* Kleisli and colimits
@ 2001-04-25 14:31 Tom Leinster
2001-04-25 16:04 ` Carsten Fuhrmann
0 siblings, 1 reply; 2+ messages in thread
From: Tom Leinster @ 2001-04-25 14:31 UTC (permalink / raw)
To: categories
Does anyone know if the Kleisli construction (sending a monad to its Kleisli
category) behaves in any decent way with respect to colimits? E.g. does it
in any sense preserve or reflect them?
The actual situation that I have is a fixed category C, and a certain
coequalizer diagram in the category of monads on C. The resulting fork in
Cat is also a coequalizer, and the proofs that both diagrams are coequalizers
have some ingredients in common, but I can't at present see how to deduce one
from the other.
Thanks,
Tom Leinster
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* Re: Kleisli and colimits
2001-04-25 14:31 Kleisli and colimits Tom Leinster
@ 2001-04-25 16:04 ` Carsten Fuhrmann
0 siblings, 0 replies; 2+ messages in thread
From: Carsten Fuhrmann @ 2001-04-25 16:04 UTC (permalink / raw)
To: categories
The construction of the Kleisli category can be turned into a
left adjoint functor whose domain is the category of Mnd monads
(on arbitrary categories) and monad morphisms. However, the
codomain of this functor is not Cat, but a category AbsKl whose
objects I call "abstract Kleisli categories". An abstract
Kleisli category is a category K together with a functor L:K->K,
a natural transformation \epsilon: L->Id, and a (not generally
natural) transformation \theta:Id->L, such that
(1) \theta_L is a natural transformation
(2) L\theta o \theta = \theta_L o \theta
(3) \epsilon o \theta = id
(3) L\epsilon o \theta_L = id
A morphism K->K' of AbsKl is a functor that preserves the
solutions of the non-naturality square
\theta o f = Lf o \theta (I)
The Kleisli construction forms a functor Mnd->AbsKl. Its right
adjoint sends an abstract Kleisli category K to the evident monad
on the subcategory given by the solutions of Equation (I). The
counit of this adjunction is in fact an iso, so AbsKl is a full
reflective subcategory of Mnd. The full subcategory of Mnd which
is equivalent to Abskl is given by those monads for which every
component of the unit is a regular mono.
Cheers,
Carsten
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