Is the tensor product of chain complexes a Day convolution? Recently, Jade Master asked whether the tensor product of chain complexes could be viewed as a special case of Day convolution. Noting that chain complexes may be viewed as $\mathsf{Ab}$-functors from a certain $\mathsf{Ab}$-category $\mathsf{C}$, Yuri Sulyma suggested¹ that maybe we could obtain the tensor product of two chain complexes as a Day convolution by endowing $\mathsf{C}$ with the monoidal structure given by $[n]\otimes_\mathsf{C}[m]\overset{\mathrm{def}}{=}[n+m-1]$.
Questions: Is this affirmation true? More precisely:


*

*Given two chain complexes $X_\bullet$ and $Y_\bullet$ on an abelian category $\mathcal{A}$, is their Day convolution as $\mathsf{Ab}$-functors from $(\mathsf{C},\otimes_\mathsf{C})$ the usual tensor product of chain complexes $\otimes_{\mathsf{Ch}(\mathcal{A})}$?

*If not, is there some other monoidal structure on $\mathsf{C}$ for which Day convolution gives $\otimes_{\mathsf{Ch}(\mathcal{A})}$?

*If this fails too, is there perhaps another way to view $\otimes_{\mathsf{Ch}(\mathcal{A})}$ as a special case of some general construction in enriched category theory?



¹Note that his account is protected and hence his reply is not public.
 A: If you use the category $C$ to represent chain complexes and you mean day convolution using a functor $C \otimes C\to C$ it is not possible. This boils down to whether you can obtain the totalization functor from bi-complexes to chain complexes,  as a left adjoint to restriction for some functor $m: C \otimes C \to C$.  
You cannot do this because the left adjoint $m_!$ will always take representable projectives to representable projectives.  I.e.  we will have that $$ m_! (C \otimes C((i,j), -))(r) = C(m(i,j),r).$$
But the totalization of a representable projective of $C \otimes C$ is a direct sum of two different principal projectives of $C$, so no choice of $m$ will work.
What is going wrong is that the totalization functor is given by a unique $(C \otimes C, C)$ bi-module, and this bi-module cannot come from a homomorphism $C \otimes C \to C$, because in some sense it is "multi-valued."  To fix this, one could change $C$ to a morita equivalent category,  $C'$ for which the bi-module is in fact given by a homomorphism.  To construct such a $C'$, we need to choose a collection of generating projectives of ${\rm Ab}^{C}$ which is closed under tensor product. I don't see a particularly nice choice. But skd's comment is that if we use derived Morita equivalence instead of ordinary Morita equivalence, there is a very nice choice of (non-projective) generators, where   $C'$ becomes the category $\mathbb N, \leq$.  
A: The answer to the question posed in the title of your post is yes, the tensor product of chain complexes is a Day convolution product. The important thing to note is that, to define a Day convolution monoidal structure on the $\mathcal{V}$-enriched functor category $[\mathcal{C},\mathcal{V}]$ (where $\mathcal{V}$ is a complete and cocomplete symmetric monoidal closed category, e.g. $\mathbf{Ab}$), we needn't demand $\mathcal{C}$ to be a monoidal $\mathcal{V}$-category: it suffices for $\mathcal{C}$ to be a promonoidal $\mathcal{V}$-category. This is the generality at which Day convolution was originally defined in Day's thesis, which may be found here (see also his earlier paper in the Reports of the Midwest Category Seminar IV, where the word "premonoidal" was used).
A promonoidal structure on a small $\mathcal{V}$-category $\mathcal{C}$ consists of tensor product and unit "profunctors", i.e. $\mathcal{V}$-functors $P \colon \mathcal{C}^\mathrm{op}\times\mathcal{C}^\mathrm{op} \times \mathcal{C} \to \mathcal{V}$ and $J \colon \mathcal{C} \to \mathcal{V}$, together with associativity and unit constraints subject to the usual two 
"pentagon" and "triangle" axioms. Given a promonoidal structure on $\mathcal{C}$, we may construct the Day convolution monoidal structure on $[\mathcal{C},\mathcal{V}]$, whose tensor product is given at a pair of $\mathcal{V}$-functors $F,G \in [\mathcal{C},\mathcal{V}]$ by the coend
$$F\ast G = \int^{A,B \in \mathcal{C}} P(A,B;-) \otimes FA \otimes GB$$ in $\mathcal{V}$, 
and whose unit object is the $\mathcal{V}$-functor $J \in [\mathcal{C},\mathcal{V}]$, and so on. This monoidal structure on $[\mathcal{C},\mathcal{V}]$ is biclosed (i.e. the tensor product $\mathcal{V}$-functor has a right $\mathcal{V}$-adjoint -- equivalently, preserves (weighted) colimits -- in each variable). In fact, every biclosed monoidal structure on $[\mathcal{C},\mathcal{V}]$ arises in this way from some promonoidal structure on $\mathcal{C}$. (For instance, one recovers the $\mathcal{V}$-functor $P$ from the tensor product $\ast$ by $P(A,B;C) = (\mathcal{C}(A,-) \ast \mathcal{C}(B,-))C$.)
So, since the $\mathbf{Ab}$-category $\mathbf{Ch}$ of chain complexes is (equivalent to) an $\mathbf{Ab}$-enriched functor category $[\mathcal{C},\mathbf{Ab}]$ (for the $\mathbf{Ab}$-category $\mathcal{C}$ described in the question to which you linked), and since the standard monoidal structure on $\mathbf{Ch}$ is $\mathbf{Ab}$-enriched and biclosed, this monoidal structure must be the Day convolution monoidal structure for some promonoidal structure on $\mathcal{C}$. And it isn't too hard to describe that promonoidal structure. For instance, (presuming I haven't bungled the calculation) the functor $P$ is defined on objects by $$P(i,j;k) = \begin{cases} \mathbb{Z} & \mathrm{if\,\,} i+j=k, \\
\mathbb{Z} \oplus \mathbb{Z} & \mathrm{if\,\,} i+j=k+1, \\
\mathbb{Z} & \mathrm{if\,\,} i+j=k+2, \\
0 & \mathrm{else}.
\end{cases}$$
