Computation 1. The co-Poisson homomorphism condition gives us that $$\begin{array}{rcl}\delta \left(xy\right)& =& \left(\delta \circ m\right)\left(x\otimes y\right)\\ & =& \left(m\otimes m\right){\delta }^{\otimes }\left(x\otimes y\right)\\ & =& \left(m\otimes m\right)\left(id\otimes \sigma \otimes id\right)\left(\Delta \left(x\right)\otimes \delta \left(y\right)+\delta \left(x\right)\otimes \Delta \left(y\right)\right)\\ & =& \Delta \left(x\right)\delta \left(y\right)+\delta \left(x\right)\Delta \left(y\right)\\ & =& \sum x_{\left(1\right)}{y}^{\left(1\right)}\otimes x_{\left(2\right)}{y}^{\left(2\right)}+x^{\left(1\right)}{y}_{\left(1\right)}\otimes x^{\left(2\right)}{y}_{\left(2\right)}\end{array}$$

Computation 2. Now let us analyse the condition that $\delta$ is a co-Poisson map. $$\begin{array}{rcl}\left(id\otimes \Delta \right)\delta \left(x\right)& =& \left(\delta \circ m\right)\left(x\otimes y\right)\\ & =& \sum x^{\left(1\right)}\otimes {\left(x^{\left(2\right)}\right)}_{\left(1\right)}\otimes {\left(x^{\left(2\right)}\right)}_{\left(2\right)}.\end{array}$$ On the other hand this must also be equal to $$\begin{array}{rcl}\left({\delta }^{\left(12\right)}+{\delta }^{\left(13\right)}\right)\circ \Delta \left(x\right)& =& \left({\delta }^{\left(12\right)}+{\delta }^{\left(13\right)}\right)\sum x_{\left(1\right)}\otimes x_{\left(2\right)}\\ & =& \sum {\left(x_{\left(1\right)}\right)}^{\left(1\right)}\otimes {\left(x_{\left(1\right)}\right)}^{\left(2\right)}\otimes x_{\left(2\right)}+{\left(x_{\left(2\right)}\right)}^{\left(1\right)}\otimes x_{\left(1\right)}\otimes {\left(x_{\left(2\right)}\right)}^{\left(2\right)}.\end{array}$$

Computation 3. The following fact will also be useful: $$\begin{array}{rcl}\Delta \left(xy\right)& =& \left(\Delta \circ m\right)\left(x\otimes y\right)\\ & =& m^{\otimes }\left(\Delta \otimes \Delta \right)\left(x\otimes y\right)\\ & =& \left(m\otimes m\right){\Delta }^{\otimes }\left(x\otimes y\right).\end{array}$$

$\Rightarrow$: Suppose that $\left(𝔤,\phi \right)$ is a Lie bialgebra. We must show that there is a unique extension $\phi :𝔤\to 𝔤\otimes 𝔤$ to a map $\delta :𝔘𝔤\to 𝔘𝔤\otimes 𝔘𝔤$ such that

1. $\delta \circ m=\left(m\otimes m\right)\circ {\delta }^{\otimes }$,
1. $\left(id\otimes \Delta \right)\circ \delta =\left({\delta }^{12}+{\delta }^{13}\right)\circ \Delta$,
3. $\delta$ is a Lie cobracket on $𝔘𝔤$.

