Analogue Bialgeras vs Lie bialgebras I once thought that the analogue of bialgebras and Lie bialegras is similar to that of (associative) algebras and Lie algebras, but it seems not that trivial. 
Recall the definitions: a) bialgebra $A$ is a algebra $A$ with a comultiplication $\delta: A \to A\otimes A$ such that $\delta$ is coassociative and a morphism of algebras. b)Lie bialgebra $\mathfrak{g}$ is a Lie algebra $\mathfrak{g}$ with a cobracket $\Delta:\mathfrak{g} \to \mathfrak{g}\otimes\mathfrak{g}$ such that $\Delta$ subjects to co-Jacobi identity and $\Delta$ is a cocycle, where the action of $\mathfrak{g}$ on $\mathfrak{g}\otimes \mathfrak{g}$ is by adjoints.
Naïvely, we may expect the cobracket $\Delta$ to be a Lie algebra morphism but not a cocycle. Why so? This is the first part of my question, the other part: Is it possible to build a Lie bialgebra out of a bialgebra via alternating? Thanks in advance.
 A: There is of course a much more profound "analogy" between bialgebras and Lie bialgebras: the quantization and dequantization functors of Etingof and Kazhdan. From this point of view, Lie bialgebras appear as first order terms of a deformation theory which ultimately assigns to every Lie bialgebra a bialgebra by deformation (quantization). If I recall correctly, the construction uses a Drinfel'd associator, so it probably works only in characteristic zero or evenover the complex numbers. You can find this with many details in a sequel of papers (4?).
A: Lie bialgebras and bialgebras cannot be analogous.  The tensor product of two Lie algebras is not a Lie algebra.  We can check this, let $L_1$ and $L_2$ be two Lie algebras, so for $a,b,c \in L_1$ and $x,y,z \in L_2$ we have 
$$[a,[b,c]] = [[a,b], c] + (-1)^{|a||b|} [b, [a,c]] $$
$$[x, [y,z]] = [[x,y], z] + (-1)^{|x||y|} [y, [x,z]].$$
We use this version of the Jacobi identity-that $[,]$ is a derivation of itself- to keep track of signs, although it won't be that important.
So we can try to define a Lie bracket on $L_1 \otimes L_2$ by $[a\otimes b , x \otimes y]= [a,b] \otimes [x,y].$  But because the Jacobi identity has three terms (and not two like in the associative identity), we get
$$[a \otimes x, [b \otimes y, c \otimes z]] = [a \otimes x , [b,c] \otimes [y,z]] $$
$$= [a, [b,c] \otimes [x,[y,z]] $$
$$=  \left([[a,b], c] + (-1)^{|a||b|} [b, [a,c]]  \right) \otimes \left( [[x,y], z] + (-1)^{|x||y|} [y, [x,z]] \right).$$
When we expand this out, we get four terms, and not two terms as needed.  So $L \otimes L$ is not a Lie algebra, and we can't ask $\Delta$ to be a Lie algebra map. 
Edit:  There seems to be some related notions.  One could ask that $\Delta: L \rightarrow L \otimes L$ be a map of Lie modules.  The analogous thing with associative algebras is an open (non-commutative) Frobenius algebra.  This is an algebra $A$ with multiplication and comultiplication $\Delta: A \rightarrow A \otimes A$, such that $\Delta$ is a map of $A$ modules.  
One could also ask that $\Delta: L \rightarrow L \otimes L$ be a derivation of $L$ with values in the bimodule $L \otimes L$.  This is the cocycle condition.  The analogous thing for associative algebras is called an infinitesimal bialgebra.  Aguiar discusses these objects in the paper "On the associative analog of Lie bialgebras."  The idea of symmetrizing the associative product of an infinitesimal bialgebra to obtain a Lie bialgebra is discussed (this process does not always yield a Lie bialgebra).  
In "Skein quantization of Poisson algebras of loops on surfaces," Turaev describes a Lie bialgebra.  He also discusses a Hopf algebra which quantizes the Lie bialgebra.  Since a Hopf algebra is a bialgebra with an antipode, maybe there is a connection to be said between Lie bialgebras and bialgebras.  
A: just a short answer:
For associative algebras, you ask $\Delta:A\to A\otimes A$ to be an algebra morphism, then for Lie algebras, you may ask to be a derivation, so that "its exponential" is an algebra morphism. You can easily interpret the 1-cocycle condition as a derivation.
This is of course the tip of the iceberg of Stefan Waldmann's answer.
