How algebraic is the holonomy map? Let $G$ be either a compact, simple, simply-connected Lie group, or a simply-connected complex reductive group (so either $SU(n)$ or $SL(n,\mathbb{C})$, for instance), or even complex affine. I don't know the correct setting for this question, so assume the case that makes it work. Please bear with the tentativeness of this question.
Consider the space $\mathcal{A}$ of polynomial or Laurent connections (again, take the option that works or makes sense) on the trivial $G$-bundle on $\mathbb{C}^\times$. This is something like an ind-scheme (perhaps only in the reductive case, and possibly only after picking a faithful representation). In the case of smooth connections on $G\times S^1$ (and probably for analytic too) we have a holonomy map $\mathcal{A}^{sm} \to G$, and this is a surjective submersion. I think there is also such a map for the case of $\mathcal{A}$, going around $S^1\subset \mathbb{C}^\times$. 

In the event this exists, how algebraic is this map? Is it surjective?

My thought is that this may be a cover in some generalised sense. Maps from ind-schemes to schemes aren't so common, though that I know how to find how they work or what they do. There are some notes by Gaitsgory that don't help me at all.
 A: Connections on the trivial $G$-bundle can be identified with maps from the base to $\mathrm{Lie}(G)$ via $x\frac{d}{dx} + f(x) \leftrightarrow f(x)$.  One way to present this as an ind-scheme is by choosing generators of $\mathcal{O}_{\mathrm{Lie}(G)}$, and considering the finite dimensional affine space of ring maps $\mathcal{O}_{\mathrm{Lie}(G)} \to \mathbb{C}[x,x^{-1}]$ for which the generators land in $\bigoplus_{n=-N}^N \mathbb{C} x^n$.
Following Ben Webster's example in the comments, consider the tangent field $x \frac{d}{dx} - a$ for $a \in \mathrm{Lie}(G)$.  This connection lies in the very first affine space of our sequence, with $N=0$.  If $a$ is semisimple, then solutions to the equation $x\frac{df}{dx} = af$ look like "$cx^a$".  The holonomy with respect to the basepoint $1$ is exponential in $a$ - in the one-dimensional case, identifying $\mathrm{Lie}(G)$ with $\mathbb{C}$, we get $e^{2 \pi i a}$.
In general, this example gives us an exponential map from $\mathrm{Lie}(G)$ to $G$.  This (and more generally, the holonomy map) exists in the category of (ind) complex analytic spaces, but outside the case of unipotent groups it is not algebraic.