2) Let $x,y=𝔘𝔤$ and let us assume that we know that $$\begin{array}{l}\left(id\otimes \Delta \right)\delta \left(x\right)=\left({\delta }^{12}+{\delta }^{13}\right)\Delta \left(x\right),\text{and}\\ \left(id\otimes \Delta \right)\delta \left(y\right)=\left({\delta }^{12}+{\delta }^{13}\right)\Delta \left(y\right),\end{array}$$ Consider the following $$\begin{array}{ccc}\left(id\otimes \Delta \right)\Delta \left(xy\right)& =& \left(id\otimes \Delta \right)\left(m\otimes m\right){\delta }^{\otimes }\left(x\otimes y\right)\\ & =& \left(m\otimes \Delta \circ m\right){\delta }^{\otimes }\left(xy\right)\\ & =& \left(m\otimes \left(m\otimes m\right){\Delta }^{\otimes }\right){\delta }^{\otimes }\left(x\otimes y\right)\\ & =& \left(m\otimes m\otimes m\right)\left(id\otimes {\Delta }^{\otimes }\right){\delta }^{\otimes }\left(x\otimes y\right).\end{array}$$ On the other hand we have that $$\begin{array}{rcl}\left({\delta }^{12}+{\delta }^{13}\right)\Delta \left(xy\right)& =& \left(\delta \otimes id+\left(\sigma \otimes id\right)\left(id\otimes \delta \right)\right)\Delta \circ m\left(x\otimes y\right)\\ & =& \left(\delta \otimes id+\left(\sigma \otimes id\right)\left(id\otimes \delta \right)\right)\left(m\otimes m\right){\Delta }^{\otimes }\left(x\otimes y\right)\\ & =& \left(\left(\delta \circ m\right)\otimes m+\left(\sigma \otimes id\right)\left(m\otimes \left(\delta \circ m\right)\right)\right){\Delta }^{\otimes }\left(x\otimes y\right)\\ & =& \left(\left(\left(m\otimes m\right)\circ {\delta }^{\otimes }\right)\otimes m+\left(\sigma \otimes id\right)\left(m\otimes \left(\left(m\otimes m\right)\circ {\delta }^{\otimes }\right)\right)\right){\Delta }^{\otimes }\left(x\otimes y\right)\\ & =& \left(m\otimes m\otimes m\right)\left({\delta }^{\otimes }\otimes {id}^{\otimes }+\left({\sigma }^{\otimes }\otimes {id}^{\otimes }\right)\left({id}^{\otimes }\circ {\delta }^{\otimes }\right)\right){\Delta }^{\otimes }\left(x\otimes y\right)\\ & =& \left(m\otimes m\otimes m\right)\left({\left({\delta }^{\otimes }\right)}^{12}+{\left({\delta }^{\otimes }\right)}^{13}\right){\Delta }^{\otimes }\left(x\otimes y\right).\end{array}$$ Thus we need only show that $$\left(id\otimes {\Delta }^{\otimes }\right){\delta }^{\otimes }\left(x\otimes y\right)=\left({\left({\delta }^{\otimes }\right)}^{12}+{\left({\delta }^{\otimes }\right)}^{13}\right){\Delta }^{\otimes }\left(x\otimes y\right).$$ The computation $$\begin{array}{rcl}\left\{a\otimes b,\left(c\otimes d\right)\left(e\otimes f\right)\right\}& =& \left\{a\otimes b,ce\otimes df\right\}\\ & =& ace\otimes \left\{b,df\right\}+\left\{a,ce\right\}\otimes bdf\\ & =& ace\otimes \left\{b,d\right\}f+ace\otimes d\left\{b,f\right\}+\left\{a,c\right\}e\otimes bdf+c\left\{a,e\right\}\otimes bdf\\ & =& \left(ac\otimes \left\{b,d\right\}\right)\left(e\otimes f\right)+\left(c\otimes d\right)\left(ae\otimes \left\{b,f\right\}\right)\\ & & +\left(\left\{a,c\right\}\otimes bd\right)\left(e\otimes f\right)+\left(c\otimes d\right)\left(\left\{a,e\right\}\otimes bf\right)\end{array}$$ shows that ${\left\{,\right\}}^{\otimes }$ is a Poisson bracket. Dualizing this computation will show that ${\delta }^{\otimes }\left(x\otimes y\right)$ satisifies the appropriate condition if $\delta \left(x\right)$ and $\delta \left(y\right)$ do.

