Product in the category of cocommutative coalgebras Let $(C,\Delta,\epsilon)$ and $(C',\Delta',\epsilon')$ be two coalgebras. Consider their tensor product $C\otimes C'$ and the two coalgebra homomorphisms
\begin{align*}
\pi:C\otimes C'\to C, \quad c\otimes c'&\mapsto\epsilon'(c')c,\\
\pi':C\otimes C'\to C', \quad c\otimes c'&\mapsto\epsilon(c)c'.\\
\end{align*}
We are then supposed to show that the tensor product is the product in the category of cocommutative coalgebras. 
I have shown that for any cocommutative coalgebra $(D,\Delta_D,\epsilon_D)$ and morphisms $f:D\to C$, $f':D\to C'$ there exists a morphism
\begin{align*} 
m :D&\to C\otimes C',\\
d&\mapsto f\otimes f' \circ \Delta_D (d) = \sum_{(d)}f(d_{(1)})\otimes f'(d_{(2)})\,,
\end{align*}
such that $\pi \circ m = f$ and $\pi'\circ m = f'$. But I can't prove that it's unique. Let $$\widehat{m}:D \to C\otimes C'$$ be another morphism with $\pi\circ\widehat{m} = f$ and $\pi'\circ\widehat{m} = f'$. A priori, $$ \widehat{m} (d) = \sum_i c_i\otimes c_i',$$
s.t. $f(d) = \sum_{i}\epsilon'(c_i') c_i = \sum_{(d)}\epsilon_D(d_{(2)}) f(d_{(1)})$ and similarly for $f'$.
How do we arrive at $m = \widehat{m}$?

Edit: Is there something wrong with the following calculation?
Let $$ \widehat{m}(d) = \sum_i \widehat{d}_i\otimes \widehat{d'}_i.$$ Then we compute
\begin{align*}
m(d) &= \sum_{(d)}f(d_{(1)})\otimes f(d_{(2)}) \\
&= \sum_{(d)}(\pi\circ\widehat{m})(d_{(1)})\otimes (\pi'\circ\widehat{m})f(d_{(2)})  \\
&= \sum_{(d),i,j} \left( \epsilon'(\widehat{(d_{(1)})'}_i) \widehat{(d_{(1)})}_i \right)
\otimes
\left( \epsilon(\widehat{(d_{(2)})}_j) \widehat{(d_{(2)})'}_j \right) 
\end{align*}
On the other hand, using counitality and the fact that $\widehat{m}$ is a homomorphism, we obtain
\begin{align*}
\widehat{m}(d) = \sum_{(d)} \left( \epsilon\otimes \epsilon' \right)\left( \widehat{m}(d_{(2)}) \right)\cdot \widehat{m}(d_{(1)}).
\end{align*}
This doesn't show anything, in fact it even looks like it would go wrong.
 A: I was just looking in the wrong direction - it was actually fairly easy. Here is the solution. We use the following notation:
$$ \widehat{m}(d) = \sum_i \widehat{d\phantom{'}}_i\otimes \widehat{d'}_i. $$
With this we compute 
\begin{align*}
m(d) &= \sum_{(d)}f(d_{(1)})\otimes f(d_{(2)}) \\
&= \sum_{(d)}(\pi\circ\widehat{m})(d_{(1)})\otimes (\pi'\circ\widehat{m})f(d_{(2)})  \\
&= \sum_{(d),i,j} \left( \epsilon'(\widehat{(d_{(1)})'}_i) \widehat{(d_{(1)})}_i \right)
\otimes
\left( \epsilon(\widehat{(d_{(2)})}_j) \widehat{(d_{(2)})'}_j \right)  \\
&= 1\otimes \epsilon'\otimes\epsilon \otimes 1 \left( 
\sum_{(d),i,j}
 \widehat{(d_{(1)})}_i 
 \otimes
 \widehat{(d_{(1)})'}_i
 \otimes
 \widehat{(d_{(2)})}_j
 \otimes 
 \widehat{(d_{(2)})'}_j
 \right)\\
&=(1\otimes \epsilon'\otimes\epsilon \otimes 1) \circ (\widehat{m}\otimes \widehat{m})\circ \Delta_D (d) \\
&\stackrel{(*)}{=}(1\otimes \epsilon\otimes\epsilon' \otimes 1) \circ (\Delta\otimes\Delta')\circ \widehat{m}(d) \\
&= \widehat{m}(d)\,,
\end{align*}
hence $m = \widehat{m}$, as desired. In step $(*)$ we used the fact that $\widehat{m}$ is a homomorphism to flip the epsilons, the last equality is then simply the counitality of both $C$ and $C'$.
