Let $C$ be a category with a zero object, i.e. an object 0 which is both initial and terminal. Then $C$ is automatically (and uniquely) enriched over the category $Set_\star$ of pointed sets with smash product, where the basepoint $0\in C(x,y)$ is the unique map which factors through the object 0. Also, a functor between such categories is $Set_\star$-enriched (i.e. preserves the basepoints of homsets) if and only if it preserves the zero object. Call these pointed categories and pointed functors. (This is the "zero-ary" version of additive categories and biproducts.)

Now let $G:C^{op} \to Set_\star$ and $F:C\to Set_\star$ be pointed functors; we can then consider the tensor product $G \otimes_C F$ (otherwise known as the $G$-weighted colimit of $F$) either (1) as ordinary unenriched functors, or (2) as $Set_\star$-enriched functors. The first is a quotient of $\bigvee_c G(c)\times F(c)$, while the second is a quotient instead of $\bigvee_c G(c) \wedge F(c)$. (I'm writing $\vee$ for the coproduct in $Set_\star$, since it is a "wedge" which identifies basepoints in the disjoint union.) I believe, however, that these two tensor products turn out to be the same, since any pair $(x,0)$ in $G(c)\times F(c)$ is equal to $(x,F(t)(0))$ where $t:0\to c$ is the unique map from 0 to $c$ in $C$, and hence gets identified in the tensor product with $(G(t)(x),0) = (0,0)$, which is the basepoint of $\bigvee_c G(c)\times F(c)$. Thus when $C$ has a zero object, the ordinary tensor product over $C$ has the effect of automatically performing the smash product as well.

My question is: does this remain true for homotopy tensor products? Suppose we consider $Set_\star$ as sitting inside the category $sSet_\star$ of pointed simplicial sets, and instead of the tensor product we take its homotopical replacement, which can be described as a two-sided bar construction. We thus get two pointed simplicial sets $B^u(G,C,F)$ and $B^p(G,C,F)$ (for unpointed and pointed), defined by $$B^u_n(G,C,F) = \bigvee_{c_0,\dots,c_n} G(c_n) \times C(c_{n-1},c_n)\times\dots\times C(c_0,c_1)\times F(c_0) $$ and $$B^p_n(G,C,F) = \bigvee_{c_0,\dots,c_n} G(c_n) \wedge C(c_{n-1},c_n)\wedge\dots\wedge C(c_0,c_1)\wedge F(c_0) $$ There is a canonical quotient map $B^u(G,C,F) \to B^p(G,C,F)$, and the above observation (assuming it is correct) means that this map induces an isomorphism on $\pi_0$. Is it a weak homotopy equivalence?

Edit: Reading over the answers, I realized that there are actually two different questions here. Greg answered a question which isn't quite what I asked, but fortunately the question he answered is the one that I meant to ask, which is comparing $B^p(G,C,F)$ with the homotopy tensor product when considering $G$ and $F$ as functors landing in unpointed simplicial sets, so that the $\bigvee$s would actually become disjoint unions $\coprod$.

  • $\begingroup$ I must be missing something obvious: it seems to me that the unenriched version is simply $F \otimes_C G = F(0) \times G(0)$. The coend can be expressed as a colimit over the twisted arrow category of $C$, which is pointed because $C$ is, so the colimit is just evaluation at $0 \to 0$. $\endgroup$ Aug 27, 2014 at 13:23
  • $\begingroup$ @OmarAntolín-Camarena, I don't think the twisted arrow category is pointed. $\endgroup$ Aug 28, 2014 at 4:46
  • $\begingroup$ You're right, @MikeShulman, I was careless: when $C$ is pointed, $0\to 0$ is only terminal (and not initial) in the twisted arrow category. Now, that would still imply the coend is evaluation at $0 \to 0$, if it weren't for my other mistake: an end is a limit over the twisted arrow category, but a coend is a limit over the opposite of the twisted arrow category. So all is good and your tensor product is not automatically trivial. $\endgroup$ Aug 28, 2014 at 17:59

2 Answers 2


I'll sketch an alternative approach. I've been on the lookout lately for ways to argue about homotopy (co)limits using universal properties instead of explicit constructions, so I like this question.

Let $Fun(C,Set_* )$ be the category of functors and let $Fun_* (C,Set_* )$ be the category of pointed functors. For a given choice of $G$, the weighted colimit functor $G\otimes -$ in the enriched sense is a composition $Fun_* (C,Set_* ) \rightarrow Fun(C,Set_* )\rightarrow Set_* $ , forgetful functor followed by the unenriched $G\otimes -$. This follows from the fact that its right adjoint is the composition of right adjoints $ Set_* \rightarrow Fun(C, Set_* ) \rightarrow Fun_* (C,Set_* )$, $X\mapsto (c\mapsto Set_* (G(c),X))$ followed by $F\mapsto (c\mapsto (fiber(F(c)\rightarrow F(0))))$.

