In Tom Leinster's [book on operads](http://arxiv.org/abs/math/0305049), he gives Ab(V), the category of abelian groups in a symmetric monoidal category V, as an example of a multicategory that doesn't arise from a monoidal category, since Ab(V) will not generally have a tensor product. Example 2.1.5 on page 37.


I can't quite understand this. Why do you need colimits in V to define tensor product in Ab(V)? What is the definition exactly of a tensor product? I know it as the universal object representing bilinear morphisms in some multicategory, and from that definition it's not clear how colimits in some category help it exist.


Actually I have a more basic question about this setup: Can you define group objects in an arbitrary symmetric monoidal category? I thought you could only define monoid objects, and you needed a Cartesian monoidal category to define a group object. Cartesian monoidal categories have the diagonal morphism which you need to define inverses, whereas general monoidal categories do not, right?



Why does tensor product in Ab(V) require colimits in V?
It is easy to *define* the tensor product as being the object that represents the bilinear maps functor, but to prove that tensor products exist requires something extra. If you have free abelian groups, then it is enough to have coequalisers of abelian groups to construct the tensor product, but even the construction of coequalisers of abelian groups can be quite non-trivial.


I'm more familiar with the case where the base category $\mathcal{V}$ is cartesian closed, so I'll briefly mention the extra structures we need to get tensor products in $\textbf{Ab}(\mathcal{V})$:


1. If $\mathcal{V}$ has countable colimits and is a well-powered regular category, then the forgetful functor $\textbf{Ab}(\mathcal{V}) \to \mathcal{V}$ is strictly monadic. (In particular, it has a left adjoint.) A proof can be extracted by chasing the references in theorem 2.4.21 of [my notes](http://zll22.user.srcf.net/writing/algebraic-theories/20121121-Main.pdf). Moreover these hypotheses suffice to show $\textbf{Ab}(\mathcal{V})$ has coequalisers for all parallel pairs: combine proposition 2.4.7 (ii) and theorem 2.4.21 (ii).
2. Once we know that $\textbf{Ab}(\mathcal{V})$ has coequalisers of reflexive pairs and that $\textbf{Ab}(\mathcal{V}) \to \mathcal{V}$ is monadic, we can construct tensor products the usual way using generators and relations.


As for your second question: you are quite right in saying that we need some kind of diagonal map $\delta\_X : X \to X \otimes X$ in order to even define inverses. But that's OK: we just build that into the definition of a group object in a symmetric monoidal category. More accurately, what we are doing is expressing the axioms for a *Hopf algebra* in diagrammatic form: so a group object in a general symmetric monoidal category will consist of


* an object $X$, and
* morphisms $\eta : I \to X$, $\mu : X \otimes X \to X$, $\epsilon : X \to I$, $\delta : X \to X \otimes X$, and $\sigma : X \to X$


satisfying the following axioms:


* $(X, \eta, \mu)$ is a monoid.
* $(X, \epsilon, \delta)$ is a comonoid.
* $\sigma$ gives left inverses: $\mu \circ (\sigma \otimes \textrm{id}\_S) \circ \delta = \eta \circ \epsilon$
* $\sigma$ gives right inverses: $\mu \circ (\textrm{id}\_S \otimes \sigma) \circ \delta = \eta \circ \epsilon$


This *is* a generalisation of the usual definition of group object in a cartesian monoidal category because every object in a cartesian monoidal category is a comonoid in a *unique* way. (This is a fun exercise.) The coordinate ring of any affine group scheme (over $\operatorname{Spec} A$) will also give examples in the category of $A$-modules.


