Let $\mathcal{C}$ be a category and $\mathcal{P}=Functors(\mathcal{C}^{op},\mathcal{S})$ the category of simplicial presheaves, where $\mathcal{S}=sSet$. I want $\mathcal{P}$ to be enriched, tensored, and cotensored over the category of simplicial sets $\mathcal{S}$ in the correct way. So I'll spell this out myself, and then the question will be if I got it right. (I'm pretty sure the answer is yes, but I'm uncertain. A reference where this is spelled out would be very welcome.)

First, to any simplicial set $K \in \mathcal{S}$, I can associate a constant simplicial presheaf, also denoted $K$. Then I can define an action of the symmetrical monoidal category $sSet$ on $\mathcal{P}$ by taking the (categorical) product with $K$ (which is computed levelwise). This, I think, is going to be the tensoring.

Next, for any two $F,G \in \mathcal{P}$, we want mapping spaces $Map_{\mathcal{P}}(F,G) \in \mathcal{S}$ so that the $0$-simplices are just the morphisms between $F,G$ in $\mathcal{P}$, and so that this is compatible with the tensoring: $Map_{\mathcal{P}}(K \times F,G) \simeq Map_{\mathcal{S}}(K, Map_{\mathcal{P}}(F,G))$. In particular, setting $K=\Delta^{n}$, we see that we must have for the $n$-simplices $Map_{\mathcal{P}}(F,G)_{n}=Hom_{\mathcal{P}}(\Delta^{n} \times F,G)$.

To the categorical product with constant simplicial presheaves and the above mapping spaces should make $\mathcal{P}$ tensored and enriched over $\mathcal{S}$.

Finally, we want $\mathcal{P}$ to be 'cotensored' or 'powered'. In fact, I think more is true. $\mathcal{P}$ should have an internal Hom whose value at $x \in \mathcal{C}$ is $\mathcal{Hom}(F,G)(x)=Map_{\mathcal{P}}(F_{| \mathcal{C}/x},G_{| \mathcal{C}/x})$, and the cotensoring $F^{K}$ should just be $\mathcal{Hom}_{\mathcal{P}}(K,G)$.


1 Answer 1


Hi, yes you are right. I don't know if you're still interested, but since I took some time to understand what happens here, I might share it.

First, $(\textbf{sSet}, \times, \ast)$ is a closed, symmetric monoidal category, where the internal-hom is $\textbf{Map}(F \times \Delta[-],G)$.

For any small category $\mathcal{C}$, the category of simplicial presheaves $[\mathcal{C}^{\text{op}}, \textbf{sSet}]$ inherits object-wise the monoidal structure, that is, $(F \times G)(c) := F(c) \times G(c) \in \textbf{sSet}$ and the unit is just the object-wise unit. This is trivially symmetric, and in fact also closed with the object-wise internal-hom $(F^G)(c) := F(c)^{G(c)} = \textbf{Map}(G(c), F(c)) \in \textbf{sSet}$. No tricks, everything is object-wise.

Now, there is a fully faithful embedding $\textbf{sSet} \hookrightarrow [\mathcal{C}^{\text{op}}, \textbf{sSet}]$ sending a simplicial set to the constant (on objects) diagram. Therefore, there is a now a notion of tensor product between a simplicial set $X_{\bullet}$ and a simplicial presheaf $F$, by doing the product in the category of simplicial presheaves after the above embedding, i.e., $(X_{\bullet} \otimes F)(c) := X_{\bullet} \times F(c) \in \textbf{sSet}$ and as you said, this is the tensor. The enriched-hom is now just $\textbf{Map}(F \otimes \Delta[-], G)$, and the cotensor similarly $(F^{X_{\bullet}})(c) := \textbf{Map}(F(c), X_{\bullet})$. This gives the $\textbf{sSet}$-enrichment of the category of simplicial presheaves which is tensored and cotensored.

So there are two structures on the category of simplicial presheaves : it is first symmetric closed monoidal and so it can be enriched over itself with a simplicial presheaf as internal-hom, and it also is enriched over simplicial sets which is just a full subcategory and gives a simplicial mapping space.

For the model structures (projective and injective) on the category of simplicial presheaves you can have a look to chapter 3 here. The projective is very good since it is in fact monoidal,simplicial and proper, while the injective is only simplicial and proper.

  • 1
    $\begingroup$ Setting $(F^G)(c):=F(c)^{G(c)}=\mathop{\mathbf{Map}}(G(c),F(c))$ does not define a simplicial presheaf $F^G$, that $\mathop{\mathbf{Map}}$ is contravariant in the first argument and covariant in the second. How is $f^\ast:\mathop{\mathbf{Map}}(G(d),F(d))\to \mathop{\mathbf{Map}}(G(c),F(c))$ defined, for a morphism $f:c\to d$ in $\mathcal{C}$? $\endgroup$
    – user337830
    Oct 10, 2016 at 14:58
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    $\begingroup$ The internal-hom $F^G$ is given by $F^G(c):=\mathop{\mathbf{Map}}(G\times \Delta^0_c,F)$, where $\Delta^0_c$ is the simplicial presheaf represented by $c$, i.e. $(\Delta^0_c)_n=\mathop{\mathrm{h}}_c$. $\endgroup$
    – user337830
    Oct 10, 2016 at 15:16

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