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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.

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.

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 $E_p:=f^{-1}(p)$. Then, if we apply the functor $f_*$ to the split 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 $\mathcal{O}_X$,

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 $\delta$ where 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.

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.

The answer is no, as the following examples shows. It is mos inspired to 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 $E_p:=f^{-1}(p)$. Then, if we apply the functor $f_*$ to the split 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 $\mathcal{O}_X$,

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 $\delta$ where 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.

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 $E_p:=f^{-1}(p)$. Then, if we apply the functor $f_*$ to the split 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 $\mathcal{O}_X$,

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 $\delta$ where 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.