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• CommentRowNumber1.
• CommentAuthordanlior2
• CommentTimeApr 30th 2010

Hello,

Let me first preemptively apologize for any protocol violations. This is my first post to nlab and I had some trouble with the math formatting. My question is simple so I don’t think the formatting will be problematic.

Suppose $B$ and $C$ are (small?) categories, $H:B \rightarrow C$ a functor and $D$ is a closed, symmetric monoidal category. Then $H$ induces functors $H^{*} : Func(C,D) \rightarrow Func(B,D)$ and $(H^*)^{op} : Func(C^{op},D) \rightarrow Func(B^{op},D)$ by precomposition. I would like to know sufficient conditions on $H$ so that:

$F \otimes_{C}G \equiv (H^*)^{op}(F) \otimes_{B} H^*(G)$

for each $F \in Func(C^{op},D)$ and $G \in Func(C,D)$.

Questions:

1) Does what I’m saying make sense?

2) Is what I am asking that $H^*$ be a “monoidal functor”?

3) What are sufficient conditions for $H$ so that $H*$ is a monoidal functor?

• CommentRowNumber2.
• CommentAuthordanlior2
• CommentTimeApr 30th 2010

I now realize that “monoidal functor” is the wrong word for H*, so ignore question 2. I’m still interested in sufficient conditions on H so that H* preserves tensor products.

dan

• CommentRowNumber3.
• CommentAuthorTodd_Trimble
• CommentTimeApr 30th 2010
• (edited Apr 30th 2010)

The question makes sense; I interpret as asking under what conditions the canonical map

$(H^{op})^*(F) \otimes_B H^*(G) \to F \otimes_C G$

is an isomorphism.

It might help to look first at the special case $D = Set$, and try to generalize later.

Look at the case where $F, G$ are representables first. If the canonical map $hom(H-, c)) \otimes_B hom(c', H-)) \to hom(c', c)$ is an isomorphism for all $c, c'$, then the same is true more generally. For surjectivity, it suffices that every morphism $f: c' \to c$ factors through an object of the form $H(b)$. For injectivity, it suffices that for any two such factorizations of the same morphism,

$c' \to H(b) \to c, \qquad c' \to H(b') \to c,$

there be some $g: b \to b'$ such that sticking in $H(g)$ as diagonal filler makes the two triangles commute.

Not sure what else to say right now. What’s the context for this problem?

Edit: the diagonal filler condition may be overkill. It would suffice to have a zig-zag of such diagonal fillers

$b \to b_1 \leftarrow b_2 \to \ldots \to b'$

which I guess can be restated as a connectivity condition on some comma category.

• CommentRowNumber4.
• CommentAuthordanlior2
• CommentTimeApr 30th 2010

Todd,

Thank you very much for your quick response.

Your interpretation of my problem is exactly right.

I’ll study your example and try to apply it to my problem.

As for context; I don’t know if this will help but :

$D$ is the category of spaces (or simplicial sets), $C$ is the category of nontrivial partitions of a fixed n element set, $B$ is a certain category of labelled trees (which I haven’t yet completely defined), $F$ is an arbitrary (fixed) functor, $G$ is the functor : $Nerve(- \downarrow C)$ and $H$ is a functor which I also haven’t yet completely defined.

dan

• CommentRowNumber5.
• CommentAuthorMike Shulman
• CommentTimeApr 30th 2010

Here’s (one version of) the abstract argument that reduces it to the case of representables that Todd considered. Tensoring $F\otimes_C G$ is equivalent to tensoring $F\otimes G : C^{op}\times C \to D$ with $C(-,-):C^{op}\times C \to Set$, and likewise for the other tensor product we have

$(H^{op})^*(F) \otimes_B H^*(G)\; \cong\; B(-,-) \otimes_{B^{op}\times B} (H^{op})^* F \otimes H^* G.$

Now by the co-Yoneda lemma, we have $H^* G \cong H_\bullet \,\times_C\, G$, and similarly $(H^{op})^* F \cong H^\bullet \,\times_C \, F$, where $H_\bullet(c,b) = C(c,H b)$ and $H^\bullet(b,c) = C(H b,c)$. Thus we get

$(H^{op})^*(F) \otimes_B H^*(G)\; \cong\; B(-,-) \otimes_{B^{op}\times B} (H^\bullet \times H_\bullet) \otimes_{C^{op}\times C} (F\otimes G).$

so clearly a sufficient condition would be

$C(-,-) \;\cong\; B(-,-) \otimes_{B^{op}\times B} (H^\bullet \times H_\bullet).$

But by the co-Yoneda lemma again, this is equivalent to

$C(-,-) \;\cong\; H^\bullet \otimes_B H_\bullet$

which is essentially where Todd started from. Note that this also implies that the result holds for arbitrary $D$.

• CommentRowNumber6.
• CommentAuthorTodd_Trimble
• CommentTimeApr 30th 2010

Thanks, Mike – that was exactly the reduction I had in mind.

It sounds (Dan) like your $H$ is concrete enough so that the sufficient conditions I stipulated could be checked by hand.

• CommentRowNumber7.
• CommentAuthordanlior2
• CommentTimeMay 1st 2010

Thanks again to both of you. I appreciate it.

dan