Consider the usual projective space $\mathbb{P}^{1} = \mathbb{C} \cup \{\infty\}$, and the Weil divisor $D = \{0\} \subset \mathbb{C} \subset \mathbb{P}^{1}$. Writing projective space as the union of the open sets, we obtain $\mathbb{P}^{1} = U_{0} \cup U_{1}$, using $U_{0} = Spec(\mathbb{C}[t]), U_{1} = Spec(\mathbb{C}[t^{-1}])$, we get $\mathbb{C}(\mathbb{P}^{1}) = \mathbb{C}(t)$.
Defining $\Gamma(\mathbb{P}^{1},O_{\mathbb{P}^{1}}(D))=\{f\in \mathbb{C}(t)^{*} | div(f) + D \geq 0\} \cup \{0 \}$, it is then claimed that it follows easily that the global sections are $1,t^{-1} \in \Gamma(\mathbb{P}^{1},O_{\mathbb{P}^{1}} (D))$ (cf. Introduction to Toric Varieties Cox, Little, and Schenck p. 247).
In addition, he further states that multiplying by $f$ gives a sheaf homomorphism $O_{\mathbb{P}^{1}}(-D) \to O_{\mathbb{P}^{1}}$ and that doing this for $1, t^{-1} \in \Gamma(\mathbb{P}^{1},O_{\mathbb{P}^{1}} (D))$ gives $$O_{\mathbb{P}^{1}}(-D) \oplus O_{\mathbb{P}^{1}}(-D) \to O_{\mathbb{P}^{1}}$$
Could someone please explain these implications in detail? I am rather new to sheaf theory and am having trouble following the arguments in this example. I know that we define $div(f) = \sum \nu_{D}(f) D_{f}$, where $\nu$ is the valuation, but I still fail to see above implication.
Thanks
The global sections $\Gamma(\mathbb{P}^1, \mathcal{O}(D))$ are, by definition, the rational functions $f(t) \in \mathbb{C}(t)$ which have at worst a simple pole at $t=0$ and are regular elsewhere, so we can write $$ f(t) = [\textrm{global regular function}] + \frac{1}{t} [\textrm{global regular function}]. $$ However, the global regular functions on $\mathbb{P}^1$ are precisely the constants, so $\Gamma(\mathbb{P}^1, \mathcal{O}(D))$ is a 2-dimensional $\mathbb{C}$-vector space with basis $\{ 1 , t^{-1} \}$.
This is in stark contrast to the global sections of $\mathcal{O}(-D)$; these are the rational functions $g(t) \in \mathbb{C}(t)$ which have a simple zero at least a simple zero at $t=0$ are regular elsewhere. In particular, such a function is everywhere regular, hence constant. The only constant function that satisfies this criteria is the zero function, so $\Gamma(\mathbb{P}^1, \mathcal{O}(-D)) = 0$.
Knowing this, can you describe the morphism $\mathcal{O}(-D) \to \mathcal{O}$?
Edit 1: Here is a more careful calculation of the global sections $\Gamma(\mathbb{P}^1, \mathcal{O}(D))$.
A global section $s \in \Gamma(\mathbb{P}^1, \mathcal{O}(D))$ consists of 2 pieces of data: a section $s|_{U_0} = f_0 \in \Gamma(U_0, \mathcal{O}(D))$ and a section $s|_{U_1} = f_1 \in \Gamma(U_1 , \mathcal{O}(D))$ which agree on the intersection $U_0 \cap U_1$. Since $0 \in U_0$, $f_0$ is a rational function in $t$ with a pole (of order at most $1$) at $t=0$. Since $0 \not\in U_1$, $f_1$ is just a polynomial in $t^{-1}$, with no specified zeros (though it could have some). The compatibility condition is precisely that $s|_{U_0 \cap U_1} = f_0(t) = f_1(t^{-1})$ on $U_0 \cap U_1$, which can only occur if $f_0(t) = a + \frac{b}{t}$ for some $a,b \in \mathbb{C}$. Therefore, the section $s$ is given by $a+ \frac{b}{t}$.
Edit 2: Let's describe the morphism $\mathcal{O}(-D) \to \mathcal{O}$ on each piece of the open cover $\mathbb{P}^1 = U_0 \cup U_1$.
On $U_0$, the sections of $\mathcal{O}(-D)$ are $\Gamma(U_0, \mathcal{O}(-D)) = t \cdot \mathbb{C}[t]$, i.e. it consists of those polynomials $g(t) \in \mathbb{C}[t]$ with (at least) a simple zero at $t=0$. If $f $ is either $1$ or $t^{-1}$, then the multiplication map $\Gamma(U_0, \mathcal{O}(-D)) \to \Gamma(U_0, \mathcal{O}) = \mathbb{C}[t]$ given by $g \mapsto gf$ is indeed a homomorphism.
On $U_1$, the sections of $\mathcal{O}(-D)$ are $\Gamma(U_1, \mathcal{O}(-D)) = \mathbb{C}[t^{-1}]$, because the divisor $D$ imposes no conditions on $U_1$ since $0 \not\in U_1$. Therefore, the multiplication map $\Gamma(U_1,\mathcal{O}(-D)) \to \Gamma(U_1,\mathcal{O}) = \mathbb{C}[t^{-1}]$, again given by $g \mapsto gf$, is a homomorphism.