Showing that $U(\mathfrak{sl}_2)$ is a coalgebra

Question

We know that there is a coalgebra structure on $U(\mathfrak{sl}_2)$ as follows for any $z\in \mathfrak{sl}_2$: $$\Delta(z)=1\otimes z+z\otimes 1, \qquad \epsilon(z)=0.$$

Can someone be so kind to explain to me that why $\Delta$ is an algebra map?

Shouldn't we have $\Delta(zz´)=\Delta(z)\Delta(z´)$, for $z,z´\in \mathfrak{sl}_2$?

As we have $\Delta(z)\Delta(z´)=1\otimes zz´+z´\otimes z+z\otimes z´+zz´\otimes 1$ which as I understand is not equal to $\Delta(zz´)$!

Thanks!

Answer 1

This is equal to $\Delta(z z')$ by definition. Remember that $z z'$ is not an element of $\mathfrak{sl}_2$. This definition of the comultiplication is hiding an extra step: it is defining the comultiplication on $\mathfrak{sl}_2$ only, and from here the comultiplication extends uniquely to $U(\mathfrak{sl}_2)$ precisely because it must be an algebra homomorphism, so must satisfy $\Delta(z z') = \Delta(z) \Delta(z')$.