3) It remains to show that $\delta$ is a Lie cobracket. Since ${\phi }^{\ast }:{𝔤}^{\ast }\otimes {𝔤}^{\ast }\to {𝔤}^{\ast }$ is a Lie bracket on ${𝔤}^{\ast }$, the skew symmetry condition implies that of $\phi \left(𝔤\right)\subseteq 𝔤\wedge 𝔤$. Now assume that $x,y\in 𝔘𝔤$ and that we know that $$\begin{array}{c}\delta \left(x\right)=\sum _x{x}^{\left(1\right)}\otimes x^{\left(2\right)}=-\sum _x{x}^{\left(2\right)}\otimes x^{\left(1\right)}\in 𝔘𝔤\otimes 𝔘𝔤,\\ \delta \left(y\right)=\sum _y{y}^{\left(1\right)}\otimes y^{\left(2\right)}=-\sum _y{y}^{\left(2\right)}\otimes y^{\left(1\right)}\in 𝔘𝔤\otimes 𝔘𝔤.\end{array}$$ Since $𝔘𝔤$ is cocommutative it follows that $$\begin{array}{c}\delta \left(x\right)=\sum _x{x}_{\left(1\right)}\otimes x_{\left(2\right)}=\sum _x{x}_{\left(2\right)}\otimes x_{\left(1\right)},\\ \delta \left(x\right)=\sum _y{y}_{\left(1\right)}\otimes y_{\left(2\right)}=\sum _y{y}_{\left(2\right)}\otimes y_{\left(1\right)}.\end{array}$$ Now it follows from computation 1 that $\delta \left(xy\right)\in 𝔘𝔤\otimes 𝔘𝔤$ satisfies the skew-symmetry condition. We must check that $\delta$ satisfies the co-Jacobi identity. Let $x,y\in 𝔘𝔤$ and let us assume that we know that $$\begin{array}{l}\left(1+\zeta +{\zeta }^2\right)\left(\delta \otimes id\right)\delta \left(x\right)=0,\text{and}\\ \left(1+\zeta +{\zeta }^2\right)\left(\delta \otimes id\right)\delta \left(y\right)=0.\end{array}$$ We need to show that $\left(1+\zeta +{\zeta }^2\right)\left(\delta \otimes id\right)\delta \left(xy\right)=0$. It follows from $$\begin{array}{rcl}\left(\delta \otimes id\right)\delta \circ m& =& \left(\delta \otimes id\right)\left(m\otimes m\right){\delta }^{\otimes }\\ & =& \left(\left(\delta \circ m\right)\otimes m\right){\delta }^{\otimes }\\ & =& \left(m\otimes m\otimes m\right)\left({\delta }^{\otimes }\otimes id\otimes {\delta }^{\otimes }\right){\delta }^{\otimes }\\ & =& \left(m\otimes m\otimes m\right)\left({\delta }^{\otimes }\otimes {id}^{\otimes }\right){\delta }^{\otimes },\end{array}$$ that $$\left(id+\zeta +{\zeta }^2\right)\delta \circ m=\left(m\otimes m\otimes m\right)\left({id}^{\otimes }+{\zeta }^{\otimes }+{\left({\zeta }^{\otimes }\right)}^2\right)\left({\delta }^{\otimes }\otimes {id}^{\otimes }\right){\delta }^{\otimes }$$ Thus it is sufficient to show that ${\delta }^{\otimes }$ satisfies the co-Jacobi identity. This follows (dually) from the fact that ${\left\{,\right\}}^{\otimes }$ satisfies the Jacobi identity.

$\Leftarrow$: Suppose that $𝔘𝔤$ is a co-Poisson Hopf algebra with the usual multiplication and comultiplication. We need to show two things:

1. $\delta \left(𝔤\right)\subseteq 𝔤\otimes 𝔤$
2. $\delta$, restricted to $𝔤$, satisfies the cocycle condition for a bialgebra.

