I want to prove that "Given a bundle $E$, for any line bundle $L$ the projectivizations of $E$ and $E$ tensor $L$ are isomorphic i.e $P(E)\cong P(E\otimes L)$".
By bundle you can as well assume it to be a smooth vector bundle and their projectivization to be a fibre bundle (in the topological sense) with fibre 1-dim subspaces of the vector space fibres
The statement can also be seen on Wikipedia page. The reference they give is for Hartshorne, Algebraic Geometry. But unfortunately, I am not aware of that.
Can anyone suggest a proof involving only fibre bundles or maybe just give a good reference with the desired approach, how can one construct a morphism between these two bundles? Or at least how we can see that morphism geometrically.
Edit 1: I found this interesting argument, I am not getting how the induced map looks like?, is it $[v]\rightarrow[u\otimes v$]? If it is, then how can I define an isomorphism at the total space level will $(x,[v])\rightarrow(x,[u\otimes v$]) work?
Any suggestions will be of great help.
Thanks and regards in advance
Best Answer
The projective bundle $\mathbb{P}_X(E)$ represents the functor $$ \Phi_E \colon \mathrm{Schemes} \to \mathrm{Sets}, \qquad S \mapsto \{(f,F,\phi) \mid f \colon S \to X,\ \phi \colon f^*E \to F \}, $$ where $F$ is a line bundle on $S$ and $\phi$ is an epimorphism of vector bundles. The functors $\Phi_E$ and $\Phi_{E \otimes L}$ are isomorphic via $$ (f,F,\phi) \mapsto (f,F \otimes f^*L, \phi \otimes \mathrm{id}_{f^*L}) $$ hence the representing schemes are isomorphic as well.