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Let $\mathbf P$ be the category with objects the natural numbers and $\hom(m,n)=Sym(n)$ if $n=m$, and the empty set otherwise. It is symmetric monoidal wrt the sum of natural numbers, and has $0$ as a unit object.

Following the lines of this, I can endow $[\mathbf{P},\mathcal V]$, for any Bénabou cosmos $\cal V$, with a monoidal structure, defining $$ F_1\star\dots\star F_m = \int^{n_1,\dots, n_m}\mathbf{P}(n_1+\dots+n_m,-)\otimes_{\cal V}F_1n_1\otimes_{\cal V}\dots\otimes_{\cal V}F_mn_m $$ Now I can define $F^{\star n}$ to be the $n$-fold convoluted product of $F$ with itself, and I can endow $[\mathbf{P},\mathcal V]$ with another (nonsymmetric) monoidal structure, define $$ T\diamond S := \int^m T(m)\otimes_{\cal V} S^{\star m} $$ My problem is to show the following identity: for any $F\in [\mathbf P,\mathcal V ]$, one has $$ (\heartsuit)\qquad F^{\star n}\cong \mathbf{P}(n,-)\diamond F = \int^m\mathbf{P}(n,m)\otimes_{\cal V}F^{\star m}. $$ Is it true? And if it is, how can I show it?

Note: I tried to keep my question self-contained to help anybody understand my request; if it seems to you that I'm implying the real origin of the problem, I'm trying to prove Lemma 3.1 here, and I'ms tuck in the row tagged with "by Yoneda". As far as I can understand, the author is exploiting the identitiy $(\heartsuit)$ to conclude his proof.

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It certainly doesn't work for arbitrary $\mathcal{V}$. At least it should be cocomplete, perhaps even a cosmos. – Martin Brandenburg Feb 1 '13 at 11:45
Woops, thanks for pointing it out. Yes, it has to be a cosmos (this is one of the initial remarks n Kelly's paper). – Fosco Loregian Feb 1 '13 at 11:47
nlab says: "There are a number of different, inequivalent, definitions of “cosmos” in the literature." Which are you using? – alancalvitti Feb 1 '13 at 15:37
I'm callng a 'cosmos' a complete and cocomplete closed symmetric monoidal category. – Fosco Loregian Feb 1 '13 at 18:11
@tetrapharmakon, the Yoneda lemma says that for any functor $K \colon \mathbb{P}^{op} \rightarrow \mathbb{V}$ there is a natural isomorphism: $K(A) \approx \int^C \hom(A, C) \otimes K(C)$. If you know that $F^*$ is a functor $n \mapsto F^{*n}$ from $\mathbb{P}^{op}$ to $\mathbb{V}$, then $F^{*n} \approx \int^m \hom(n, m) \otimes F^{*m}$ is an instance of the Yoneda lemma. – Michal R. Przybylek Feb 2 '13 at 12:28

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