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Maybe this is too obvious, but every adjunction gives a comonad. If $(F,G)$ is a pair of adjoint functors, then $F \circ G$ defines a comonad, just as $G \circ F$ defines a monad.
Maybe this is too obvious, but every adjunction gives a comonad. If $(F,G)$ is a pair of adjoint functors, then $F \circ G$ defines a comonad, just as $G \circ F$ defines a monad
Maybe this is too obvious, but every adjunction gives a comonad. If $(F,G)$ is a pair of adjoint functors, then $F \circ G$ defines a comonad, just as $G \circ F$ defines a monad.
Maybe this is too obvious, but every adjunction gives a comonad. If $(G,F)$$(F,G)$ is a pair of adjoint functors, then $F \circ G$ defines a comonad, just as $G \circ F$ defines a monad
Maybe this is too obvious, but every adjunction gives a comonad. If $(G,F)$ is a pair of adjoint functors, then $F \circ G$ defines a comonad, just as $G \circ F$ defines a monad
Maybe this is too obvious, but every adjunction gives a comonad. If $(F,G)$ is a pair of adjoint functors, then $F \circ G$ defines a comonad, just as $G \circ F$ defines a monad
Maybe this is too obvious, but there are at least as many comonads as there are adjunctionsevery adjunction gives a comonad. If $(G,F)$ is a pair of adjoint functors, then $F \circ G$ defines a comonad, just as $G \circ F$ defines a monad
Maybe this is too obvious, but there are at least as many comonads as there are adjunctions. If $(G,F)$ is a pair of adjoint functors, then $F \circ G$ defines a comonad, just as $G \circ F$ defines a monad
Maybe this is too obvious, but every adjunction gives a comonad. If $(G,F)$ is a pair of adjoint functors, then $F \circ G$ defines a comonad, just as $G \circ F$ defines a monad
