this is my first question on stackexchange. So, please don't be too hard to me. I am listening to a master course in Algebraic Geometry.
My question belongs to Proposition II.5.4 in Hartshorne's "Algebraic Geometry". I want to show that the following statements are equivalent:
Proposition:
Let $(X, \mathcal{O}_X)$ be a prescheme and $\mathcal{F}$ be an $\mathcal{O}_X$-module. Then the following statements are equivalent:
1) For all open affine $U \subseteq X$ there is an $\Gamma(U, \mathcal{O}_X)$-module M, sucht that $F_{|U} \cong \tilde{M}$.
2) There is an open affine covering $\{U_i\}_{i\in I}$ of $X$ and some $\Gamma(U_i, \mathcal{O}_X)$-modules $M_i$, such that $\mathcal{F}_{|U_i} \cong \tilde{M_i}$.}
The implication from 1) to 2) is clear. To prove the other direction, I oriented at the proof of Hartshorne. I hope, you can fill the gaps:
First, I proved the following lemma:
Lemma:
Let $M$ be an $R$-module and $\mathcal{F}=\tilde{M}$. Then $\mathcal{F}_{|D(f)} \cong \tilde{M_f}$ for any $f \in R$.
Proof of the proposition:
Let $U =SpecR \subseteq X$ be open and affine. Assume, the second statement holds. We denote the corresponding ring of $U_i$ by $R_i$. Then $\{U \cap U_i\}_{i \in I}$ is an open, not necessarily affine covering of $U$.
Step 1: We want to show, that $\mathcal{F}{|U}$ satisfies the second statement.
Since $U \cap U_i \subseteq U_i$ is open,
\begin{align*} U \cap U_i = \bigcup_{j \in J_i} D(g_{ij}) \end{align*}
for some $g_{ij} \in R_i$.
Therefore, $\{D(g_{ij}) ~|~ i \in I, j \in J_i\}$ is an open affine coveriung of $U$. Using the lemma and the assumption, we have:
\begin{align*} \mathcal{F}_{|D(g_{ij})} = {\mathcal{F}_{|U_i}}_{|D(g_{ij})} \cong \widetilde{{(M_i)}_{g_{ij}}} \end{align*}
Step 2: We want to finish the proof.
We define $M = \Gamma(U, \mathcal{F}_{|U})$. So we have the identity map
\begin{align*} M \to \Gamma(U, \mathcal{F}_{|U}). \end{align*}
Since $\widetilde{}$ and $\Gamma$ are adjoint functors, this leads to a map \begin{align*} \alpha \colon \tilde{M} \to \mathcal{F}_{|U}. \end{align*}
Now, I want to use that $\alpha_{|D(g_{ij})}$ is the isomorphism, which we received in step 1. Because then, since the $D(g_{ij})$'s are an open covering, $\alpha$ has to be an isomorphism.
But why is this the case?
Thank you in advance for your help. And please tell me, if I am abusing notation. I want to be as formal as possible.
Felix