 Solution

For a smooth projective algebraic curve, a linear system of divisors comes from a finite-dimensional space $V\subseteq H^0(C,\mathcal{O}(D))$. A basis of sections $s_0,\dots ,s_n$ defines $f(P)=[s_0(P):\cdots :s_n(P)]$, using any local trivialization; a common scalar change does not alter the projective point. A base-point-free linear system makes this defined everywhere, and $f^{\ast }{\mathcal{O}}_{{\mathbb{P}}^n}(1)\cong \mathcal{O}(D)$. Pulling back a hyperplane gives the zero divisor of a linear combination of the sections, an element of the original system.

A system with base divisor $B$ can instead define a map after removing $B$, but its hyperplane pullbacks have class $D-B$. Thus base-point-freeness is needed for the stated original-system property. For a closed immersion, the system must also separate distinct points and tangent directions; equivalently it is a very ample linear system. On a smooth curve the latter tangent condition is separation of the length-two subscheme at each point.