I am trying to understand the transition functions for $\mathcal{O}(1)$ on $\mathbb{P}^1$. I've been stuck on this while reading these notes (Proposition 1.8, page 3) on divisors and invertible sheaves (for simplicity I'm sticking with $n=1$ here).
So suppose we have $X:=\mathbb{P}^1=\operatorname{Proj}k[x,y]$ for some field $k$. Set $S=k[x,y]$. We cover $X$ with $D(x)\cong\operatorname{Spec}k[y/x]$ and $D(y)\cong\operatorname{Spec}k[x/y]$. By definition of $\mathcal{O}(1)$, we have that $\mathcal{O}(1)(D(x))=S_{(x)}=k[y/x]x$, a free $\mathcal{O}(D(x))=k[y/x]$-module of rank $1$. Similarly, $\mathcal{O}(1)(D(y))=k[x/y]y$, a free $\mathcal{O}(D(y))=k[x/y]$-module of rank $1$.
Now, the map from $\mathcal{O}(1)(D(x))$ to $\mathcal{O}(D(x))$ is given by division by $x$, and the map from $\mathcal{O}(1)(D(y))$ to $\mathcal{O}(D(y))$ is given by division by $y$. The claim made in the notes is that the transition function between $D(x)$ and $D(y)$ is multiplication by $x/y$, but I'm having trouble seeing why this is true. We have maps $\mathcal{O}(1)(D(x))\to\mathcal{O}(D(x))\to\mathcal{O}(D(xy))=k[x/y,y/x]$, given by division by $x$ and inclusion, respectively. But how do I get a map from here to $\mathcal{O}(1)(D(y))$?