$\newcommand{\C}{\mathbb{C}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\R}{\mathbb{R}} \newcommand{\P}{\mathbb{P}} \newcommand{\db}[1]{D^b(\mathrm{Coh}(#1))} \newcommand{\O}{\mathcal{O}} $ I am reading through Meet Homological Mirror Symmetry by Matthew Ballard and am trying to understand his description of how the vector bundle $\mathcal{E}=\underline{\mathbb{C}}\oplus \O(-1)$ generates the category of B-branes on $\mathbb{P}^1_k$, i.e. $D^b(\mathrm{Coh}(\mathbb{P}^1_k))$, as a triangulated category.
In order to do this, he introduces the following short exact sequence: $$0\to \O(n-2)\stackrel{(y,-x)}{\to} \O(n-1)\oplus \O(n-1)\stackrel{(x,y)^t}\to \O(n)\to 0$$ for each $n\in \Z$. Since I am not experienced in algebraic geometry, I do not know what is meant by $(y,-x)$ and $(x,y)^t$. What are the maps in this short exact sequence supposed to be?
A standard result from algebraic geometry says that sections of $\mathcal{O}(n)$ over $\mathbb{CP}^k$ may be identified with homogeneous polynomials of degree $n$ in $k+1$ variables. Thus, for $\mathbb{CP}^1$, these sections are homogeneous polynomials of $2$ variables. Now, given a section of $\mathcal{O}(n-2)$ as above, we identify it with a homogeneous degree $n-2$ polynomial $p(x,y)$. Then we multiply it by $-x$, to get another homogeneous polynomial of degree $n-1$, and hence a section of $\mathcal{O}(n-1)$. Similarly for multiplication by $y$.
Now, this is not really a good way to think about the morphism $(x,y)$, because you have to define it locally, but you do this in the same way: it's just multiplying sections by $x$ and $y$, respectively. The above explanation tells you why the result lands in $\mathcal{O}(n-1)\oplus\mathcal{O}(n-1)$. Then the next map, the transpose, just takes a direct sum and multiplies by $x$ and $y$ respectively, then adds them. The result once again lands in those polynomials which are one degree higher and adding two of them gives a homogeneous polynomial of the same degree, i.e. a section of $\mathcal{O}(n)$. That's slightly informal, but hopefully clear.