Here is the question I am trying to solve:
(Tensor product of coalgebras) Let $(C, \Delta, \varepsilon)$ and $(C', \Delta ', \varepsilon ')$ be coalgebras. Show that the linear maps $\pi: C \otimes C' \to C$ and $\pi': C \otimes C' \to C'$ defined by $\pi(c \otimes c') = \varepsilon' (c') c$ and $\pi'(c \otimes c') = \varepsilon (c) c'$ are morphisms of coalgebras and that the coalgebra $C \otimes C'$ satisfies the following universal property: for any cocommutative coalgebra $D$ and any pair $f: D \to C$ and $f': D \to C'$ of coalgebra morphisms, there exists a unique morphism of coalgebras $f \otimes f': D \to C \otimes C'$ such that $$\pi \circ (f \otimes f') = f \text{ and } \pi' \circ (f \otimes f') = f'.$$
My attempt:
I was able to prove all the first part. For the second part, the map I used is $g(d) = \sum_{(d)} f(d') \otimes f'(d'')$ (I am using Sweedler notation actually ) and everything went well and I was able to prove the required relation $$\pi \circ (f \otimes f') = f \text{ and } \pi' \circ (f \otimes f') = f'$$(I beleive that this relation is just the universal property of the product i.e., tensor product is the product) But I do not know how to prove that this map is unique. Could someone help me in this please?
Also, I wanted to point out in my solution I did not use the commutativity at all and so I am guessing that it will be used in the proof of uniqueness. If not, can someone point out where will it be used?
It seems you use the cocommutativity of $D$ while proving $g = (f \otimes f' ) \circ \Delta_D$ is a coalgebra morphism. Here $\Delta_D$ is a coalgebra morphism if and only if $D$ is cocommutative, and you may need this.
For the uniqueness part, suppose $g' : D \rightarrow C \otimes C'$ is another coalgebra morphism with the given property. Then \begin{align} g' &= ( \text{id}_C\otimes \text{id}_{C'} ) \circ g' \\ &= (\pi \otimes \pi')\circ \Delta_{C \otimes C'} \circ g' \\ & = (\pi \otimes \pi')\circ (g' \otimes g') \circ \Delta_D \\ & = (f \otimes f ') \circ \Delta_D \\ & = g.\end{align}
Here the identity $\text{id}_C \otimes \text{id}_{C'} = (\pi \otimes \pi') \circ \Delta_{C \otimes C'} $ holds because \begin{align} (\pi \otimes \pi') \circ \Delta_{C \otimes C'}(c \otimes c') &= (\pi \otimes \pi')( c_{(1)} \otimes c'_{(1)} \otimes c_{(2)} \otimes c'_{(2)} ) \\ & = \varepsilon'(c'_{(1)})c_{(1)} \otimes \varepsilon(c_{(2)}) c'_{(2)} \\ &= \varepsilon(c_{(2)}) c_{(1)} \otimes \varepsilon'(c'_{(1)})c'_{(2)} \\ &= c \otimes c'. \end{align} Recall that by the counity axiom we have $(\text{id} \otimes \varepsilon) \circ \Delta = \text{id} = (\varepsilon \otimes \text{id} ) \circ \Delta .$
The equality $\Delta_{C \otimes C'} \circ g' = (g' \otimes g') \circ \Delta_D$ holds since $g'$ is assumed to be a morphism of coalgebras.
