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 $0\rightarrow f_*F\rightarrow f_*E\rightarrow f_*G\rightarrow 0$?

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. 


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 nonsplit 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. 


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 nontrivial 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. 

