On 6/12/08 10:21 AM, "Bockermann Bockermann" <tonymeman1@googlemail.com> wrote:
Dear mathematicians,
could anybody give me a hint if the following assertion is true? Let V be a complete and co-complete symmetric monoidal closed category. The category sV of simplicial objects in V is also complete and co-complete symmetric monoidal closed with the pointwise tensor. There is a V-adjunction D:V<-->sV:Z of the V-functor Z which evaluates in 0 and the discrete V-functor D. Does this induce a V-Isomorphism of V-categories V-Fun(K,ZC)~sV-Fun(DK,C) for any small V-category K and any sV-category C?
Please note that a similar statement is true for the non-enriched case [e.g. Borceux2, Proposition 6.4.8.].
Thank you for any help.
Tony
Dear Tony, Yes, it is true. More generally, let F-|U:W-->V be a monoidal adunction. This means that V and W are monoidal categories, F and U are monoidal functors, monoidal natural transformations 1-->UF and FU-->1 satisfying the triangle equations. (A monoidal functor F:V-->W involves maps FX\otimes FY-->F(X\otimes Y) and I_W-->F(I_V), not necessarily invertible, but satisfying coherence conditions. In a monoidal adjunction, as above, the monoidal functor F is necessarily strong, so that the comparison maps are invertible. The comparison maps for U need not be invertible.) For a small V-category K and a W-category C we do indeed have an isomorphism V-Fun(K,UC) = U(W-Fun(FK,C)) of V-categories. I'll do my best to explain this via ascii. V-functors from K to UC are in bijection with W-functors from FK to C; this takes care of the object-part. For V-functors M,N:K-->UC, the hom-object V-Fun(K,UC)(M,N) is the equalizer of the evident maps ---> Pi_k UC(Mk,Nk) ---> Pi_{k,l} [K(k,l), UC(Mk,Nl)] in V, where the products run over all objects k and l of K. On the other hand, U(W-Fun(FK,C)(M,N)) is given by the equalizer of --> U(Pi_k C(Mk,Nk)) --> U Pi_{k,l} [FK(k,l),C(Mk,Nl)] or equivalently, since U is a left adjoint, the equalizer of --> Pi_k UC(Mk,Nk) --> Pi_{k,l} U[FK(k,l),C(Mk,Nl)] So we are now left to prove Lemma: U[FX,Y]=[X,UY], for X in V and Y in W. Proof: V(Z,U[FX,Y]) = W(FZ,[FX,Y]) = W(FZ\otimes FX,Y) = W(F(Z\otimes X),Y) = V(Z\otimes X,UY) = V(Z,[X,UY]) naturally in Z and so U[FX,Y]=[X,UY] as required. Regards, Steve Lack.