direct image functor

Question

Let $f: X\rightarrow Y$ be a morphism of varieties. Let $0\rightarrow F\rightarrow E\rightarrow G\rightarrow 0$ be a short exact sequence of locally free sheave of finite rank. If direct images of above sheaves are locally free, then is it true that it induces a short exact sequence $0\rightarrow f_*F\rightarrow f_*E\rightarrow f_*G\rightarrow 0$?

Answer 1

Let $X,Y$ be defined over the field $k$ and take $f$ to be the structure map $f:X\to {\rm Spec}\, k$. Then let $E\to G$ be a surjective morphism of sheaves that is not surjective on global sections, e.g., $$E=\mathcal O_{\mathbb P^1}(-1)\oplus \mathcal O_{\mathbb P^1}(-1)\to G=\mathcal O_{\mathbb P^1}.$$ Then $f_*$ is just $H^0$ and the desired statement is false.

EDIT: (to have an example mapping to an arbitrary scheme) Consider the base change of $f$ via $Y\to {\rm Spec}\, k$: $$g: X\times_{{\rm Spec}\, k} Y \to Y.$$ and let $\mathcal E:=p^*E$ and $\mathcal G:=p^*G$ where $p:X\times_{{\rm Spec}\, k} Y \to X$ is the projection to $X$. Then $g_*\mathcal E\simeq H^0(X, E)\otimes_k \mathcal O_Y$ and $g_*\mathcal G\simeq H^0(X, G)\otimes_k \mathcal O_Y$, so again the desired statement is false.

Answer 2

Two simple examples where $Y=\mathrm{Spec}k$ a point:

Consider $0\to\mathcal{O}(-2)\to\mathcal{O}(-1)^2\to\mathcal{O}\to 0$ on $\mathbb P^1$, where the two maps $\mathcal{O}(-1)\to\mathcal{O}$ are given by multiplication by $x$ and $y$ respectively, then the sequence of global sections is $0\to0\to0\to k$ and hence is not exact.

On an elliptic curve there is a non-split exact sequence $0\to\mathcal{O}\to\mathcal E\to\mathcal{O}\to0$ as $\mathrm{Ext}^1(\mathcal{O},\mathcal{O})$ is $1$-dimensional. Then $H^0(\mathcal E)\to H^0(\mathcal O)$ can not be surjective as otherwise the sequence would be split.

Answer 3

The answer is no, as the following examples shows. It is inspired by Sandor's answer to this question of mine.

Let $Y$ be an elliptic curve, $X=Y \times Y$ and $f \colon X \to Y$ the projection onto the first factor.

Since $f$ has connected fibres, we have $f_* \mathcal{O}_X= \mathcal{O}_Y$. On the other hand, since $X$ is a product, for all $p \in Y$ we may identify canonically $H^1(f^{-1}(p), \mathcal{O}_{f^{-1}(p)})$ with $H^1(Y, \mathcal{O}_Y) \cong \mathbb{C}$, hence $R^1f_* \mathcal{O}_X=\mathcal{O}_Y$, and by projection formula $R^1f_* \mathcal{O}_X(-p)=\mathcal{O}_Y(-p)$.

Set $E_p:=f^{-1}(p)$, and apply the functor $f_*$ the exact sequence

$0 \to \mathcal{O}_X \to \mathcal{F} \to \mathcal{O}_X(-E_p) \to 0$,

where $\mathcal{F}$ is the unique non-trivial extension of $\mathcal{O}_X(-E_p)$ by $\mathcal{O}_X$ (one can check that $\mathcal{F}$ is an indecomposable rank $2$ vector bundle on $X$).

We obtain

$0 \to \mathcal{O}_Y \to f_* \mathcal{F} \to \mathcal{O}_Y(-p) \stackrel{\delta}{\to} \mathcal{O}_Y \to R^1 f_* \mathcal{F} \to \mathcal{O}_Y(-p) \to 0$.

The sheaf $f_* \mathcal{F}$ is reflexive on a smooth curve, hence locally free. On the other hand, if $\delta$ were the zero map then $f_* \mathcal{F}=\mathcal{O}_Y \oplus \mathcal{O}_Y(-p)$ and so, by funtoriality of $f_*$, the vector bundle $\mathcal{F}$ would be decomposable, contradiction.