Let $M$ be a Riemann surface and $V$ a two-dimensional holomorphic vector bundle over $M$. On this paper (page 72), the author states that if $V$ has a subbundle $L$ with $\deg L \geq \frac{1}{2}\deg\wedge^{2}V$, then $V$ is an extension $$ 0\rightarrow L \rightarrow V \rightarrow L^{*}\otimes \wedge^{2}V \rightarrow 0 $$
Does anyone know how can I construct the map from $V$ to $L^{*}\otimes \wedge^{2}V$?
Just take the quotient $V/L$ and note that since its rank is 1, one has $$ V/L \cong \det(V/L) \cong \det(V) \otimes \det(L)^{-1} \cong \wedge^2V \otimes L^*. $$