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I'm sorry for the following elementary question.

Things are algebraic/holomorphic over $\mathbb{C}$.

I'm reading the book "Vector bundles on Complex Projective Spaces" by Okonek et al. and the following is from Theorem 4.1.14. Chapter II.

Let $q:P\rightarrow M$ be a principal $G$-bundle (the action is called $\rho$) where $G$ is a product of $GL_k$'s and $M$ and $P$ are smooth. Let $X$ be a smooth surface. Let $E$ be a vector bundle on $X\times M$ such that the action $id_X\times \rho$ of $G$ on $X\times M$ can be linearized for $E$.

(In the book, $X=\mathbb{P}^2$ and $P$ parametrizes certain monads on $X$ and it descends via the (free) action of $G$ to the smooth moduli of rank $2$ stable sheaves with given Chern classes on $X$.)

Why is $E/G$ (more precisely, $((id_X\times q)_*E)^G$) locally free?

Is $q$ being locally trivial necessary here? Or is the above still true if $M$ is simply a free quotient by $G$ of $P$ ($M$ and $P$ are still smooth)?

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    $\begingroup$ This is true much more generally: if $p:X\rightarrow Y$ is a flat surjective morphism, any coherent sheaf $\mathcal{F}$ on $Y$ such that $p^*\mathcal{F}$ is locally free is itself locally free. See e.g. SGA I, Exp. 8, Prop. 1.10. $\endgroup$
    – abx
    Commented Dec 11, 2016 at 5:23
  • $\begingroup$ @abx Thank you! Also, should we have that the natural morphism $p^*(p_*E)\rightarrow E$ is an isomorphism for any flat surjective morphism $p:X\rightarrow Y$ and any coherent sheaf $E$ on $X$? Some explanation would be very helpful, too. Thank you. $\endgroup$
    – HLC
    Commented Dec 11, 2016 at 6:05
  • $\begingroup$ @abx Think of the affine case. An $R$-module $M$ is finitely presented and flat if and only if the associated sheaf on $Spec(R)$ is locally free of finite rank. Now let $R\to S$ be faithfully flat. It is rather easy to see that each of the properties holds for $M$ over $R$ if and only if it holds for $M\otimes S$ over $S$. $\endgroup$
    – anon
    Commented Dec 11, 2016 at 8:27
  • $\begingroup$ No, that's not true: as soon as $E$ is negative enough (e.g. replacing $E$ by $E\otimes L^{-N}$, for $L$ ample and $N\gg 0$), $f_*E$ is $0$. $\endgroup$
    – abx
    Commented Dec 11, 2016 at 8:51
  • $\begingroup$ @abx I'm confused now. What about Lemma A.6 in this book: math.uzh.ch/index.php?file&key1=5171 $\endgroup$
    – HLC
    Commented Dec 11, 2016 at 17:25

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