Let $V$ be a complex manifold. I am trying to approach the subject of divisors through complex geometry since I am not familiar enough with algebraic geometry to come that way.
A divisor $D$ on $V$ is defined as a system of meromorphic place functions $\{ f_i \}$ on an open cover $\{ U_i \}$ such that on $U_i \cap U_j$, $f_i/f_j$ has neither zeros nor poles.
My confusion might stem from not properly understanding when two divisors are equivalent. Provisionally, I am assuming this definition:
Two divisors $D, D'$ are defined as equivalent if there is an open cover $\{ U_i \}$ for both and a global meromorphic function $f$ such that $f_i/f_i' = f$ on each $U_i$.
(Auxiliary) Question 0. Is this definition correct?
The source I have considers the sheaf $\mathcal M$ of local meromorphic not identically $0$ functions and defines divisors up to equivalence as global sections in the sheaf $\mathcal M/ \mathcal H^*$ where $\mathcal H^*$ is the sheaf of never $0$ local holomorphic functions. I suspect this definition is not the same as the one I gave above, but it is cumbersome and I have no idea how to work with it, as you will see below. It is also said that $H^0(V, \mathcal M)$ is the set of all meromorphic functions on $V$ which are not identically $0$ on any connected component of $V$, which I don't truly understand why.
Moving to what I am trying to understand, two divisors $D, D'$ are called linearly equivalent if $D - D'$ is the divisor $(g)$ of $g \in H^0(V, \mathcal M)$. For an arbitrary divisor $D$ given by place functions $\{ f_i \}$, $L(D)$ is defined as the vector space of all global meromorphic functions $g$ (not necessarily satisfying $g \in H^0(V, \mathcal M)$) such that $gf_i$ are holomorphic on each $U_i$.
The claim I am trying to understand is that if $V$ is compact and connected then the set of holomorphic divisors which are linearly equivalent to $D$ is in bijection with the projectivisation of the vector space $L(D)$.
I get that the gist of this claim is that if $D' = \{ f_i' \} $ is a holomorphic linearly equivalent divisor to $D = \{ f_i \}$, then there exists $g \in H^0(V, \mathcal M)$ such that $f_i (1/g) = f_i'$, and since $D'$ is holomorphic, it follows that $1/g \in L(D)$.
This gives me a tentative map from the set of holomorphic linearly equivalent divisors to $D$ into $L(D)$. I am not sure this map is well defined. For example, according to my definition of when divisors are equivalent, if $c \in \mathbb C^*$, then the divisor given by $f_i' = f_i (c/g)$ would be equivalent to the one given by $f_i' = f_i (1/g)$, thus this map only has a chance of being well defined if we take indeed the projectivisation of $L(D)$ i.e. compress all multiples of a meromorphic function in $L(D)$ to a point. But even then, what guarantees that it is well defined? In other words:
Question 1 If the divisor given by $f_i (1/g)$ is equivalent to $f_i (1/h)$ where $g, h \in H^0(V, \mathcal M)$ (whatever "divisor equivalence" ends up meaning), can we deduce that $g = c h$ with $c \in \mathbb C$ and, if so, how can we deduce this?
Then regarding surjectivity, I suspect what happens if $V$ is not compact or not connected is that we can find some meromorphic function $g$ such that $1/g \not\in H^0(V, \mathcal M)$, but nevertheless $gf_i$ is holomorphic; in other words $gf_i$ is holomorphic, but the holomorphic divisor $D+(g)$ is not linearly equivalent to $D$. I suspect in the end this point has nothing to do with holomorphicity also, but just with the existence of meromorphic functions which are not in $H^0(V, \mathcal M)$, right?
Question 2 Why if $V$ is compact connected then for any meromorphic $g$, $D+(g)$ is linearly equivalent to $D$?
Examples to work through in illustration of my points of confusion would certainly help. Also some references which explain these things more clearly would be beneficial.
That is the formally correct definition. Quotient sheaves might be a bit scary at first (they are defined by sheafification), but essentially this means what you write before: Given a global section $D \in \Gamma(V, \mathcal{M/H}^*)$, there exists an open cover $\{U_i\}_i$, and for each $i$ a preimage $f_i \in \Gamma(V, \mathcal M)$ of the section $D|_{U_i} \in \Gamma(U_i, \mathcal M/\mathcal H^*)$. Of course on $U_{ij} = U_j \cap U_j$, $f_i$ and $f_j$ should give the same section $D|_{U_{ij}}$, which translates in the condition that $f_i / f_j \in \Gamma(U_{ij}, \mathcal H^*)$. So the tuple $(U_i, f_i)_i$ represents the divisor $D$. Naturally, because we are taking the quotient by $\mathcal H^*$, two tuples $(U_i, f_i)$ and $(U_i', f_i')$ define the same divisor $D$, if $f_i / f_i'$ is holomorphic and nonzero on $U_i \cap U_i'$.
If you allow $f$ to be meromorphic, you get the notion of linear equivalence, which you also write later. However if $f$ is in fact holomorphic, we say the two divisors are equal (because they define the same global section of $\mathcal M / \mathcal H^*$.
Now to the map $$\{ \text{holomorphic divisors linearly equivalent to } D\} \to \mathbb P(L(D)), D' \mapsto [1/g]$$ where $(g) = D - D'$.
Asking if this map is well-defined boils is the following problem: If $g, h \in \Gamma(V, \mathcal M)$ are meromorphic with $(g) = (h)$, then $g = c \cdot h$ for some $c \in \mathbb C$. Note that $(g) = (h)$ means that their quotient $\frac g h$ is holomorphic and nonzero on $V$ (see the above discussion about equality of divisors). As $V$ is compact, this quotient has to be a nonzero constant.
I'm not sure what you are talking about here. Isn't $H^0(V, \mathcal M)$ the set of meromorphic functions by definition? Also, a meromorphic function is locally the quotient of two holomorphic functions. So for each $g \in \Gamma(V, \mathcal M)$, the function $1/g$ is also contained in $\Gamma(V, \mathcal M)$. This also has to be the case so that linear equivalence is symmetric, otherwise it wouldn't be an equivalence relation. Then it is self-evident that $D$ is linearly equivalent to $D + (g)$.