This is a refinement of my (naive, poorly asked) question [here](https://mathoverflow.net/questions/17951/what-tensor-product-of-chain-complexes-satisfies-the-usual-universal-property). The reference for my question is [Baez and Crans, HDA6](http://math.ucr.edu/home/baez/hda6.pdf).


Background: category objects, etc.
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Let $\mathcal V$ be a category. A **category object** internal to $\mathcal V$ consists of the following data and properties:


1. Objects $C\_0,C\_1 \in \mathcal V$ and morphisms $s,t: C\_1 \to C\_0$ and $i: C\_0 \to C\_1$.
2. Such that $s\circ i = t\circ i = \text{id}\_{C\_0}$ and the pull-back $C\_1 \underset{C\_0}{\times} C\_1 = C\_1 \underset{\displaystyle ^{\searrow\!\!^{\scriptstyle s}} {C\_0} ^{^{\scriptstyle t} \!\!\swarrow} }{\times} C\_1 $ exists.
3. A morphism $m: C\_1 \underset{C\_0}\times C\_1 \to C\_1$ such that $s\circ m = s\_R$ and $t\circ m = t\_L$, where $s\_R: C\_1 \underset{C\_0}\times C\_1 \to C\_0$ is the "$s$" projection from the right factor, and similarly for $t\_L$.
4. And such that the obvious "associativity" square (two ways to get from $C\_1 \underset{C\_0}\times C\_1 \underset{C\_0}\times C\_1$ to $C\_1$) and "identity" triangles (three ways to get from $C\_1 = C\_1 \underset{C\_0}\times C\_0 = C\_0 \underset{C\_0}\times C\_1$ to $C\_1$) commute.


For example, a category object in $\mathcal V = {\rm SET}$ is a small category.


For this post, I will be interested in $\mathcal V = {\rm VECT}$, the category of vector spaces over your favorite field. I will call your favorite field "$\mathbb R$".
A category object in ${\rm VECT}$ is a **2-vector space**.


2-vector spaces are relatively mild things. Indeed, it turns out that in $\rm VECT$ the data and properties of 1-2 above uniquely determine a map $m$ satisfying 3-4.


By the general yoga known as "commutativity of internalization", a 2-vector space is the same as a "vector space object in $\rm CAT$". More precisely, let $\rm CAT$ be the category of small categories. Then it makes sense to talk about "field objects" — like a category object, a field object consists of some objects, some maps, some pull-backs, and some more maps, and some commuting diagrams. In particular, by thinking of $\mathbb R$ as a discrete category ($\mathbb R\_0 = \mathbb R = \mathbb R\_1$ and $s = t = i = {\rm id}$), it is in fact a field object. Then with some more diagrams, we can talk about "vector space objects over $\mathbb R$" internal to $\rm CAT$, and it is straightforward to check that these are the same as 2-vector spaces.


Background: tensor products
---------------------------


I know of two natural approaches to define "tensor products":


1. Define a notion of "bilinear map", such that the assignment $X,Y,Z \mapsto \{\text{bilinear maps }X\times Y \to Z\}$ is contravariant in the first two spots and covariant in the last. Then set $X\otimes Y$ to be the object (if it exists) that represents the function $Z \mapsto \{\text{bilinear maps }X\times Y \to Z\}$.
2. Define a notion of "internal hom", i.e. a (nice) functor $\underline{\rm Hom}: \mathcal V^{\rm op} \times \mathcal V \to \mathcal V$. For each $X\in \mathcal V$, define the functor $-\otimes X$ by declaring that it is left adjoint to $\underline{\rm Hom}(X,-)$.


