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I asked a very similar question but it was suggested that I rephrase and re-ask it.

Let $B$ be a monoidal category with multiplication $\Box$. Let $P$ be a category and let $T \colon P^\mathrm{op} \to B$ and $S \colon P \to B$ be functors. MacLane [CWM, p226] says that these two functors have a "tensor product"

$$ T \Box_P S = \int^{p\colon P} (Tp) \Box (Sp) .$$

It is not obvious to me that this coend exists. What are the correct assumptions for this to be true? Should it be

If $B$ is cocomplete and $P$ is small then...

or maybe

If $B$ is cocomplete and $P$ is any category...

I am very interested in the second case and I was wondering if this is known to be true or not. Note that it does not make sense to assume $B$ has all (co)limits since necessarily this would force $B$ to be a preorder. So cocomplete of course means "has all small colimits".

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    $\begingroup$ $B$ cocomplete and $P$ small is a sufficient condition. $P$ can't be arbitrary. If $P$ is discrete then you end up with a giant coproduct, I think. $\endgroup$ – Zhen Lin Oct 1 '12 at 17:43
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You are maybe interested in observing that the functor tensor product $$ T*S =\int^{p\in P} Tp\square Sp $$ arises as a particular instance of the "nerve-realization" paradigm, namely the fact that there is an equivalence $$ [\mathbf P^\text{op},\mathbf B](T, N_S(M))\cong {\bf B}(T*S,b) $$ where $N_S(-)\colon b\mapsto \Big(p\mapsto \mathbf B(Sp,b)\Big)$.

In fact this is not the most general situation in which the aforementioned isomorphism holds: you can suppose $T,S$ do not share the same codomain, and instead ${\rm trg}(T)=\bf V$, a monoidal category where ${\rm trg}(S)=\mathbf B$ is enriched. If $\bf B$ is cocomplete and cotensored, namely there is a bifunctor $$ -\boxtimes-\colon \bf V\times B\to B $$ then you can define $T*S :=\int^p Tp\boxtimes Sp$.

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    $\begingroup$ You have neglected to answer the question: when does the coend exist? As I said, $\mathbb{P}$ can't be arbitrary! $\endgroup$ – Zhen Lin Sep 18 '13 at 15:54

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