Universal enveloping algebra as bialgebra

Question

If $\mathfrak{g}$ is a Lie algebra (over $k$ ), then we can construct its universal enveloping algebra $U(\mathfrak{g})$. We can define $\Delta:U(\mathfrak{g})\rightarrow U(\mathfrak{g})\otimes U(\mathfrak{g}) $ as a natural extension of $\mathfrak{g}\ni x\mapsto 1\otimes x + x\otimes 1\in U(\mathfrak{g})\otimes U(\mathfrak{g})$. We can also define $\varepsilon:U(\mathfrak{g})\rightarrow k$ by $\varepsilon(x)=0, \mathfrak{g}\ni x\neq 1, \ \varepsilon(1)=1$. There is theorem that $U(\mathfrak{g})$ with $\Delta,\varepsilon$ form a bialgebra. $\Delta(1)$ is from definition equal to $1\otimes 1 + 1\otimes 1$, but if $U(\mathfrak{g})$ forms a bialgebra it should be equal $1\otimes 1$, because in bialgebra unit element should be group-like. Where is the mistake ?

Answer 1

The mistake is that the definition you give only works for $x\in\mathfrak{g}$. A general element is not of this form. The coproduct will be the unital algebra homomorphism having this form for each element of the Lie algebra, which is the generating set, and the coproduct of a general element is a sum of products of these.