[guided]The proof rests on a single observation: condition $(*)$ — the equality $\ell(D - p - q) = \ell(D) - 2$ — encodes *exactly two units of separation* by functions in $\mathcal{L}(D)$ at each pair of points, and these two units correspond to the two geometric requirements for an embedding.
**The two universal inequalities.** For any divisor $D$ on a smooth projective curve $C$ and any closed point $r$, the inclusion $\mathcal{L}(D - r) \subseteq \mathcal{L}(D)$ has codimension at most $1$ — the codimension is $1$ if some function in $\mathcal{L}(D)$ does not vanish at $r$ (so $r$ is not a base point), and $0$ if every function in $\mathcal{L}(D)$ vanishes at $r$ (so $r$ is a base point). Iterating gives
\begin{align*}
\ell(D) - 2 \leq \ell(D - p - q) \leq \ell(D),
\end{align*}
with strict inequality at most $2$.
**The forcing structure.** Condition $(*)$ requires the *minimum* drop $\ell(D) - 2$. By the chain $\ell(D - p - q) \geq \ell(D - p) - 1 \geq \ell(D) - 2$, the minimum is achieved only if both inequalities are equalities — meaning both intermediate codimensions are exactly $1$. This forces base-point-freeness at $p$ (first inequality) AND separation of $p$ from $q$ by some function vanishing at $p$ (second inequality). When $p = q$, the second inequality becomes "the function in $\mathcal{L}(D - p)$ has order exactly $1$ at $p$", i.e. tangent-space injectivity.
**Why "exactly two" is the right number.** Geometrically, $\phi_D$ embeds $C$ into $\mathbb{P}^N_k$ if it injects the closed points (one unit of separation per pair $p \neq q$) and the tangent vectors at each point (one more unit at $p = q$). Each unit of separation costs one dimension in $\mathcal{L}(D)$. So an embedding requires "two units of $\mathcal{L}(D)$ to be expended" at every pair $\{p, q\}$ — exactly the content of condition $(*)$.
**Why the converse direction works.** The hard part of the converse is interpreting the equality $\ell(D - p) = \ell(D) - 1$ and $\ell(D - p - q) = \ell(D - p) - 1$ as geometric properties of $\phi_D$. The first equality says "evaluation at $p$" $\mathcal{L}(D) \to k$ is surjective, equivalently, $p$ is not a base point and $\phi_D$ is defined at $p$. The second says "evaluation at $q$" $\mathcal{L}(D - p) \to k$ is also surjective, equivalently, there is a function in $\mathcal{L}(D)$ vanishing at $p$ but not at $q$ — separating $\phi_D(p)$ from $\phi_D(q)$. The case $p = q$ replaces "evaluation at $q$" by "first-order Taylor coefficient at $p$", giving tangent-space injectivity.
**The closed-immersion criterion in projective space.** The theorem we use in Step 1 — that a morphism from a smooth projective curve to $\mathbb{P}^N_k$ is a closed embedding iff it is injective on closed points and on tangent spaces — is a special case of the general criterion: a morphism of finite type between schemes over an algebraically closed field is a closed immersion iff it is proper (automatic for $C$ projective), injective on closed points, and injective on tangent spaces (equivalently, formally unramified). The smooth-curve hypothesis on $C$ guarantees that the tangent-space injectivity is exactly the differential-injectivity condition; for non-smooth $C$, one needs the formal-cotangent version of the criterion.
**The converse argument in Step 3 (tangent-space injectivity).** Locally near $p$, after a choice of uniformiser $t$ at $p$ and a local trivialisation of $\mathcal{O}_C(D)$, each $f_i \in \mathcal{L}(D)$ can be written as a Laurent expansion in $t$ near $p$. Adapting the basis $\{f_0, \ldots, f_N\}$ to the filtration $\mathcal{L}(D) \supset \mathcal{L}(D - p) \supset \mathcal{L}(D - 2p)$ — say by Gauss elimination — we may assume $f_0$ has order $0$ at $p$ (using $\ell(D) - \ell(D - p) = 1$, base-point-freeness), $f_1$ has order exactly $1$ at $p$ (using $\ell(D - p) - \ell(D - 2p) = 1$), and the remaining $f_2, \ldots, f_N$ all have order $\geq 2$. In affine coordinates near $p$, divide by $f_0$:
\begin{align*}
\phi_D(t) = \left[1 : \frac{f_1(t)}{f_0(t)} : \frac{f_2(t)}{f_0(t)} : \cdots : \frac{f_N(t)}{f_0(t)}\right].
\end{align*}
The $f_1$-coordinate behaves like $c_1 t + O(t^2)$ for some $c_1 \neq 0$ (since $f_1$ has order $1$ at $p$ and $f_0$ does not vanish), so the differential is nonzero in the $f_1$-direction — proving $d(\phi_D)_p$ is injective.
**A summary table.**
| Condition $(*)$ at $\{p, q\}$ | Geometric content |
|---|---|
| $p \neq q$, base-point-freeness at $p$ | $\phi_D$ defined at $p$ |
| $p \neq q$, separation step | $\phi_D(p) \neq \phi_D(q)$ |
| $p = q$, base-point-freeness | $\phi_D$ defined at $p$ |
| $p = q$, tangent step | $d(\phi_D)_p$ injective |[/guided]