In Huybrechts book there is an exercise 2.2.5 :
Let $X$ be a complex compact manifold and $L$ a holomorphic line bundle on $X$. Show that $L$ is trivial if and only if $L$ and its dual $L^*$ admit a non trivial global sections.
Huybrechts didn't mention that the manifold is connected. Is it possible to do the exercise without this condition? This is my solution:
Consider a global section $t$ of the dual bundle and $s$ is the section of $L$, considering local descriptions $L ↔ \{U_α, g_{α\beta}\}$, $L^{-1} ↔ \{U_α, g_{α\beta}^{-1}\}$, $s ↔ \{U_α, s_α\}$ and $t ↔ \{U_α, t_α\}$ he observes that on every $U_\alpha \cap U_\beta$ we have $s_\alpha t_\alpha = s_\beta t_\beta$, so we can glue these $s_\alpha t_\alpha$ in a "global" function $F: X \rightarrow \mathbb{C}$. By the maximum principle, since $X$ is compact, $F \equiv c \in \mathbb{C}$. (we shouldn't suppose $X$ is connected?).
Best Answer
There is a subtle point here. If $X=X_1\amalg X_2$ is a disjoint union of compact complex manifolds admitting non-trivial line bundles $L_1$ and $L_2$ with non-zero sections $s_i\in H^0(X_i,L_i)$, then $M=L_1\amalg L_2^\vee$ is a line bundle on $X$ with non-zero* sections $s_1\amalg 0\in H^0(X,M)$ and $0\amalg s_2\in H^0(X, M^\vee)$ (identifying $L_2=L_2^{\vee\vee}$ via the canonical isomorphism). However, $M$ is not trivial.
*) "non-zero" meaning "not totally zero". Note that these sections vanish on (different) components.
If, however, both sections are non-zero on each component, then the statement becomes true, e.g., by carrying out the argument on each component separately, as was suggested in the comments to the OP.