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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.

Thank you in advance!

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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
    
@only: Thank you, I wasn't aware of the fact that morphisms of vector bundles have to do that. Imho, it's 'mathematical bad manners' to define vector bundles in the lecture but not define what morphisms are :) Thanks for clarifying! Next time I encounter a problem like this, I should probably google 'morphisms of vector bundles' first ;) –  InvisiblePanda Nov 16 '12 at 10:04
    
@only: Just for the sake of completeness, if you might want to write this as an answer (maybe including the definition of morphisms of vector bundles), I'd be glad to accept it to close this question :) –  InvisiblePanda Nov 22 '12 at 9:24

1 Answer 1

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.

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