A coproduct $\varphi: \mathbb{C}\_q[U] \to \mathbb{C}\_q[U] \otimes \mathbb{C}\_q[U]$ is given by: $x \mapsto 1 \otimes x + x \otimes 1$, where $x$ is a generator of $\mathbb{C}\_q[U]$. There is a coproduct $\mathbb{C}[U] \to \mathbb{C}[U] \otimes \mathbb{C}[U]$ which is the pull-back of the multiplication map: $U \times U \to U$. The coproduct $\psi: \mathbb{C}[U] \to \mathbb{C}[U] \otimes \mathbb{C}[U]$ is given by: $x\_{ik} \mapsto \sum\_{j} x\_{ij} \otimes x\_{jk}$.


Are there some relation between the map $\varphi$ and the map $\psi$? Thank you very much.


Edit: $U$ is the group of unipotent upper triangular matrices.



The coproducts $\mathbb{C}_q[U] \to \mathbb{C}_q[U] \otimes \mathbb{C}_q[U]$ and $\mathbb{C}[U] \to \mathbb{C}[U] \otimes \mathbb{C}[U]$
The algebra $ \mathbb{C}[U] $ is a polynomial ring on generators $ x\_{ij} $ where $ i < j $. The coproduct is given (as you say) by $ \psi(x\_{ik})= \sum\_j x\_{ij} \otimes x\_{jk} $ where where interpret $ x\_{ij} = 0 $ if $ i > j $ and $ x\_{ii} = 1 $. In particular, we see that 
$$
\psi(x\_{i i+1})= 1 \otimes x\_{i i+1} + x\_{i i+1} \otimes 1.
$$
Of course, these $ x\_{i i+1} $ do not generate $ \mathbb{C}[U] $, since they are only some of the generators of this polynomial ring. 


However, when we move to the quantization $ \mathbb{C}\_q[U] $, then $ x\_{i i+1} $ are now generators of this non-commutative algebra and as you say $ \phi(x\_{i i+1}) = 1 \otimes x\_{i i+1} + x\_{i i+1} \otimes 1$. So we see that $ \phi $ and $ \psi $ do agree on these elements.


