-functors functors natural-transformations
asked Nov 24 at 9:38
Roland
19311
In the Kleisli adjunction
$Gvarepsilon F = mu$ where $varepsilon$ is a natural transformation called the counit.
By definition $F(X) = X$ so isn't $F$ superfluous in the definition of the counit? I.e.:
$Gvarepsilon F = Gvarepsilon = mu$
category-theory adjoint-functors functors natural-transformations
category-theory adjoint-functors functors natural-transformations
asked Nov 24 at 9:38
Roland
19311
asked Nov 24 at 9:38
Roland
19311
asked Nov 24 at 9:38
Roland
19311
asked Nov 24 at 9:38
Roland
19311
asked Nov 24 at 9:38
Roland
19311
19311
1
No, because $F(X) = X$ is only half the definition of functor. Look at how it acts on morphisms, it's not identity.
– Roll up and smoke Adjoint
Nov 24 at 9:56
@RollupandsmokeAdjoint Understood. But what concerns natural transformations does it matter what $F$ does to morphisms?
– Roland
Nov 24 at 9:59
1
@Roland It matters because $X$ is an object of the base category, which is not equal to the object named $X$ in the Kleisli category, just as happens in the case of opposite categories.
– Kevin Carlson
Nov 24 at 21:47
add a comment |
1
No, because $F(X) = X$ is only half the definition of functor. Look at how it acts on morphisms, it's not identity.
– Roll up and smoke Adjoint
Nov 24 at 9:56
@RollupandsmokeAdjoint Understood. But what concerns natural transformations does it matter what $F$ does to morphisms?
– Roland
Nov 24 at 9:59
1
@Roland It matters because $X$ is an object of the base category, which is not equal to the object named $X$ in the Kleisli category, just as happens in the case of opposite categories.
– Kevin Carlson
Nov 24 at 21:47
1
1
No, because $F(X) = X$ is only half the definition of functor. Look at how it acts on morphisms, it's not identity.
– Roll up and smoke Adjoint
Nov 24 at 9:56
No, because $F(X) = X$ is only half the definition of functor. Look at how it acts on morphisms, it's not identity.
– Roll up and smoke Adjoint
Nov 24 at 9:56
@RollupandsmokeAdjoint Understood. But what concerns natural transformations does it matter what $F$ does to morphisms?
– Roland
Nov 24 at 9:59
@RollupandsmokeAdjoint Understood. But what concerns natural transformations does it matter what $F$ does to morphisms?
– Roland
Nov 24 at 9:59
1
1
@Roland It matters because $X$ is an object of the base category, which is not equal to the object named $X$ in the Kleisli category, just as happens in the case of opposite categories.
– Kevin Carlson
Nov 24 at 21:47
@Roland It matters because $X$ is an object of the base category, which is not equal to the object named $X$ in the Kleisli category, just as happens in the case of opposite categories.
– Kevin Carlson
Nov 24 at 21:47
add a comment |
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No, because $F(X) = X$ is only half the definition of functor. Look at how it acts on morphisms, it's not identity.
– Roll up and smoke Adjoint
Nov 24 at 9:56
@RollupandsmokeAdjoint Understood. But what concerns natural transformations does it matter what $F$ does to morphisms?
– Roland
Nov 24 at 9:59
1
@Roland It matters because $X$ is an object of the base category, which is not equal to the object named $X$ in the Kleisli category, just as happens in the case of opposite categories.
– Kevin Carlson
Nov 24 at 21:47