1) First let us show that $\delta \left(1\right)=0$. Since $\Delta \left(1\right)=1\otimes 1$, the co-Poisson homomorphism condition gives that $$\begin{array}{ccc}\delta \left(1\right)& =& \delta \left(1\cdot 1\right)\\ & =& \sum 1^{\left(1\right)}\otimes 1^{\left(2\right)}+1^{\left(1\right)}\otimes 1^{\left(2\right)}\\ & =& 2\delta \left(1\right).\end{array}$$ It follows that $\delta \left(1\right)=0$ if $char\left(k\right)\ne 2$.

Now suppose that $x\in 𝔤$. Then $\Delta \left(x\right)=1\otimes x+x\otimes 1$, and computation 2 becomes $$\begin{array}{ccc}\left({\delta }^{12}+{\delta }^{13}\right)\Delta \left(x\right)& =& \left({\delta }^{12}+{\delta }^{13}\right)\left(x\otimes 1+1\otimes x\right)\\ & =& \sum x^{\left(1\right)}\otimes x^{\left(2\right)}\otimes 1+0+0+x^{\left(1\right)}\otimes 1\otimes x^{\left(2\right)}\\ & =& \left(id\otimes \Delta \right)\delta \left(x\right)\\ & =& \sum x^{\left(1\right)}\otimes \delta \left(x^{\left(2\right)}\right)\end{array}$$ It follows that $x^{\left(2\right)}$ is a primitive element of $𝔘𝔤$ and thus is an element of $𝔤$. The result follows by noting that $\delta \left(x\right)\in 𝔘𝔤\otimes 𝔘𝔤$ since ${\delta }^{\ast }$ must satisfy the skew symmetry condition for a Lie algebra.

2) Suppose that $x,y\in 𝔤$. Then $\Delta \left(x\right)=x\otimes 1+1\otimes x$ and $\Delta \left(y\right)=y\otimes 1+1\otimes y$. It follows from computation 1 above that $$\begin{array}{ccc}\delta \left([x,y]\right)& =& \delta \left(xy-yx\right)\\ & =& \sum x_{\left(1\right)}{y}^{\left(1\right)}\otimes x_{\left(2\right)}{y}_{\left(2\right)}+x^{\left(1\right)}{y}_{\left(1\right)}\otimes x^{\left(2\right)}{y}_{\left(2\right)}-y_{\left(1\right)}{x}^{\left(1\right)}\otimes y_{\left(2\right)}{x}^{\left(2\right)}-y^{\left(1\right)}{x}_{\left(1\right)}\otimes y^{\left(2\right)}{x}_{\left(2\right)}\\ & =& \sum x{y}^{\left(1\right)}\otimes y^{\left(2\right)}+y^{\left(1\right)}\otimes x{y}^{\left(2\right)}+x^{\left(1\right)}y\otimes x^{\left(2\right)}+x^{\left(1\right)}\otimes x^{\left(2\right)}y\\ & & -y{x}^{\left(1\right)}\otimes x^{\left(2\right)}-x^{\left(1\right)}\otimes y{x}^{\left(2\right)}-y^{\left(1\right)}x\otimes y^{\left(2\right)}-y^{\left(1\right)}\otimes y^{\left(2\right)}x\\ & =& \sum [x,y^{\left(1\right)}]\otimes y^{\left(2\right)}+y^{\left(1\right)}\otimes [x,y^{\left(2\right)}]-[y,x^{\left(1\right)}]\otimes x^{\left(2\right)}-x^{\left(1\right)}\otimes [y,x^{\left(2\right)}]\\ & =& \sum \left({ad}^{\otimes 2}x\right)\left(y^{\left(1\right)}\otimes y^{\left(2\right)}\right)-\left({ad}^{\otimes 2}y\right)\left(x^{\left(1\right)}\otimes x^{\left(2\right)}\right)\\ & =& \left({ad}^{\otimes 2}x\right)\delta \left(y\right)-\left({ad}^{\otimes 2}y\right)\delta \left(x\right).\end{array}$$ So $\delta$ satisfies the $1$-cocycle condition.