A cosmos where coproduct injections are not monic The injections (coprojections) of a coproduct in a category are very often monomorphisms.  For instance, this happens in any extensive category (essentially by definition) and also in any category with zero morphisms (since in that case they are split monos).  However, there are examples of (complete and cocomplete) categories where this is not always the case, e.g. in the category of commutative rings the coproduct is the tensor product, and the injection $\mathbb{Z} \to \mathbb{Z} \otimes \mathbb{Z}/2 \cong \mathbb{Z}/2$ is not monic.
My question is, can you give an example of a complete and cocomplete closed monoidal category (a "Benabou cosmos") in which coproduct injections are not always monic?  (Or, I suppose, a proof that no such example exists, but that would surprise me.)
 A: Let me recall that injectivity of coproduct's injections follows from the distributivity of products over coproducts (rather than full extensivity of coproducts). Since every cartesian closed category is (obviously) distributive we cannot find counterexamples in cartesian closed categories.
However, the proof of injectivity of coproduct's injections highly relies on the cartesian structure of $\times$, and one should not expect to carry it to the context where product $\times$ is substituted by a general tensor $\otimes$.
One class of categories which cannot be cartesian closed (unless degenerated) are self-dual categories. I claim that very many of such categories do not have injective coproduct's injections, and this fact is (almost) unrelated to the existence of any closed monoidal structure. Here is an explicit example.
There are various notions of Chu spaces, but the underlying idea is common --- a Chu space is thought of as a "non-standard relation", and morphisms of Chu spaces are thought of as "adjoint pairs" between relations. Let me describe the category that is usually denoted by $\mathit{Chu}(\mathbf{Set}, \Omega)$. Its objects consist of typed binary relations: $$A = \langle A_!, A^*, A_! \times A^* \overset{A}\rightarrow \Omega \rangle$$ in $\mathbf{Set}$, and its morphisms $A \rightarrow B$ consist of pairs of functions (notice opposite directions!): $$f = \langle f_! \colon A_! \rightarrow B_!,  f^* \colon B^* \rightarrow A^* \rangle$$ in $\mathbf{Set}$ that satisfy the following adjoint-like condition:
$$B(b, f^*(a)) = A(f_!(b), a)$$
One may easily check that this category is self-dual, where the dualization swaps the domain with the codomain of a relation:
$$(A^\bot)_! = A^*$$
$$(A^\bot)^* = A_!$$
$$(A^\bot)(b, a) = A(a, b)$$
and complete (thus, also cocomplete) --- limits are constructed point-wise: the first component of a Chu space inherits limits from $\mathbf{Set}$, and the second from $\mathbf{Set}^{op}$.
Like in many self-dual categories, objects may have many (co)global coelements --- i.e. there are non-trivial morphisms to the initial object $0 = \langle \emptyset, \{ {*} \}, \emptyset \rangle$ (just pick any Chu space $A$ whose $A_! = \emptyset$ and whose $A^*$ is non-trivial). Therefore, the unique morphism $0 \rightarrow 1$ is not mono. So the canonical coproduct's injection $0 \rightarrow 0 \sqcup 1 \approx 1$ is not a monomorphism in $\mathit{Chu}(\mathbf{Set}, \Omega)$.
There is a closed monoidal structure on $\mathit{Chu}(\mathbf{Set}, \Omega)$ given by the following tensor:
$$(A \otimes B)_! = A_! \times B_!$$
$$(A \otimes B)^* = \{\langle h \colon A_! \rightarrow B^*, k \colon B_! \rightarrow A^* \rangle \colon B(b, h(a)) = A(a, k(b))\}$$
$$(A \otimes B)(a, b, h, k) = B(b, h(a)) = A(a, k(b))$$
In fact, $\mathit{Chu}(\mathbf{Set}, \Omega)$ is $\star$-autonomous (with linear implication $A \multimap B = (A \otimes B^\bot)^\bot$).
In particular, the construction is valid for any $\Omega$ (not necessarily the subobject classifier), and taking $\Omega = 1$ yields category $\mathbf{Set} \times \mathbf{Set}^{op}$ with linear exponentiation:
$$\langle A_!, A^*\rangle \multimap \langle B_!, B^*\rangle = \langle {B_!}^{A_!} \times {A^*}^{B^*}, A_! \times B^*\rangle$$
A: I am following up on my comment. 
The category of algebras over a commutative algebraic theory is monoidal closed. This is explained here http://ncatlab.org/nlab/show/commutative+algebraic+theory. 
One can "commutativize" any theory by adding axioms expressing commutativity of any of its two operations (as in the nlab article). So, let us try to find a counterexample among categories of algebraic theories first, and then try to commutativize them.
Here is an example of a theory whose category of algebras may have non-injective coproduct injections: Take a theory which has $n$-ary terms $f, g, l$, $m$-ary terms $h$, $k$, and a binary term $p$, which satisfy
$$f(x_1, ..., x_n) = p(l(x_1, ..., x_n), h(y_1, ..., y_m))$$
$$g(x_1, ..., x_n) = p(l(x_1, ..., x_n), k(y_1, ..., y_m)).$$
Take a free algebra in this theory $F\{a_1, ..., a_n\}$ (shortly $F\{a\}$) on a set $\{a_1, ..., a_n\}$. Take a quotient of a free algebra $F\{b_1, ..., b_m\}/h(b_1, ..., b_m) \sim k(b_1, ..., b_m)$. In the coproduct of these two algebras we have
$$f(a) \sim p(l(a), h(b)) \sim p(l(a), k(b)) \sim g(a).$$
Thus, the coproduct injection from $F\{a\}$ is not injective (unless for some reason f(x) and g(x) equal to each other in the theory itself).
The theory of rings falls under this example by taking $f, g, l, h, k$ all the following zero terms $f = 2, g = 0, l = 1, h = 2, k = 0$, and $p(x, y) = xy$.
If we try to commutativize the theory of rings the constants get identified. In particular $2$ and $0$ get identified in the original theory, and we stay without a counterexample :).
In fact commutativization destroys the counterexample if the theory has any constant. Because, for a constant $c$, we should have $h(c, c, ...) = c = k(c, c, ...)$ by commutativity, and this forces $f(x)$ to equal $g(x)$ in the theory itself. This should be like this too, because the category of algebras of a commutative theory with a constant (necessarily unique) has a zero object, and thus has injective coproduct injections.
However, in other situations commutativization should be a harmless process. For example one can take a theory given by unary operation $f, g, l, h, k$ and a binary operation $p$, with the only axioms the above given two equations. Commutativazation of this theory should give a counterexample to the question. 