Approach 1 is the way that tensor products are introduced in grade school. Approach 2 is I think more standard in the real world. We can implement each in the case of 2-vector spaces:


1. The trick for approach 1 is that the notion of "bilinear" depends on more than just the category. So realize the category of 2-vector spaces as the category of vector spaces objects in $\rm CAT$. Recall that a morphism of 2-vector spaces is simply a morphism of underlying categories so that some diagrams commute. Then we can say the following. Let $X,Y,Z$ be 2-vector spaces. Then a morphism of underlying categories $X \times Y \to Z$ is **bilinear** if a bunch of diagrams commute (these diagrams refer to the vector-space-object structures of $X,Y,Z$, and are precisely the diagrams that you learned in grade school). By reproducing the proofs from vector spaces internal to $\rm SET$, this in fact defines a functor $\otimes$.
2. Given 2-vector spaces $X,Y$, there is a category whose objects are linear functors $X \to Y$ and whose morphisms are linear natural transformations of functors, and this category has a natural structure as a 2-vector space. Moreover, the corresponding notion of $\underline{\rm Hom}$ is correctly functorial, and has an adjoint. So this approach defines a functor $\otimes$.


However:



> 
> The two "tensor products" defined in 1-2 above do not agree.
> 
> 
> 


Writing 2-vector spaces $X = (X\_1 \rightrightarrows X\_0)$ and $Y = (Y\_1 \rightrightarrows Y\_0)$ as category objects in $\rm VECT$, approach 1 gives $(X\otimes Y)\_a = X\_a \otimes Y\_a$ for $a=0,1$, with the tensor products of the structure maps. Approach 2 also has $(X\otimes Y)\_0 = X\_0 \otimes Y\_0$, but $(X\otimes Y)\_1 \cong $
$$ X\_0 \otimes Y\_0 \oplus \text{coker}\Bigl( \bigl(\ker(X\_1 \overset s \to X\_0) \otimes \ker(Y\_1 \overset s \to Y\_0)\bigr) \overset{t\otimes{\rm id} - {\rm id}\otimes t}\longrightarrow \bigl( X\_0 \otimes \ker(Y\_1 \overset s \to Y\_0) \oplus \ker(X\_1 \overset s \to X\_0) \otimes Y\_0\bigr) \Bigr)$$
The structure maps are: $s$ is the projection onto the first factor, and $t$ is the sum of the same projection and the map ${\rm id}\otimes t + t\otimes {\rm id}$ from the second factor (it is well-defined out of the cokernel).


(There is probably a way to simplify the above description. The trick is that, as explained in HDA6, the category of 2-vector spaces is equivalent as a category to the category of 2-term chain complexes, and this is the natural "internal Hom" over there.)


Anyway, a dimension count shows that these two "tensor" constructions are inequivalent in general.


Hence:


Question
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Does the tensor product of 2-vector spaces given in "approach 1" above — the tensor product defined as representing "bilinear functors" — under this tensor product, does the functor $\otimes X$ have a right adjoint?



What functor is adjoint to the tensor product of 2-vector spaces?
I'll denote your category of 2-vector spaces by 2Vect. By your preliminary remarks, 2Vect is actually the category of Vect-valued presheaves on Δ≤1 where Δ≤1 denotes the full subcategory of Δ on the objects [0] and [1]. Therefore, colimits in 2Vect are computed objectwise under this identification. So the functor – ⊗ X certainly has a right adjoint **Hom**(X, –) (by the adjoint functor theorem for locally presentable categories). This adjunction also respects the Vect enrichment.


To compute this adjoint, we can use the "Vect-enriched Yoneda lemma": writing Δi for the 2-vector space (Δi)j = HomΔ≤1([j], [i]) • ℝ, we have hom2Vect(Δi, X) = Xi as vector spaces, where hom denotes the Vect-enriched Hom. So


**Hom**(X, Y)0 = hom(Δ0, **Hom**(X, Y)) = hom(Δ0 ⊗ X, Y) = hom(X, Y)


since Δ0 happens to be the unit for this ⊗, but


**Hom**(X, Y)1 = hom(Δ1 ⊗ X, Y)


will have a more complicated formula which you'll have to work out (my guess is it will look similarly complicated to your expression for the other tensor product).


