Let $X$ be a normal (possibly singular) projective surface over $\mathbb{C}$. Consider the set $M\_X$ of all coherent sheaves $F$ on $X$ such that there exists a finite subset $Y\subset X$ such that $F$ restricted to $X\setminus Y$ is a line bundle. $M\_X$ becomes a monoid via the tensor product. Now let $G\_X$ be the set of equivalence classes of $M\_X$ where two sheaves $F\_1,F\_2\in M\_X$ are equivalent if there is a finite subset $Y\subset X$ such that $F\_1$ and $F\_2$ are isomorphic on $X\setminus Y$. This equivalence relation is compatible with the tensor product and so $G\_X$ is a group.


In general, one has the group homomorphism $\textrm{Pic}(X)\to G\_X$ that sends a line bundle to its equivalence class. If $X$ is smooth, then this is an isomorphism and thus $G\_X$ is just the usual Picard group.


My hope is that in general we can understand $G$ in terms of a desingularisation $f: X'\to X$. Because $X$ is normal, it has only finitely many singularities. Away from these singularities $f$ is an isomorphism and the group $G\_{X'}$ is just the Picard group of $X'$. So my hope is that we can identify $G\_X$ with $\textrm{Pic}(X')$. Is something like that true?



Picard group modulo codimension 2
The group $G\_X$ can be identified with the group of rank 1 reflexive sheaves on $X$ ($F$ is reflexive if the canonical morphism $F \to F^{\vee\vee}$ is an isomorphism) by taking a sheaf $F$ to the reflexive sheaf $F^{\vee\vee}$. The monoidal structure on the set of all reflexive sheaves is given by 
$$
(F,G) \mapsto (F \otimes G)^{\vee\vee}.
$$
Furthermore, the group $G\_X$ can be identified with the class group $\operatorname{Cl}(X)$ of Weil divisors on $X$. 


The relation of $\operatorname{Pic}(X')$ to $\operatorname{Cl}(X)$ is given by the following exact sequence
$$
0 \to \bigoplus \mathbb{Z}[E\_i] \to \operatorname{Pic}(X') \to \operatorname{Cl}(X) \to 0,
$$
where $E\_i$ are the components of the exceptional divisor of $X' \to X$.