Now make a derived-functor version of the same argument.

  • $\begingroup$ I should have thought of that. The composition $Fun_*(C,sSet_*) \overset{forget}{\to} Fun(C,sSet_*) \overset{G\otimes -}{\to} sSet_*$ is isomorphic to $Fun_* (C,sSet_*) \overset{G \otimes_* -}{\to} sSet_*$, so since all functors involved have left derived functors, functoriality tells you that their derived functors agree. (The left (= right) derived functor of "forget" is just itself, since it preserves all weak equivalences.) $\endgroup$ Jun 10, 2010 at 18:08
  • $\begingroup$ OK, I see it now. I was unnecessarily worried that the forgetful functor does not seem to preserve cofibrant objects. But it actually does not matter. $\endgroup$ Jun 10, 2010 at 19:05
  • $\begingroup$ Hmm, now I don't see it any more. It is possible to have functors f and g such that g and gf have left derived functors, and f preserves all weak equivalences, but $L(gf) \ncong Lg \circ Lf$. $\endgroup$ Jun 11, 2010 at 14:36
  • $\begingroup$ I don't think I want to try to deduce $Lh\simeq Lg\circ Lf$ from $h\simeq g\circ f$. Instead, by "a derived-functor version of the same" I mean use adjointness in the homotopy categories instead of the original categories. $f$, $g$, and $h$ were forgetful functor, $G$-weighted colimit, and enriched $G$-weighted colimit. Denote their right adjoints by $f'$, $g'$, $h'$. I obtained $h\cong g\circ f$ from $h'\cong f'\circ g'$. Now I want to obtain $Lh\simeq Lg\circ Lf$ from $Rh'\simeq Rf'\circ Rg'$ (which I think is clear), using that $Rf'$, $Rg'$, $Rh'$ are right adjoint to $Lf$, $Lg$, $Lh$. $\endgroup$ Jun 12, 2010 at 4:29
  • $\begingroup$ I don't think that what I originally called the "unpointed homotopy tensor product" is quite the same as the G-weighted colimit of the diagram of pointed simplicial sets F; wouldn't that involve $G_+ \wedge F$ rather than $G\times F$? So I don't know what its right adjoint is. And for the other version of the question (which I meant to ask), where we consider G and F as both landing in unpointed simplicial sets, the tensor product in question is the G-weighted colimit, but in that case I don't see what the right adjoint of the forgetful functor is. $\endgroup$ Jun 12, 2010 at 12:15

I think the answer is yes. Here is an attempt at an argument.

Let $SS_*$ and $SS$ be the categories of pointed and unpointed simplicial sets. Let $[C, SS]$ be the category of all functors from $C$ to $SS$ and let $[C, SS_*]_*$ be the category of all pointed functors from $C$ to $SS_*$. Define similarly the functor categories $[C,Sets]$ and $[C, Sets_*]_*$. Consider $[C, Sets]$ and $[C,Sets_*]_*$ to be subcategories of $[C, SS]$ and $[C, SS_*]_*$ respectively.

The functor categories $[C, SS]$ and $[C, SS_*]_*$ have well-known model structures where weak equivalences and fibrations are defined pointwise. It is not difficult to describe the cofibrations explicitly. The cofibration in $[C, SS]$ are generated by maps of the form $$I\times \hom(x_0, -) \longrightarrow J\times \hom(x_0, -)$$ where $I, J$ are simplicial sets, $x_0$ is an object of $C$, $\hom(x_0, x)$ denotes the (pointed) set of morphisms in $C$, and the map is induced from a cofibration of simplicial sets $I\hookrightarrow J$. Similarly, the cofibrations in $[C, SS_*]_*$ are generated by maps of the form $$I_+\wedge \hom(x_0, -) \longrightarrow J_+\wedge \hom(x_0, -).$$

One can define homotopy tensor product using cofibrant replacement in this model structure. Namely, if F and G are two functors (either pointed or unpointed), then $B(G, C, F)\simeq cG \otimes cF$, where $c$ denotes a cofibrant replacement in the appropriate functor category. In fact, it is enough to take a cofibrant replacement of either $F$ or $G$. That is, $cG\otimes F\simeq G\otimes cF\simeq cG\otimes cF$.

