Let $f:X\rightarrow Y$ be a finite morphism of smooth projective varieties over the field of complex numbers. Let $E$ be a locally free sheaf over $X$. We have the natural morphism $\phi:f^*f_*E\rightarrow E$. Does the finiteness of $f$ imply that $\phi$ is surjective? If $f$ is a closed immersion, then the above morphism is an isomorphism. If it is an arbitrary finite map, then is $\phi$ onto?
Pullback of the direct image of a vector bundle surjects to the vector bundle
algebraic-geometry
Related Solutions
No. I think you might be misinterpretting what 'locally trivial fibration' means.
Namely, what Ehresmann's theorem implies is the following:
Corollary to Ehresmann's theorem: Let $f:X\to Y$ be a smooth proper surjection of complex manifolds. Then, there is an open cover $\{U_i\}$ of $Y$ such that $f^{-1}(U_i)$ is diffeomorphic (as a manifold over $U_i$) to $U_i\times B$ for some smooth manifold $B$.
One cannot replace diffeomorphic by 'biholomorphic' or, if $X$ and $Y$ are algebraic and the $U_i$ is a Zariski cover, 'algebraic'.
Here's a silly counterexample to your literal claim.
Consider $D(\Delta)\subseteq \mathbb{A}^2_k=\mathrm{Spec}(k[a,b])$ where, for simplicity, $k$ is an algebraically closed field of characteristic greater than $3$ where
$$\Delta(a,b)=-16(4a^3+27b^2)$$
We then have the universal family of Weierstrass equations living over $D(\Delta)$
$$\mathcal{W}^\text{univ}=V(y^3-axz^2-bz^3)\subseteq \mathbb{P}^2_k\times D(\Delta)$$
We then note that
$$\mathcal{W}^\text{univ}\to D(\Delta)$$
is a family of elliptic curves and so, in particular, smooth proper and surjective. Moreover, each $\mathcal{W}^\text{univ}$ and $D(\Delta)$ is smooth.
That said it's not an etale local fibration since this, in particular, would imply that the fibers of $\mathcal{W}^\text{univ}\to D(\Delta)$ are Zariski locally isomorphic. Indeed, let $\{U_i\to D(\Delta)\}$ be an etale cover such that $\mathcal{W}^\text{univ}\times_{D(\Delta)}U_i\cong U_i\times B_i$ for some $k$-space $B_i$. Then, in particular we see that for all $x\in V_i:=\text{im}(U_i\to D(\Delta))$ (this is an open set) we have that
$$\mathcal{W}^\text{univ}\times_{D(\Delta)}U_i\times_{V_i}x\cong \bigsqcup_{y\mapsto x}B_i$$
which since $k(x)\to k(y)$ is an isomorphism implies that $\mathcal{W}^\text{univ}_x\cong B_i$. Thus, $\mathcal{W}^\text{univ}_x\cong B_i$ for all $x\in V_i$.
Note that we have an algebraic map $j:D(\Delta)\to \mathbb{A}^1_k$, the $j$-function map, which sends, in particular, $x$ to the $j$-invariant $\mathcal{W}^\text{univ}_x$. Since $\mathcal{W}^\text{univ}_x$ is locally isomorphic this implies that $j$ is locally constant which implies, since $D(\Delta)$ is connected, that $j$ is constant and thus $\mathcal{W}^\text{univ}_x$ is actually constant for $x$ in $D(\Delta)$.
Of course, this is ludicrous since $\mathcal{W}^\text{univ}_x$ contains isomorphism classes of every elliptic curve over $k$.
Of course, it is true that the fibration $W^\text{univ}\to D(\Delta)$, when $k=\mathbb{C}$, is analytically locally on the target DIFFEOMORPHIC to the trivial fiber bundle. The operative point being that any two elliptic curves are diffeomorphic but are rarely isomorphic/biholomorphic.
No, there are many counterexamples. Any surjective map of line bundles on a locally ringed space is an isomorphism, because any surjective module endomorphism of the regular module over a local ring is in fact an isomorphism (if $f:R\to R$ is the endomorphism, then $f(r)=r\cdot f(1)$, and if $f(u)=1$, then $uf(1)=1$ so $f$ is multiplication by a unit). Therefore if $S$ has any line bundles not isomorphic to $\mathcal{O}_S$, you have found a counterexample.
Best Answer
Since $f$ is finite, the functor $f_*$ is exact and conservative, so surjectivity of $f^*f_*E \to E$ is equivalent to surjectivity of $$ f_*f^*f_*E \to f_*E. $$ On the other hand, by adjunction there is also a natural morphism $$ f_*E \to f_*f^*f_*E $$ and the composition $f_*E \to f_*f^*f_*E \to f_*E$ is the identity. This proves required surjectivity.