Is the category commutative monoids cartesian closed? Is the category commutative monoids cartesian closed?
I understand it breaks down at the point at which the evaluation map cannot be confirmed, however this is not enough to show it is not cartesian closed.
Thanks!
 A: My category theory is a bit rusty, but my copy of Basi Category Theory seems to say that $ (-)\times G $ functor does not preserve the initial object, it cannot have a right adjoint. Actually, it says this for Grp and Abgp, but then proceeds to mention that the same argument holds for Mon (and I see no reason why this should not hold for commutative monoids).
A: Here's a couple of useful references:
P. Freyd: Algebra valued functors in general and tensor products in particular (MR0195920)
This paper shows that a commutative variety of algebras (such as commutative monoids) is closed monoidal.  This is the construction that Peter mentions in his answer.
I don't have access to MathSciNet right now, but if you go to the MSN page for the above and click on the "in reviews" button, then you will (I think) get a paper that considers when such an algebraic theory has a cartesian closed structure.  I'm not sure whether or not the condition is merely sufficient, or is an if-and-only-if, but if I remember correctly, the condition had to do with diagonals.  That is, you take an arbitrary operation and feed the same thing in to every input.  But I'm not entirely sure what the condition was, off the top of my head, except that it was very strong and not satisfied by commutative monoids!
(Apologies for the vagueness of the above; when I get MathSciNet access again I'll fill in the details)

Update: A long bus ride with little to do provided ample opportunity to fill in the details.  We want an algebraic theory that is cartesian closed.  Let $\mathcal{V}$ be our theory.  Then we need an internal hom, $\underline{\mathcal{V}}(B,C)$, and an adjunction $\mathcal{V}(A,\underline{\mathcal{V}}(B,C)) \cong \mathcal{V}(A \times B,C)$.  Now, having the internal hom is, via the paper of Freyd above, tantamount to the algebraic theory being commutative.  That means that all the operations of the theory commute with each other (so, for example, a binary operation commuting with itself has to satisfy $(ab)(cd) = (ac)(bd)$, more on the nlab page http://ncatlab.org/nlab/show/commutative+theory).  This implies, as Peter says above and Freyd proves in his paper, that $\mathcal{V}$ is closed monoidal, but the monoidal structure is not necessarily the cartesian product.  So we want to look at when it is the cartesian product.  So assume that it is.  Then a $\mathcal{V}$-morphism $A \to \underline{\mathcal{V}}(B,C)$ corresponds to a morphism $A \times B \to C$.
Consider an operation in the algebraic theory, say $\nu$, and suppose that it has arity $n$.  Let $f \colon A \times B \to C$ be a $\mathcal{V}$-morphism.  As this is a morphism, we have that
$$
\nu\big(f(a_1,b_1),...,f(a_n,b_n)\big) = f(\nu\big((a_1,b_1),...,(a_n,b_n)\big)) = f(\nu(a_1,...,a_n),\nu(b_1,...,b_n))
$$
We also insist that the map $a \mapsto (b \mapsto f(a,b))$ is a $\mathcal{V}$-morphism.  Now the $\mathcal{V}$-structure on $\underline{\mathcal{V}}(B,C)$ is given as follows: $\nu$ applied to $g_1,...,g_n$ is the map $b \mapsto \nu(g_1(b),...,g_n(b))$.  That is, it factors as $B \to B^n \to C^n \to C$ where the first map is the diagonal, the second the product of the $g_i$, and the third is $\nu$.  So to say that $a \mapsto (b \mapsto f(a,b))$ is a $\mathcal{V}$-morphism, we mean that it commutes with our typical operation, $\nu$.  Thus
$$
b \mapsto f(\nu(a_1,...,a_n),b)
$$
is the same as
$$
\nu\Big(b \mapsto f(a_1,b), b \mapsto f(a_2,b), ..., b \mapsto f(a_n,b)\Big)
$$
This latter is
$$
b \mapsto \nu(f(a_1,b), ..., f(a_n,b))
$$
But since $f$ itself was a morphism of $\mathcal{V}$-algebras, this simplifies to
$$
b \mapsto f(\nu(a_1, ..., a_n), \nu(b,...,b))
$$
So we conclude that
$$
f(\nu(a_1,...,a_n),\nu(b,...,b)) = f(\nu(a_1,...,a_n),b)
$$
As everything was generic, we conclude that we must have the identity
$$
\nu(b,...,b) = b
$$
for every $\mathcal{V}$-operation.  This is a necessary condition.  I am not sure if it is sufficient (I have a vague recollection that it is not, but still don't have MathSciNet access so can't check).
For commutative monoids, we have two operations: $0$ and $+$.  For addition, we get the condition $b + b = b$ which is pretty strong!  For $0$, we get the even stronger condition that $0 = b$.  So if we take "commutative monoids" and impose identities to try to get a cartesian closed category, we end up with a pretty trivial algebraic theory!
A: Here's a simple observation: the category of commutative monoids has an object that is both initial and terminal (just like the category of groups or abelian groups), so it cannot be cartesian closed. Indeed, if a cartesian closed category $\mathcal{C}$ has the property that $0 \cong 1$, then
$$\mathcal{C}(X, Y) \cong \mathcal{C}(1, Y^X) \cong \mathcal{C}(0, Y^X) \cong 1$$
and so $\mathcal{C}$ is equivalent to the trivial category with one object and one morphism.
A: No, the category of commutative monoids is not cartesian closed, exactly for the reason given by Jan (left adjoints preserve colimits, in particular the initial object). It is however monoidal closed, i.e. there is a "tensor product" $\otimes$ of monoids with the property that
$$Hom(M \otimes N, K)\cong Hom (M, \underline{Hom}(N,K))$$
where $\underline{Hom}(N,K)$ denotes the monoid of monoid homomorphisms with product given  by applying the multiplication of $K$ pointwise.
You can construct $M \otimes N$ as the commutative monoid generated by pairs $(m,n),\  m\in M, n \in N$, and subject to the relations $(m \cdot m',n)=(m,n)\cdot(m',n)$,  $(m,n\cdot n')=(m,n)\cdot (m,n')$ and $(1,n)=(m,1)=(1,1)$. It has the property that monoid homomorphisms $M \otimes N \rightarrow K$ are in bijection to maps $M \times N \rightarrow K$ which are monoid homomorphisms in each variable, i.e. if you fix an element on one side, the function from the other factor to $K$ is a monoid homomorphism.
