Symmetric monoidal category which is not closed? A monoidal category is symmetric if its tensor product is commutative up to natural isomorphism.  And a symmetric monoidal category is closed if the tensor product functor has a right adjoint.  We call this right adjoint the internal Hom functor.
My question is, what is an example of a symmetric monoidal category which is not closed?
 A: A simple example is vector spaces, or abelian groups, but with the direct sum as the monoidal product, instead of the tensor. If there existed a vector space $B^A$ with maps $C\to B^A$ in natural bijection with maps $A\oplus C\to B$, then we could write, for every $A,B,C,D$: 
$$\mathrm{Hom}(A\oplus B\oplus C, D)\cong \mathrm{Hom}(B\oplus C,D^A)\cong \mathrm{Hom}(B,D^A)\times \mathrm{Hom}(C,D^A)\cong \mathrm{Hom}(A\oplus B,D)\times \mathrm{Hom}(A\oplus C,D)\cong \mathrm{Hom}(A\oplus B\oplus A\oplus C,D),$$
which is easily seen to be absurd (for instance, set $B=C=0$.) This is an example of asdq's proposal: a closed monoidal product distributes over coproducts by the argument above, but direct sum certainly does not distribute over itself. 
A: You actually don't need to know anything about limits or colimits to answer this question. Any commutative monoid $M$ gives rise to a symmetric monoidal category whose underlying category is discrete, meaning it has no non-identity morphisms; the set of objects is $M$ and the monoidal operation is the operation in $M$. This symmetric monoidal category is closed iff $M$ has inverses (meaning it is actually an abelian group), in which case the internal hom $[x, y]$ is just $x^{-1} y$ (or $y - x$ in additive notation).
