The universal enveloping algebra of a Lie algebra has a canonically defined Hopf algebra structure. Is the same true of the universal enveloping of a super Lie algebra? A presentation in terms of the universal properties would be of most interest.



Hopf structure on the universal enveloping of a super Lie algebra
No, it is not true unless the odd part is zero or the characteristic is 2. More precisely, there is no natural Hopf algebra structure outside these conditions, but there may be some odd examples. It is a Hopf superalgebra instead, which is what Peter May means in his answer.


You can see this on the easiest example: $L\_0=0$, $L\_1=F$, the field. Its universal enveloping is $F[x]/(x^2)$. It is a Hopf algebra if and only if the characteristic of $F$ is 2.


In general, you have difficulty with comultiplying the odd element $x\in L\_1$ because $2x^2=[x,x]$. If you try $\Delta (y)=1\otimes y + y \otimes 1$ for $y\in L\_0$ as suggested by Gods, no formula for $\Delta (x)$ will accommodate this relation.


**PS** Note that $C[x]/(x^2)$ is the cohomology $H^\ast (SU\_2, C)$. The cohomology of a compact Lie group is the original example of "a Hopf algebra" studied by Hopf. However, the modern terminology follows Sweedler's 1969 book, which I also follow in my answer and urge everyone to follow as well.


**PPS** There is no difference between associative ($Z\_2$-graded) algebras and associative superalgebras. Why is there a difference between Hopf algebras and superalgebras? It happens because the interaction axiom between multiplication and comultiplication requires braiding. Take any associative superalgebra $A=A\_0\oplus A\_1$. Then $A\otimes^{sup} A = A \otimes A$ as a vector space but their algebra structures are different. Given homogeneous $w,x,y,z$, the products are different:
$$
(w \otimes^{sup} x) \cdot (y \otimes^{sup} z) = (-1)^{|x||y|}wy \otimes^{sup} xz, \ \
(w \otimes x) \cdot (y \otimes z) = wy \otimes xz.
$$
Now the universal enveloping $U=U(L)$ is an associative $Z\_2$-graded algebra or superalgebra. For each $x\in L$, the God-given $\Delta (x) = 1\otimes x + x \otimes 1$ extends to an algebra homomorphism
$$
\Delta: U \rightarrow U\otimes^{sup}U
$$ 
but not to an algebra homomorphism to $U\otimes U$. This is why it is Hopf superalgebra. It has the obvious universal property:


*the unique algebra map $U \rightarrow U\otimes^{sup}U$ given on $L$ by $x\mapsto 1\otimes x + x \otimes 1$.*


