Lie Algebras and Simple Connectivity for general algebraic groups In the representation theory of Lie groups (say, over $\mathbb{R}$ or $\mathbb{C}$), one can show that a Lie algebra homomorphism between the Lie algebras of two algebraic groups $G$ and $H$ always results in a Lie group homomorphism if $G$ is simply connected.
It seems like one might be able to make this argument over a general field. In place of fundamental group, we could ask that the etale fundamental group of $G$ be trivial. Does this allow us to show that a homomorphism of Lie algebras results in a homomorphism of groups? Then one could prove that the semisimplicity of the algebra and of the group are equivalent using this argument.
 A: No, it does not. The additive $G$ and the multiplicative $H$ groups have isomorphic Lie algebras but only trivial group homomorphisms between them. 
The Lie algebra of $G$ has a restricted structure! You may wonder what happens if there is a homomorphism of restricted Lie algebras. This would uniquely lift to a homomorphism of the first Frobenius kernels but not whole groups. 
The easiest example does not have trivial $\pi_1$: consider a semidirect product of multiplicative and additive groups with the action of the mulitplicative group twisted by Frobenius. It will have an abelian Lie algebra, isomorphic as a restricted Lie algebra to the Lie algebra of the direct product. There is no corresponding homomorphism of groups.
With trivial $\pi_1$, I can think of $SL_2 (K)$ where $K$ is algebarically closed of characteristic 2. The Lie algebra will have a nontrivial homomorphism to the Lie algebra of the additive group $K_a$ but no such homomorphism exists for groups.
A: Just to make explicit BCnrd's comment that the Lie algebra is not-so-great in non-zero characteristic, consider the special linear group $G=\operatorname{SL}_2$ of $2 \times 2$ matrices of det 1 in char. 2, a simply connected semisimple group. The Lie algebra $L = \mathfrak{sl}_2$ contains a 1-dimensional ideal $I$ spanned by the identity matrix, and the quotient algebra $L/I$ is isomorphic to the (restricted) Lie algebra $M = \operatorname{Lie}(\mathbf{G}_a \times \mathbf{G}_a)$.
But the natural quotient mapping $L \to M$ does not result in a non-trivial homomorphism of alg groups $\operatorname{SL}_2 \to \mathbf{G}_a \times \mathbf{G}_a$.
For what it is worth, consider $G = \operatorname{SL}_p$ in char p>0. Again the Lie algebra $L$ of $G$ contains a 1 dimensional ideal $I$ spanned by the identity matrix. 
For $p>2$, I'm not aware that the Lie algebra $L/I$ is the Lie algebra of any algebraic group, though $L/I$ is isomorphic to an invariant subalgebra of 
the Lie algebra of the adjoint group $G_1 = \operatorname{PGL}_p$, and the mapping
$L \to L/I \subset \operatorname{Lie}(G_1)$ is the tangent mapping of the standard (inseparable) isogeny $G \to G_1$.
[oops: just noticed that much of this is redundant w/ last bit of Bugs Bunny's answer...]
