TITLE: Maps to the universal punctured elliptic curve
QUESTION [5 upvotes]: I have just started reading Hain's paper On the Universal Elliptic KZB Connection. I am a bit confused about a comment made there about base points on orbifolds. I am still very new to the idea of orbifolds so I apologise in advance for this likely being very trivial. 
Let $\mathcal{E}\to\mathcal{M}_{1,1}$ denote the universal elliptic curve over the moduli space of genus 1 curves with 1 marked point (over the complex numbers). Following Hain's Lectures on moduli spaces of elliptic curves, I am thinking about these as Riemann surfaces in the category of orbifolds, although they are also treated as stacks in the present paper. Let $\mathcal{E}'$ denote $\mathcal{E}$ with its identity section removed (i.e. the "universal punctured elliptic curve"). On page 5 Hain writes:

A non-zero point $x$ on an elliptic curve $E$ determines (and is
  determined by) an orbifold map $[E,x]: \mathbb{C}\to\mathcal{E}'$.

I don't really understand why this is. The picture in my head is that a point of $\mathcal{E}'$ is (roughly) a nonzero point $x$ on some elliptic curve $E$. Therefore constant maps $\mathbb{C}\to \mathcal{E}'$ correspond to pairs $[E,x]$ as above. But I don't understand why other possibly nonconstant maps $\mathbb{C}\to\mathcal{E}'$ should also correspond to such a pair $[E,x]$. Or are all such maps constant?

REPLY [6 votes]: It is indeed true that a holomorphic map $\mathbb C \to \mathcal E'$ is constant. This is because it must factor through the universal cover of $\mathcal E'$, since $\mathbb C$ is simply connected. But the universal cover is a product of two copies of the complex unit disk, so the result follows by the maximum principle.
Nevertheless I don't think that this is what he meant. My guess is that he accidentally left out the word "Spec", so that he meant to say that maps $\mathrm{Spec}\,\mathbb C \to \mathcal E'$ are canonically in bijection with pairs $(E,x)$ of an elliptic curve and a non-identity point.