I am trying to show that there is an exact sequence: The exact sequence $0 \rightarrow L(-1) \rightarrow L(0)^2 \rightarrow L(1) \rightarrow 0$, where $L(-1)$ is the tautological line bundle of $\mathbb{P}^1_{\mathbb{C}}$ and $L(0)$ is the trivial line bundle.
As I understand it I need to show that $L(1)$ is the quotient bundle of the other two, or, equivalently, that the transition matrix of $L(0)^2$, which is $h_{i,j}(x)=\bigl(\begin{matrix} 1&0\\ 0&1 \end{matrix} \bigr)$ for each $i,j = {0,1}, x \in \mathbb{P}^1$, is equal to the matrix $h_{i,j}(x)=\bigl(\begin{matrix} g_{i,j}(x)&0\\ 0&t_{i,j}(x) \end{matrix} \bigr)$ , where $g_{i,j}(x) = \frac{x_i}{x_j}$ - the transition function of $L(-1)$, and $t_{i,j}=\frac{x_j}{x_i}$ - the transition function of $L(1)$. The transition functions are considered with respect to the standard covering of $\mathbb{P}^1$. I am not sure where to go from here, or entirely sure that I am correct so far.
Well, $L(1)$ is generated by the sections $x$ and $y$, so the degree-$1$ map $L(0)^2 \to L(1)$ given by $(a,b)\mapsto ax+by$ is surjective. The kernel is the set of all $(a,b)$ such that $a=fy, b=-fx$ for some $f$; so the degree-$1$ map from $L(-1)$ to $L(0)$ given by $c\mapsto (cy,-cx)$ maps isomorphically to the kernel. I hope that helped...