Finally, we can derive the fourth equality in the line of the proof for $g' = g$ as follows: \begin{align} (\pi \otimes \pi') \circ (g' \otimes g') &= (\pi \otimes \text{id}) \circ (\text{id} \otimes \text{id} \otimes \pi') \circ (\text{id} \otimes \text{id} \otimes g') \circ (g' \otimes \text{id})\\ & = (\pi \otimes \text{id}) \circ (\text{id} \otimes \text{id} \otimes f') \circ (g' \otimes \text{id})\\ & = (\pi \otimes \text{id}) \circ (g' \otimes \text{id}) \circ ( \text{id} \otimes f') \\ & = (f \otimes \text{id}) \circ ( \text{id} \otimes f') \\ & = f \otimes f' . \end{align}
Here I mean $\Delta_{D \otimes D}$ by the comultiplication of $D \otimes D$. For two coalgebras $(A, \Delta_A, \varepsilon_A)$ and $(B, \Delta_B, \varepsilon_B)$, the tensor product $A \otimes B$ has a coalgebra structure $(A \otimes B, \Delta_{A \otimes B}, \varepsilon_{A \otimes B})$ given by $\Delta_{A \otimes B}(a \otimes b) = a_{(1)}\otimes b_{(1)} \otimes a_{(2)} \otimes b_{(2)}$ and $ \varepsilon_{A \otimes B} ( a \otimes b) = \varepsilon_{A}(a) \varepsilon_B (b)$. It is often written that $\Delta_{A \otimes B} = (\text{id} \otimes \tau \otimes \text{id}) \circ (\Delta_A \otimes \Delta_B)$, where $\tau : A \otimes B \rightarrow B \otimes A$ is the flip map sending each $a \otimes b$ to the flipped item $b \otimes a$.
This is because $g = (f \otimes f' ) \circ \Delta_D$ is the composition of two coalgebra morphisms. First you can verify that the composition of two coalgebra morphisms are also a coalgebra morphism. Next, given two coalgebra morphisms $h : A \rightarrow B$ and $h' : A' \rightarrow B'$ , the map $h \otimes h' : A \otimes A' \rightarrow B \otimes B'$ is a coalgebra morphism. It left to show $ \Delta_D : D \rightarrow D \otimes D $ is a coalgebra morphism, under the cocommutativity assumption on $D$. According to the definition of coalgebra morphisms, you should prove $\Delta_{D \otimes D } \circ \Delta_D = (\Delta_D \otimes \Delta_D) \circ \Delta_D $. Let us simply write $\Delta = \Delta_D$. After writing $(\Delta \otimes \Delta ) \circ \Delta = (\text{id} \otimes \text{id} \otimes \Delta) \circ (\Delta \otimes \text{id} ) \circ \Delta $, we can apply the coassociativity law $(\Delta \otimes \text{id} ) \circ \Delta = (\text{id} \otimes \Delta) \circ \Delta $ to obtain the following calculations:
\begin{align} (\Delta \otimes \Delta ) \circ \Delta & = (\text{id} \otimes \text{id} \otimes \Delta) \circ \color{red}{(\Delta \otimes \text{id} ) \circ \Delta} \\ & = (\text{id} \otimes \color{blue}{\text{id} \otimes \Delta} ) \circ \color{red}{(\text{id} \otimes \Delta ) \circ \Delta} \\ & = (\text{id} \otimes \color{blue}{\Delta \otimes \text{id} } ) \circ (\text{id} \otimes \color{blue}{\Delta} ) \circ \Delta. \end{align} Since $D$ is cocommutative we know $\Delta = \tau \circ \Delta$ where $\tau : D\otimes D \rightarrow D \otimes D$ is the flip map. Therefore,
\begin{align} (\Delta \otimes \Delta ) \circ \Delta & = (\text{id} \otimes \color{blue}{\Delta} \otimes \text{id} ) \circ (\text{id} \otimes \Delta ) \circ \Delta. \\ & = (\text{id} \otimes \color{blue}{\tau} \otimes \text{id}) \circ (\text{id} \otimes \color{blue}{\Delta} \otimes \text{id} ) \circ (\text{id} \otimes \Delta ) \circ \Delta \\ &= (\text{id} \otimes \tau \otimes \text{id}) \circ (\Delta \otimes \Delta ) \circ \Delta \\ & = \Delta_{D \otimes D} \circ \Delta ,\end{align} because $\Delta_{D \otimes D} = (\text{id} \otimes \tau \otimes \text{id}) \circ (\Delta \otimes \Delta ) $. As you can prove $\varepsilon_{D \otimes D} \circ \Delta_D = \varepsilon_D$ (without the cocommutativity assumption), $\Delta_D$ turns out to be a morphism of coalgebras.