# Isomorphy of vector bundles over a variety $X$

As in my other question, let $X$ be a variety over a field $k$, and let $\pi:F\to X$, $\psi:G\to X$ be vector bundles of rank $r$ over $X$ defined on the same open cover $\{U_i\}$. That is, we have isomorphisms $f_i:\pi^{-1}(U_i)\to U_i\times\mathbb{A}^r$ with $pr_1\circ f_i=\pi|$, and the transition maps $f_{ij}=f_j\circ f_i^{-1}:(U_i\cap U_j)\times\mathbb{A}^r\to(U_i\cap U_j)\times\mathbb{A}^r$ are of the form $(p,v)\mapsto(p,A(p)\cdot v)$ for matrices $A_{ij}\in\operatorname{GL}_r(\mathcal{O}_X(U_i\cap U_j))$. The same holds for $G$, respectively.

Now the thing to prove is that $F$ and $G$ are isomorphic as vector bundles if and only if there exist $M_i\in\operatorname{GL}_r(\mathcal{O}_X(U_i))$ such that

$M_i(p)A_{ji}(p)=B_{ji}(p)M_j(p)$ for all $p\in U_i\cap U_j$.

My problem is already quite at the start of the proof. If there is such an isomorphism $\phi$, we have $\pi^{-1}(U_i)\xrightarrow{\phi}\psi^{-1}(U_i)$, and by utilizing the isomorphisms $f_i$ and $g_i$, we get a map $U_i\times\mathbb{A}^r\to U_i\times\mathbb{A}^r$ of the form $(p,v)\mapsto(p,M_i(p)\cdot v)$.

I see that we can get such a map, but why doesn't it do anything on the first coordinate here? I still need to get used to this 'big' definition of a vector bundle.

Well, you want maps of vector bundles to map fibers above $p$ to fibers above $p$, which is why the first coordinate doesn't change. – only Nov 13 '12 at 11:51
I think that the transition matrixs are in $\operatorname{GL}_r(\mathcal{O}_X(U_i))$, this is the matrix whose interiors are in $\mathcal{O}_X(U_i)$. So the isomorphic maps are locally $A_i$-linear maps,where $A_i$ means the corresponding ring of $\mathcal{O}_X(U_i)$ when $U_i$ is so small that it is affine. The point $p$ lies on the base space i.e. it is a prime ideal of $A_i$, since $A_i$ is unchanged, so is point $p$. Hope this is useful.