 Solution

Pullback and tensor product preserve holomorphic line bundles and their isomorphisms. Interpret negative tensor powers of $L$ as powers of $L^{\ast }$. Thus the formula defines a group homomorphism on isomorphism classes.

To prove injectivity, suppose $p^{\ast }M\otimes L^{\otimes n}$ is trivial. Restrict it to any fibre. A line pulled back from its base point is trivial there, so its fibre restriction is ${\mathcal{O}}_{{\mathbb{P}}^1}(n)$. By the Picard group classification, $n=0$. It remains to show that $p^{\ast }M$ trivial implies $M$ trivial.

Let $\sigma$ be a nowhere-zero holomorphic section of $p^{\ast }M$. On a sufficiently small open set where both $E$ and $M$ are trivial, write $\sigma =a_i(x,[v])p^{\ast }{e}_i$ for a frame $e_i$ of $M$. For each fixed $x$, $a_i(x,\cdot )$ is a holomorphic function on the compact complex projective line, hence is constant. It is therefore $b_i(x)$; evaluating at a constant projective point in the local trivialization shows $b_i$ is holomorphic in $x$. The section is nowhere zero, so each $b_i$ is nowhere zero. Its overlap laws are exactly those of a section of $M$, and $b_i{e}_i$ glue to a global holomorphic frame. Thus $M$ is trivial.

The kernel consists only of $({\mathcal{O}}_X,0)$, provingThis is the Picard injection for holomorphic projective bundles. The argument uses local fibrewise constancy, so no global section of the projective bundle is required.