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> **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?
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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...



In a closed monoidal abelian category, are the compact projectives a monoidal subcategory?
In general, the answer to this question is no. Counterexamples can be found by using Day's notion of a promonoidal category, see <http://ncatlab.org/nlab/show/promonoidal+category>.


To give a promonoidal structure on a small (additive) category is equivalent to giving biclosed monoidal structure on the category of (additive) presheaves. The compact projective objects in the latter are well understood, they are precisely the retracts of finite direct sums of representables.


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.


