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Let $\mathcal{V}$ a symmetric monoidal category. A commutative comonoid is a triple $(A, \delta, \epsilon)$ whit $\delta: A \to A\otimes A$, $\epsilon: A\to I$ by the dual of monoid axioms. Given two commutative comonoids $(A_k, \delta_k, \epsilon_k)\ k=1, 2$ its monoidal product $(A_1\otimes A_2)$ ha a natural structure of a comonoid $(A_1\otimes A_2, \delta, \epsilon)$. We have the projection $\pi_1: A_1\otimes A_2\xrightarrow{1\otimes \epsilon_2} A_1\otimes I \cong A_1 $, $\pi_2: A_1\otimes A_2\xrightarrow{\epsilon_1\otimes 1} I\otimes A_2 \cong A_2 $. I ask if this could be a product (the limit on two-elements discrete set) on the category of commutative comonoids of $\mathcal{V}$.

EDIT: I specify that I mean commutative comonoids

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This part of the answer refers to the original question. By duality, you ask if the tensor product of two monoids (with the usual monoid structure, which can be remembered by $(a \otimes b)(a' \otimes b') := aa' \otimes bb'$) is the coproduct of the two monoids. This is well-known to be false. If $A_1 \to M$ and $A_2 \to M$ are monoid homomorphisms, we get a map $A_1 \otimes A_2 \to M$, but this doesn't have to be a homomorphims of monoids unless $M$ is commutative. This also shows that everything works out fine if we work with commutative monoids (or dually cocommutative comonoids).

Proof for the commutative case. One checks that $A_1 \otimes A_2$ with the multiplication $(A_1 \otimes A_2) \otimes (A_1 \otimes A_2) \xrightarrow{S} (A_1 \otimes A_1) \otimes (A_2 \otimes A_2) \to A_1 \otimes A_2$ (where $S$ stands for the obvious symmetry) is a commutative monoid, I won't spell that out. The inclusions are given by $\iota_1 : A_1 \cong A_1 \otimes 1 \to A_1 \otimes A_2$ (induced by the unit $1 \to A_2$) and $\iota_2 : A_2 \cong 1 \otimes A_2 \to A_1 \otimes A_2$; they are easily seen to be homomorphisms. Let $f_1 : A_1 \to M$ and $f_2 : A_2 \to M$ be homomorphisms of commutative monoids. Define $f : A_1 \otimes A_2 \to M$ as the composition of $f_1 \otimes f_2 : A_1 \otimes A_2 \to M \otimes M$ with the multiplication map $M \otimes M \to M$. Then $f$ is a homomorphism: It is unital because

$$\begin{array}{c} 1 & \rightarrow & 1 \otimes 1 & \rightarrow & 1\\ \downarrow & & \downarrow & & \downarrow \\ A_1 \otimes A_2 & \rightarrow & M \otimes M & \rightarrow & M \end{array}$$

commutes. It is multiplicative because

$$\begin{array}{c} (A_1 \otimes A_2) \otimes (A_1 \otimes A_2) & \rightarrow & (M \otimes M) \otimes (M \otimes M) & \rightarrow & M \otimes M \\ ~~ \downarrow S & & ~ \downarrow S && \\ (A_1 \otimes A_1) \otimes (A_2 \otimes A_2) & \rightarrow & (M \otimes M) \otimes (M \otimes M) && \downarrow\\ \downarrow & & \downarrow & & \\ A_1 \otimes A_2 & \rightarrow & M \otimes M & \rightarrow & M \end{array}$$

commutes (the right square because $M$ is commutative, the left squares by naturality of the symmetry and because of functoriality of $\otimes$). We have $f \circ \iota_1 = f_1$, since $f_2$ is unital:

$$\begin{array}{c} A_1 & \rightarrow & A_1 \otimes 1 & \rightarrow & M \otimes 1 & \\ \downarrow & & \downarrow && \downarrow & \searrow \\ A_1 \otimes A_2 & \rightarrow & A_1 \otimes M & \rightarrow & M \otimes M & \rightarrow & M \end{array}$$

Now let $g : A_1 \otimes A_2 \to M$ be a homomorphism of monoids with $g \circ \iota_1 = f_1$ and $g \circ \iota_2 = f_2$. We claim $g=f$. This follows from the following commutative diagram:

$$\begin{array}{c} A_1 \otimes A_2 & \rightarrow & (A_1 \otimes 1) \otimes (1 \otimes A_2) & \rightarrow & (A_1 \otimes A_2) \otimes (A_1 \otimes A_2) & \rightarrow & M \otimes M \\ && ~~~~ \downarrow S=\mathrm{id} & & ~~ \downarrow S & & \\ \parallel && (A_1 \otimes 1) \otimes (1 \otimes A_2) & \rightarrow & (A_1 \otimes A_1) \otimes (A_2 \otimes A_2) & & \downarrow \\ & \nearrow & & & \downarrow & & \\ A_1 \otimes A_2 && = & & A_1 \otimes A_2 & \rightarrow & M \end{array}$$

Namely, this shows that $g$ is the composition of $(g \circ \iota_1) \otimes (g \circ \iota_2)$ with the multiplication of $M$.

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Sure, I forgot to specify that the comonoids are commutative. –  Buschi Sergio Aug 14 '13 at 12:05
    
Ok then your question is answered? –  Martin Brandenburg Aug 14 '13 at 12:31
    
Sorry no: given $f_k: A_k \to B$ (comm.monoid morphisms) there is $(f_1, f_2): A_1\otimes A_2 \to M\otimes M\to M$ such that $(f_1, f_2)\circ \epsilon_k= f_k$ (where $\epsilon_1: A_1 \cong A_1\otimes I\to A_1\otimes A_2$ similarly for $\epsilon_2$). My problem is to show the uniqueness: given $g: A_1\otimes A_2 \to M$ such that $g\circ \epsilon_k= f_k$ how to prove that $g=(f_1, f_2)$ ? –  Buschi Sergio Aug 14 '13 at 14:42
    
I've added the proof. Is it clear now? –  Martin Brandenburg Aug 15 '13 at 14:38
    
Thank you very much! –  Buschi Sergio Aug 15 '13 at 17:39
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