Contents 1. Introduction 2 2. Preliminaries 4 2.1. Some results on ...

180− ⊗ T x ⊗ R 1 A ⊗ A f ⊗ T y ⊗ A g ↦→ − ⊗ T x ⊗ R 1 A ⊗ A f (y 0 ) ⊗ R y 1 ⊗ A gW A can AA F Q : − ⊗ T Σ R ⊗ R Σ ∗ T ⊗ T Σ ⊗ A Σ ∗ → − ⊗ T Σ R ⊗ R A ⊗ R C ⊗ A Σ ∗− ⊗ T x ⊗ R f ⊗ T y ⊗ A g ↦→ − ⊗ T x ⊗ R f (y 0 ) ⊗ R y 1 ⊗ A g.If we apply W A can AA F Q both to θ l and θ r we get the following[(W A can AA F Q) ◦ θ l] (− ⊗ T x ⊗ R f)= (W A can AA F Q) ( − ⊗ T x 0 ⊗ R x 1 1 ⊗ T x 2 1 ⊗ A f )(( ) ) ( )= − ⊗ T x 0 ⊗ R x 1 1 x21 ⊗R0 x2 ⊗ 1 1 A f= − ⊗ T x 0 ⊗ R 1 A ⊗ R x 1 ⊗ A fand[(W A can AA F Q) ◦ θ r ] (− ⊗ T x ⊗ R f)= (W A can AA F Q) (− ⊗ T x ⊗ R f ⊗ T x i ⊗ A x ∗ i )= − ⊗ T x ⊗ R f ((x i ) 0) ⊗ R (x i ) 1⊗ A x ∗ i .Since A can A is an isomorphism, we get that(D, j) = Equ Fun((W A can AA F Q) ◦ θ l , (W A can AA F Q) ◦ θ r)Hence D ⊆ P Q = − ⊗ T Σ ⊗ R Σ ∗ . At this point we stop because it is not so clearwhat is the comonad D.We try to compute the functor Q which does not require the comonad D, but wecannot do it as well. In fact, we have the following:We calculate the equalizerQ q →P C(θ l P)◦(P can −11 )◦(P Cu A )⇒(θ r P )◦(P can −11 )◦(P Cu A )(Q, q)= EquFun((θ l P ) ◦ (P i) , (θ r P ) ◦ (P i) )= Equ Fun((θ l P ) ◦ ( P can −11BP QP)◦ (P CuA ) , (θ r P ) ◦ ( )P can −11 ◦ (P CuA ) ) .We have(θ l P ) ◦ ( )P can −11 ◦ (P CuA ) : − ⊗ R C ⊗ R Σ ∗ T −→ − ⊗ R Σ ∗ T ⊗ T Σ R ⊗ R Σ ∗ T ⊗ T B− ⊗ R c ⊗ R f ↦→ − ⊗ R 1 A ⊗ R c ⊗ R f ↦→ − ⊗ R c 1 ⊗ T c 2 ⊗ R f(↦→ − ⊗ R c 1 ⊗ ) T c2 ⊗ ( )0 R c2 1⊗ ( )1 T c2 2· f 1(= − ⊗ R c 1 ⊗ ) T c2 ⊗ 0 R β 1 ⊗ T β 2 · fwhereso thatMoreover we havec 1 ( c 2 0)⊗R c 2 1 = 1 A ⊗ R c1 A ε C (c) = c 1 ( c 2 0ε C ( c 2 1))= c1 ( c 2) .β 1 ( β 2 0)⊗R β 2 1 = 1 A ⊗ R c 2 <strong>1.</strong>