A pencil of functions with poles at one point defines a finite map to the projective line

The two sections and their divisors. Since $f$ is nonconstant, $1$ and $f$ are linearly independent in $k(C)$; since $f\in L(np)$ and $1\in L(np)$ for $n\ge m$ by [F2] (as $\mathrm{div}(f)+np\ge 0$ and $np\ge 0$), they span a two-dimensional subspace of $L(np)$ and of $L(D_0)$. By [F1] the pole divisor of $f$ is $(f{)}_{\mathrm{\infty }}=m[p]$ with $m\ge 1$, and the divisor attached to $1$ and $f$ in $L(np)$ is $\mathrm{div}(1)+np=np$ and $\mathrm{div}(f)+np=(f{)}_0-m[p]+n[p]=(f{)}_0+(n-m)[p]$.

The same divisor with two different subspaces. On ${\mathbb{P}}_k^1$ let $D=2[\mathrm{\infty }]$. The two subspaces $V_1=k\cdot 1+k\cdot t^2$ and $V_2=k\cdot 1+k\cdot (t^2+t)$ of the single space $L(2[\mathrm{\infty }])$ are two-dimensional. Both are base-point-free: by [F2] and [F6], $\mathrm{div}(1)+2[\mathrm{\infty }]=2[\mathrm{\infty }]$ and $\mathrm{div}(t^2)+2[\mathrm{\infty }]=2[0]$, and $\mathrm{div}(t^2+t)+2[\mathrm{\infty }]=[0]+[-1]$ (as $\mathrm{div}(t^2+t)=[0]+[-1]-2[\mathrm{\infty }]$), so in each pair the constant $1$ is a unit away from infinity and the second section is a unit at infinity; by [F3] there is no base point. By [F4] each subspace attaches a morphism $C\to {\mathbb{P}}_k^1$, and the pullback of the coordinate is the ratio of the two basis sections: ${\phi }_{V_1}^{\mathrm{♯}}(t)=t^2$ and ${\phi }_{V_2}^{\mathrm{♯}}(t)=t^2+t$. If the two morphisms were related by the projective-linear action of ${\mathrm{P}\mathrm{G}\mathrm{L}}_2(k)$ on the target, some $M(u)=\frac{au+b}{cu+d}$ would satisfy $M(t^2)=t^2+t$; clearing denominators, $a{t}^2+b=(t^2+t)(c{t}^2+d)=c{t}^4+c{t}^3+d{t}^2+dt$, so comparing coefficients gives $c=0$, then $d=0$, then $a=b=0$, contradicting that $M$ is invertible. Hence ${\phi }_{V_1}$ and ${\phi }_{V_2}$ are not related by the target action: the morphism attached to a base-point-free subspace of $L(D)$ depends on the subspace, not only on $D$.

Base-point-freeness at the pole divisor. Take $n=m$ and $D_0=m[p]$. By step 1.1, $\mathrm{div}(f)+D_0=(f{)}_0$, which does not contain $p$, so the section $f$ does not vanish at $p$; and $\mathrm{div}(1)+D_0=D_0=m[p]$ is supported at $p$, so the constant section $1$ does not vanish at any other point. By [F3] no point of $C$ lies in both divisors, so $V_0=k\cdot 1+k\cdot f\subseteq L(D_0)$ is base-point-free of dimension two.

