Classification of vector bundles of rank $n$ on $\mathbb{P}_k^{1}$ (Gortz's Algebraic Geometry)

Question

I'm reading the Gortz's Algebraic Geometry, proof of the Theorem 11.51 and stuck at understanding existence part :

We will use the his book, Lemma 11.50 :

Here, $R^{+} := k[T], R^{-}:=k[T^{-1}], R^{\pm}:=k[T,T^{-1}]$.

Note that in his book, p.313, (11.17.1) there is a bijection.

$$A:= \{ \operatorname{isomorphism classes of vector bundles of rank n on} \mathbb{P}_k^{1} \} \xrightarrow{\psi} B:=GL_{n}(R^{+}) \backslash GL_{n}(R^{\pm})/GL_n(R^{-1})$$ (double quotient)

C.f. p.313 is as follows :

Now, to show the underlined statement, fix a vector bundle $\mathcal{E}$ of rank $n$ on $\mathbb{P}_k^{1}$. By the above Lemma 11.50, there exists a $\mathbf{d}:=(d_1 \ge d_2 \ge \cdots \ge d_n) \in (\mathbb{Z}^{n})_{+}$ such that $\psi(\mathcal{E})=GL_n(R^{+})T^{\mathbf{d}}GL_n(R^{-1})$.

Let $X:= \mathbb{P}_k^{1}$ and $G:=GL_n(\mathcal{O}_X) : U \mapsto \operatorname{Aut}_{\mathcal{O}_U}(\mathcal{O}_X^{n}|_U) = GL_n(\Gamma(U,\mathcal{O}_U))$. Note that $\mathcal{U}:=(U_0, U_1)$, where $U_0:=\operatorname{Spec}k[T]$, $U_1:=\operatorname{Spec}k[T^{-1}]$, is an affine open cover of $X$ ( c.f. Gortz's Book, p.313 )

Q. Then $\psi ( \bigoplus_{i=1}^{n} \mathcal{O}_{X}(d_i)) = GL_n(R^{+})T^{\mathbf{d}}GL_n(R^{-1})$ ? If so, then $\psi(\mathcal{E}) = \psi ( \bigoplus_{i=1}^{n} \mathcal{O}_{X}(d_i))$ and $ \mathcal{E} \cong \bigoplus_{i=1}^{n} \mathcal{O}_{X}(d_i)$.

To show this question, I think that we can describe the above map $\psi$ more in detail.

First, let $A_{\mathcal{U}}:= \{\operatorname{isomorphism classes of locally free} \mathcal{O}_X-\operatorname{modules} \mathcal{E} \operatorname{of rank n such that} \mathcal{E}|_{U_i} \cong \mathcal{O}_{U_i}^{n} \operatorname{for all} i=0,1 \}$

Then $A = A_{\mathcal{U}}$ as in his book, p.313, (11.17) (True?).

Then I think that the $\psi$ is as follows :

$$ \psi : A=A_{\mathcal{U}} \to H^{1}(\mathcal{U},G) \xrightarrow{c_{G,\mathcal{U}}} \check{H}^{1}(\mathcal{U},G) \to B:=GL_{n}(R^{+}) \backslash GL_{n}(R^{\pm})/GL_n(R^{-1}) $$

Here, $H^{1}(\mathcal{U},G) := \{T\in H^{1}(X,G) : T \operatorname{is trivialized by} \mathcal{U} ; i.e., T(U_i) \neq \emptyset \operatorname{for all} i \}$ (Here $H^{1}(X,G)$ is the set of isomorphism classes of $G$-torsors (c.f. his book p.292)). And the first map is $\mathcal{G} \mapsto \mathcal{Isom}(\mathcal{O}_X^{n},\mathcal{G})$ (c.f. his book p.295, (11.6)). And the second map $c_{G,\mathcal{U}}$ is in p.294 above. And the third map is as p.313 (above image).

Second, let $\mathcal{G}:= \bigoplus_{i=1}^{n}\mathcal{O}_X(d_i)$. Then by the first map, $\mathcal{G}$ sends to $\mathcal{Isom}(\mathcal{O}_X^{n} , \mathcal{G})$. And this sends to an element of $\check{H}^{1}(\mathcal{U},G)$ via the second map $c_{G,\mathcal{U}}$ as follows.

First, by the definition/or construction of $c_{G,\mathcal{U}}$ (c.f. his book p.294 above. If needed, I will upload the page), choose canonical(?) isomorphisms

$$t_0 : \mathcal{O}_{X}^{n}|_{U_0} \to \mathcal{G}|_{U_0} \in \mathcal{Isom}(\mathcal{O}_X^{n}, \mathcal{G}))(U_0) \neq \emptyset$$ $$t_1 : \mathcal{O}_{X}^{n}|_{U_1} \to \mathcal{G}|_{U_1} \in \mathcal{Isom}(\mathcal{O}_X^{n}, \mathcal{G}))(U_1) \neq \emptyset $$

Then since $G$ acts simply transitively (c.f. his book p.294 above and p.295, (11.6) ), there exists an element $g=g_{01} \in G(U_0 \cap U_1) =GL_{n}(R^{\pm})$ such that

$$ g_{01}t_1 = t_0$$

More precisely, $t_1|_{U_0 \cap U_1} \circ (g_{01})^{-1} =: g_{01}t_1|_{U_0 \cap U_1} = t_0|_{U_0 \cap U_1}$ (c.f. As the action of $G(U)$ on $\mathcal{Isom}(\mathcal{O}_X^{n}, \mathcal{G})(U)$, refer to his book p.295).

Note that $\{ g_{01}\}$ forms a $\check{\operatorname{C}}$ech $1$-cocycle of $G$ over $\mathcal{U}$ (c.f. his book, p.313). Then $c_{G, \mathcal{U}} (\mathcal{Isom}(\mathcal{O}_X^{n}, \mathcal{G})) = [ g_{01} ] \in \check{H}^{1}(\mathcal{U},G)$ (the cohomology class ; c.f. his book p.293 ).

Third, my question is, the cohomology class $[g:=g_{01}] \in \check{H}^{1}(\mathcal{U},G)$ sends via the third map of $\psi$ to $GL_{n}(R^{+})T^{\mathbf{d}}GL_{n}(R^{-})$ ? ; i.e.,

$$GL_n(R^{+})g_{01}GL_{n}(R^{-}) = GL_n(R^{+})T^{\mathbf{d}}GL_{n}(R^{-}) $$

in $B:= GL_n(R^{+}) \backslash GL_n(R^{\pm}) /GL_{n}(R^{-})$? If so, then $\psi ( \mathcal{G}) = GL_n(R^{+})T^{\mathbf{d}}GL_n(R^{-1})$ and we are done.

Note that to show the equality, it suffices to show that there exist $h^{+} \in GL_n(R^{+})$ and $h^{-} \in GL_n(R^{-})$ such that $h^{+}g_{01}h^{-} = T^{\mathbf{d}}$. And is it true?

It seems that we need more information about $t_0, t_1$, and $g_{01}$.

Can anyone help?