# In a closed monoidal abelian category, are the compact projectives a monoidal subcategory?

Question: In a closed monoidal abelian category such that the unit object is compact projective, must the tensor product of compact projective objects be compact projective?

Recall that an object $X\in \mathcal C$ for an abelian category $\mathcal C$ is compact projective if $\hom(X,-) : \mathcal C \to \mathrm{AbGp}$ preserves colimits. If $\mathcal C$ is monoidal with unit $\mathbb 1$, then $\hom(\mathbb 1,-)$ is a "global sections" functor, so I am assuming that it is exact.

Note that if you do not assume that $\mathbb 1$ is compact projective, then there are plenty of examples of closed monoidal categories in which compact projectives do not tensor to compact projectives. For example, let $H$ be any infinite-dimensional Hopf algebra (over a field for convenience), and $\mathcal C = \mathrm{Mod}_H$ its category of (right, say) modules with the usual tensor structure coming from Hopf comultiplication. Then the rank-one free module $H$ itself is compact projective, but Hopfness provides an isomorphism $H \otimes H \cong H^{\oplus \dim H}$, and the latter is not compact projective if $\dim H = \infty$. On the other hand (at least in the case of infinite group algebras and universal enveloping algebras) the trivial module is not compact projective.

For comparison, in the category of comodules for a Hopf algebra, if the trivial is compact projective, then the compact projectives are precisely the dualizables.

It feels like there should be some trivial argument using the closedness of the monoidal functor. For example, call an object $X \in\mathcal C$ "compact projective over $\mathcal C$" if the inner hom functor $\underline{\hom}(X,-) : \mathcal C \to \mathcal C$ is cocontinuous. (For example, $\underline{\hom}(\mathbb 1,-) = \mathrm{id}$, so $\mathbb 1$ is always compact projective over $\mathcal C$.) Then a hom-tensor adjunction verifies that the compact-projectives-over-$\mathcal C$ are closed under $\otimes$, and that if $X$ is compact projective over $\mathcal C$ and $Y$ is compact projective in the usual sense (i.e. over $\mathrm{AbGp}$), then $X \otimes Y$ is compact projective in the usual sense. In particular, if $\mathbb 1$ is compact projective, then compact projectivity over $\mathcal C$ implies usual compact projectivity, but the converse can fail.

Perhaps there is some counterexample...

• If your tensor category is finite, then compact projective is the same as projective. In that setting, tensoing a projective by anything is projective, see Prop 2.1 of arxiv.org/abs/math/0301027 – Noah Snyder Feb 15 '15 at 0:10

For an explicit example, take your domain category to consist of two objects and no non-trivial morphisms. Its presheaf category is simply the category $\mathrm{Ab} \times \mathrm{Ab}$ of pairs of abelian groups $(A,B)$. The representables are $(\mathbb{Z},0)$ and $(0,\mathbb{Z})$, so the compact projective objects in this category are precisely the pairs $(P,Q)$ where both $P$ and $Q$ are projective.
For any fixed abelian group $M$ there is a promonoidal structure whose resulting tensor product is given by $(A,B) \otimes (A^{\prime},B^{\prime})=(A\otimes A^{\prime},A\otimes B^{\prime} + B \otimes A^{\prime} + B \otimes B^{\prime} \otimes M)$. This has unit $(\mathbb{Z},0)$ and the associativity isomorphism is built from associator and symmetry of the standard symmetric monoidal structure on $\mathrm{Ab}$ (you can even get it to be strictly associative if you take a skeleton of $\mathrm{Ab}$). Now if you take $M$ to be any abelian group that is not finitely generated projective, then $(0,\mathbb{Z}) \otimes (0,\mathbb{Z}) \cong (0,M)$ is not compact projective.
• Of course. I should have thought of that. I had spent a while playing with cases where $(0,\mathbb Z)\otimes (0,\mathbb Z) \cong (\mathbb Z,M)$, in which an associator puts very strong restrictions on $M$. Here I guess there might be some choice for the associator for general $M$, but really only one choice for, say, $M = \mathbb Z / (n)$. – Theo Johnson-Freyd Feb 15 '15 at 6:39
• I have not thought about the uniqueness of this monoidal structure. I guess picking an associator boils down to choosing an automorphism of $M \otimes M$ (the three-fold tensor product of $(0,\mathbb{Z})$), subject to some sort of cocycle condition on $M\otimes M \otimes M$ (the four-fold tensor product of $(0,\mathbb{Z})$). The remaining conditions should follow by writing general presheaves as colimits of representables. The trick with just choosing a skeleton of $\mathrm{Ab}$ is sort of a cop out, but it shows existence of a counterexample with the least amount of effort. – Daniel Schäppi Feb 15 '15 at 21:52