What is the meaning of canonical isomorphism in the cocycle construction?

In the definition of pull back of a smooth vector bundle, one usually use the cocycle description of vector bundle. But this method only decide the object up to isomorphism. Then the statement "the fibers of pull back is canonically ismorphic to the fibers of image points" seem cant give a precise meaning about canonically. Or there is some uniqueness statement about the cocycle description? Thx!

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–  user2035 Nov 15 '11 at 14:29
"In the definition of pull back of a smooth vector bundle, one usually use the cocycle description of vector bundle." I hope not! Pullbacks are defined via a universal property (also in this context), not via some uncanonical choices. Unfortunately, there are lots of people who are working with this "wrong definition" of a vector bundle as an equivalence class of cocycles. –  Martin Brandenburg Nov 15 '11 at 15:33
I agree with Martin that, hopefully, you don't use cocycles. Say you have $f:X\rightarrow Y$ and $E$ a vector bundle with base $Y$, with projection $p$. Then $f^*(E)=\{(x,v)\in X\times E:p(v)=f(x)\}$ (the pull-back!), and it's clear that the fiber over $x$ is the fiber over $f(x)$. –  Alain Valette Nov 15 '11 at 15:46
Thank the comments above all. I know the other definitions of pullback, but I was confused by the cocycle method. But even in the categorical case, I feel there is analogous dilemma. For example, in the definition of differential sheaf of scheme X over Y no one say that it is independent of the representation of fibre product and diagonal map. When we must make an operation sequence, why every one talk the word "natural" and never confirm the independence of representation? Of course I believe the essential uniqueness will solve the question, but for some complex constructions, it isnt clear. –  MZWang Nov 22 '11 at 12:57