Is there an algebraic invariant for complex curves that's mapped to injectively? Consider the functor $\pi_1: \text{Closed Surfaces} \rightarrow \textbf{Grp}$. This is homotopy invariant; every closed topological surface has a unique fundamental group. In the reverse direction, by classification, if two closed surfaces have isomorphic fundamental group, then the surfaces themselves are homeomorphic. This fails if we move up to $3$-manifolds, and fails spectacularly for dimension $\geq 4$. So I'll restrict my questions to small dimensions. 
Question: is there a functor $\text{Complex Curves} \rightarrow \textbf{Grp}$ that has the same properties as before? (In this case, it should be conformally invariant, but the preimages of a given group should be conformally equivalent.) I doubt it, if only because there are continuum many complex structures on a torus, and it seems unlikely that something which can be continuously varied like this will give me a nice algebraic invariant that can differentiate these structures. Bonus question: if this is provably false, is it provably false for rings as well? (Feel free to let rings have unity or not and morphisms take unity to unity or not.)
Edit: Zhen Lin noted in the comments below that the Yoneda embedding to PreSheaf works fine. I've restricted now to groups and rings. I want there to be no algebraic invariant in the spirit of, say, homotopy groups or the cohomology ring or the group of line bundles that does the job for us.
 A: With a closed Riemann surface $\Sigma$ of genus $g>0$ there is naturally  associated a pair $(\Lambda, H)$, where $H$ is a $g$-dimensional complex vector space and $\Lambda\subset H$ is a full rank lattice (an abelian subgroup isomprphic to $\mathbb Z^{2g}$). 
For $H$ you can take $H^1_{DR}(\Sigma; \mathbb R)$ (the degree 1 DeRham cohomology of $\Sigma$ with real coefficients). This is a $2g$-dimensional real vector space, if you just consider the differential structure on $\Sigma$. The complex structure on $\Sigma$ gives $H$ the structure of a complex vector space of dimension $g$. For $\Lambda\subset H$ you can take the  cohomology classes with integer "periods" (=integrals along oriented closed curves in $\Sigma$).
Another natural definition of $H$ is the space of holomorphic 1-forms on $\Sigma$ (but it is then less obvious why it is a $g$-dimensional complex vector space).  
If $f:\Sigma\to\Sigma'$ is a conformal diffeomorphism, then the induced map $f^*:H^1_{DR}(\Sigma';\mathbb R)\to H^1_{DR}(\Sigma;\mathbb R)$ is a complex linear isomorphism such that $f^*(\Lambda_{\Sigma'})= \Lambda_{\Sigma}.$
It follows  that the map $[\Sigma]\mapsto [(\Lambda,H)]$ is well-defined (the brackets idicate equivalence classes with respect to the obvious equivalence relations). 
The map $[\Sigma]\mapsto [(\Lambda, H)]$ is injective (a complete invariant), but for $g>1$ it is not surjective. The image is an open subset of the set of all equivalence classes of pairs $(\Lambda, H)$, given by the so-called Riemann bilinear relations. 
Reference: Griffith-Harris, Principles of Algebraic Geometry.  