The attached morphism is ${\phi }_f$. By [F5] there is a finite locally free morphism ${\phi }_f:C\to {\mathbb{P}}_k^1$ with ${\phi }_f^{\mathrm{♯}}(t)=f$, of degree $[k(C):k(f)]$, whose fibre over infinity is $(f{)}_{\mathrm{\infty }}$, and an isomorphism ${\mathcal{O}}_C(D_0)\cong {\phi }_f^{\ast }\mathcal{O}(1)$. Under the section dictionary [F2] the pullbacks ${\phi }_f^{\ast }{x}_0$ and ${\phi }_f^{\ast }{x}_1$ correspond to the rational functions $1$ and $f$, whose span is $V_0$; the converse clause of [F4], applied with $r=1$, $D=D_0$ and that isomorphism, gives that $V_0$ is base-point-free — recovering step 2.1 — and that the morphism attached to the data $({\mathcal{O}}_C(D_0);1,f)$ is ${\phi }_f$. Hence ${\phi }_{V_0}={\phi }_f$, so ${\phi }_{V_0}$ is finite of degree $[k(C):k(f)]$ with fibre over infinity equal to $(f{)}_{\mathrm{\infty }}$.

The enlarged divisors have the base point $p$. Let $n>m$. By step 1.1 both $\mathrm{div}(1)+np=np$ and $\mathrm{div}(f)+np=(f{)}_0+(n-m)p$ contain $p$, since $n\ge 1$ and $n-m\ge 1$; by [F3] the point $p$ is a base point of $k\cdot 1+k\cdot f\subseteq L(np)$, so the hypothesis of [F4] fails and no morphism is attached. Together with step 2.1 this shows that among the divisors $np\ge (f{)}_{\mathrm{\infty }}$ the pair $1,f$ is base-point-free exactly at the pole divisor $D_0=(f{)}_{\mathrm{\infty }}$ ($n=m$).

The projective-line identity. Let $C={\mathbb{P}}_k^1$ and $f=t$, so that $(t{)}_{\mathrm{\infty }}=[\mathrm{\infty }]=D_0$ by [F6] (the coordinate section $x_0$ has divisor $[\mathrm{\infty }]$). By step 3.1 the attached morphism ${\phi }_{V_0}$ is ${\phi }_t$, finite of degree $[k({\mathbb{P}}_k^1):k(t)]=1$ with fibre over infinity the single point $[\mathrm{\infty }]$; and the converse clause of [F4], applied with $\phi ={\mathrm{i}\mathrm{d}}_{{\mathbb{P}}^1}$, $D=[\mathrm{\infty }]$ and the isomorphism $\mathcal{O}(1)\cong \mathcal{O}([\mathrm{\infty }])$ of [F6] carrying $x_0\mapsto s_1$ and $x_1\mapsto s_t$ (matching divisors $[\mathrm{\infty }]$ and $[0]$ by [F2]), shows that the morphism attached to the data $(\mathcal{O}([\mathrm{\infty }]);1,t)$ is the identity $[1:t]$. For $n\ge 2$, the same two elements of $L(n[\mathrm{\infty }])$ have $\mathrm{div}(1)+n[\mathrm{\infty }]=n[\mathrm{\infty }]$ and $\mathrm{div}(t)+n[\mathrm{\infty }]=[0]+(n-1)[\mathrm{\infty }]$, both containing $\mathrm{\infty }$, so by [F3] $\mathrm{\infty }$ is a base point of $k\cdot 1+k\cdot t\subseteq L(n[\mathrm{\infty }])$ and no morphism is attached to the larger system.

Realization and conclusion. By [F7] the example is nonempty: for every closed point $p$ of residue degree $d$ and every $n\ge 1$ with $nd+1-g\ge 2$ there is a nonconstant $f\in L(np)$ with all poles at $p$ of order at most $n$, and the construction above attaches to the pole divisor $(f{)}_{\mathrm{\infty }}$ the base-point-free pencil $k\cdot 1+k\cdot f$ with morphism exactly ${\phi }_f$; on ${\mathbb{P}}_k^1$ the case $f=t$ is the simplest instance, in which the only pole is at infinity and ${\phi }_f$ is the identity. This realizes the construction of Finite morphisms from a curve to the projective line in the single-pole case. The Axiom of Choice declared in [F8] is inherited only through the suppliers of [F2], [F4], [F5] and [F7], and nothing is selected beyond the given curve, point and function.