Let $X$ be a smooth projective curve of genus $g\geq2$, we can construct rank $2$ vector bundles on $X$ with determinant $\omega\_X$ by extension $$0\to \mathcal{O}\to E\to\omega\_X\to 0,$$ does such $E$ necessarily embeds in $\omega\_X\oplus\omega\_X$? (for general $E$?)



Embedding extension of sheaves in direct sum
Ok, let me merge my comments into an answer. I claim that for a general extension $0\rightarrow \mathscr{O}\_X\rightarrow E\rightarrow \omega \_X\rightarrow 0$, we have $h^0(E)=1$; therefore $h^0(\mathscr{H}om(E,\omega \_X))=1$ by Riemann-Roch and Serre duality, so there is only one way (up to a scalar) to map $E$ to $\omega \_X$.


The claim follows from a linear algebra lemma:


**Lemma**$.-$ *Let $b:V\times W\rightarrow U$ be a bilinear map between vector spaces, with $\dim V\leq \dim U$ and $b(v,-): W\rightarrow U$ surjective for each $v\neq 0$ in $V$. Then $b(-,w): V\rightarrow U$ is injective for $w$ general in $W$*.


*Proof* : For $v\neq 0$ in $V$, the kernel of $b(v,-)$ has codimension $\dim U$. These kernels are parameterized by $\mathbb{P}(V)$, of dimension $\dim V-1\leq \dim U-1$, and therefore they do not fill $W$. So for $w\in W$ general $b(v,w)=0$ implies $v=0$.


Now we apply the lemma with $V=H^0(\omega \_X)$, $W= \operatorname{Ext}^1(\omega \_X,\mathscr{O}\_X)\cong H^1(\omega \_X^{-1}) $, $U=H^1(\mathscr{O}\_X)$. We must check that $\times \,\alpha : H^1(\omega \_X^{-1}) \rightarrow H^1(\mathscr{O}\_X)$ is surjective for $\alpha \neq 0$ in $H^0(\omega \_X)$. This follows from the exact sequence $$0\rightarrow \omega \_X^{-1}\xrightarrow{\ \alpha \ }\mathscr{O}\_X\rightarrow \mathscr{O}\_Z\rightarrow 0$$
with $Z=\operatorname{div}(\alpha ) $, and from $H^1(\mathscr{O}\_Z)=0$.


