Let $X = \mathbb{C} \setminus \{0\}$ and let $Y$ denote the complex parabola $$Y = \{ (x,y) \mid y^2 = x \}.$$ Let $\pi : Y \to X$ be the projection onto the first factor. Consider the presheaf on $X$ given by $$\mathcal{G}(U) = \{f : U \to Y \mid f \text{ is holomorphic, and } \pi \circ f = \operatorname{id}_U \}.$$ Show that $\mathcal{G}$ is a sheaf of sets.
I'm stuck with showing that this satisfies the gluability axiom. If we take $U$ open in $X$ and a cover $\{U_i\}_i$ for it such that for $s,t \in \mathcal{G}(U)$ it holds that $s|_{U_i} = t|_{U_i}$ how can I show that $s = t$? We know that $s$ and $t$ are holomorphic maps with $\pi \circ s$ and $\pi \circ t$ identities on $U$, but I don't see how to use this to conclude that the maps are equal?