There is an obvious forgetful functor that I will denote by $R$. $$R\colon [C, SS_*]_* \longrightarrow [C, SS].$$ Your question is equivalent to the following: does $R$ preserve homotopy coends? You only ask the question for set-valued functors, but I think the answer is yes in general. Let me formulate it a little more precisely. Let $F\colon C\to SS_*$ and $G\colon C^{op}\to SS_*$ be pointed functors. There is an evident natural map from the (unpointed) homotopy coend $RG\otimes^h RF$ to the pointed homotopy coend $G\otimes^h F$. We want to show that this map is an equivalence. Let us first check it when $F$ has the form $F(-)=I_+\wedge \hom(x_0, -)$ for some simplicial set $I$ and object $x_0$ of $C$. In this case, $F$ is cofibrant in $[C, SS_*]_*$, so the pointed homotopy coend of $F$ and $G$ is equivalent to the pointed strict coend which, by Yoneda Lemma, is equivalent to $I_+\wedge G(x_0)$. Now let us consider $RF$ and $RG$. It is not immediately obvious whether $RF$ is cofibrant in $ [C, SS]$. On the other hand, $RF$ is objectwise equivalent to the following homotopy pushout $$*\times \hom(0, -)\longleftarrow I\times \hom(0, -) \longrightarrow I\times \hom(x_0, -) .$$ Taking homotopy coend with $RG$ preserves objectwise homotopy pushouts. It follows that $RF \otimes^h RG$ is equivalent to the following homotopy pushout $$*\times \hom(0, -)\otimes^h RG\longleftarrow I\times \hom(0, -)\otimes^h RG\longrightarrow I\times\hom(x_0, -) \otimes^h RG.$$ Which, again using Yoneda Lemma, together with the fact that $RG(0)=*$, implies that $RF \otimes^h RG$ is equivalent to $I_+\wedge G(x_0)$. So we obtain that $RF\otimes^h RG$ is equivalent to $F\otimes^h G$. With a little more careful diagram-chasing it should not be hard to see that the canonical map $RF\otimes^h RG\longrightarrow F\otimes^h G$ induces this equivalence.

For a general pointed $F$, one can present $F$ as a homotopy colimit along generating cofibrations in $[C, SS_*]_*$ (take a cofibrant replacement of $F$), and one obtains the result using a similar calculation plus induction.

Edit: When I was writing this post, I actually changed my mind in the middle about how I wanted to do this, so I think it came out a bit unfocused. The idea is straightforward. Fix a contravariant pointed functor $G$. We want to check that the natural map from $F\otimes^h G$ (unpointed derived tensor product) to $F\otimes^h_* G$ (pointed derived tensor product) is an equivalence for functors $F$ of the form $F=I_+\wedge\hom(x_0, -)$. This is good enough, because all other homotopy types of pointed functors can be built as repeated homotopy pushouts of functors of this type. So, I need to calculate both the pointed and unpointed derived tensor products of $F$ and $G$ for this type of $F$. The pointed tensor product is easy, because $F$ is cofibrant in the pointed models structure, so the derived product is equivalent to the strict product, which can be calculated using the YL. The unpointed tensor product is slightly less obvious, because it is not clear that $F$ is cofibrant in the unpointed model structure, and this is why the derived case does not follow immediately from the strict case. But, $F$ can be presented as a homotopy pushout of free (in the unpointed sense!) functors, and an elementary little calculation shows that the unpointed derived tensor product agrees with the pointed one.

This is a proof by calculation. Since the "calculation" is extremely easy, I feel it is not too bad. But it would be nice to see a conceptual reason why it ought to be true. I believe such a reason exists, but I have not been able to nail it down.

  • $\begingroup$ Regarding Tom's answer below: I would be interested in seeing the derived argument fleshed out. If I am not mistaken, the functor from $Fun(C,Set_*)$ to $Fun_*(C,Set_*)$ does not preserve fibrations or acyclic fibrations in the standard model structure, so it is a little unclear what its right derived functor is. I don't doubt that it can be worked out, but I would like to learn how to do it the right way. $\endgroup$ Jun 10, 2010 at 14:03
  • $\begingroup$ I expect that your answer works, but I like Tom's better because it's easier and more conceptual. See my reply to Tom's answer re: your question. $\endgroup$ Jun 10, 2010 at 18:10
  • $\begingroup$ Now I'm looking back at this because I unconvinced myself that Tom's answer works, and it seems plausible but I don't have a feel for what's really going on. Does it have something to do with the fact that all (pointed) simplicial sets are cofibrant, so that those pushouts are automatically homotopy pushouts? It would seem unpleasant to me if a general fact like this depended on such a technical condition. $\endgroup$ Jun 11, 2010 at 20:33
  • $\begingroup$ I don't think that the fact that all simplicial sets are cofibrant is important here. The argument can be modified to work for functors from $C$ to topological spaces (or for that matter any cofibrantly generated model category) rather than simplicial sets. In that case the point would be that for every cofibrant object $I$ and any objects $x_0, x$ of $C$, the map $I\times\hom(0, x)\to I\times \hom(x_0, x)$ would be a cofibration. Well, this map is the inclusion of $I$ into a coproduct of a bunch of copies of $I$, so it is a cofibration, isn't it? $\endgroup$ Jun 12, 2010 at 1:26
  • $\begingroup$ I added a paragraph to my original post, trying to summarize the argument. $\endgroup$ Jun 12, 2010 at 1:47

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