[proofplan] The projective map is controlled locally by ratios of sections. Near a point where a section $s_0$ does not vanish, the affine coordinates of $\Phi_V$ are the functions $s_j/s_0$, so injectivity is exactly separation of points and injectivity of the differential is exactly separation of tangent vectors. Once injectivity and immersion are established, compactness of $X$ upgrades the injective holomorphic immersion into a holomorphic embedding by the standard topological and local holomorphic inverse-function arguments. [/proofplan]

[step:Express the projective map in affine coordinates by ratios of sections]Let $r=\dim_{\mathbb{C}}V-1$, and choose a complex basis $s_0,\dots,s_r$ of $V$. Let $x \in X$. Since $V$ is base point free, after reordering the basis if necessary we may assume $s_0(x)\neq 0$. Choose an open neighbourhood $U \subset X$ of $x$ such that $s_0$ has no zero on $U$, and choose a holomorphic local trivialisation \begin{align*} \tau: L|_U &\to U \times \mathbb{C}. \end{align*} For each $j \in \{0,\dots,r\}$, define the [holomorphic function](/page/Holomorphic%20Function) \begin{align*} f_j: U &\to \mathbb{C} \end{align*} by the identity $\tau(s_j|_U)=(\cdot,f_j)$. Since $f_0$ has no zero on $U$, define \begin{align*} g_j: U &\to \mathbb{C} \\ z &\mapsto \frac{f_j(z)}{f_0(z)} \end{align*} for $j=1,\dots,r$. On the affine chart of $\mathbb{P}(V^*)$ where the coordinate dual to $s_0$ is nonzero, the map $\Phi_V$ is represented by \begin{align*} \Phi_V|_U(z) = \bigl(g_1(z),\dots,g_r(z)\bigr). \end{align*} Indeed, under the basis $s_0,\dots,s_r$, the functional $\operatorname{ev}_z$ is represented up to the common scalar determined by the trivialisation by \begin{align*} (f_0(z),f_1(z),\dots,f_r(z)), \end{align*} and division by the nonzero first coordinate gives the affine coordinates above.[/step]

[guided]The map $\Phi_V$ sends a point $z \in X$ to the projective class of the evaluation functional at $z$. To compute it locally, we choose a basis $s_0,\dots,s_r$ of $V$ and a local trivialisation of the line bundle. In that trivialisation, each section $s_j$ becomes an ordinary holomorphic function \begin{align*} f_j: U &\to \mathbb{C}. \end{align*} Because $V$ is base point free, some section in $V$ does not vanish at the chosen point $x$. We put that section first and call it $s_0$. Shrinking the neighbourhood $U$ if necessary, $s_0$ remains nonvanishing on $U$, so $f_0(z)\neq 0$ for every $z \in U$. Projective space is described locally by affine charts. In the chart where the first homogeneous coordinate is nonzero, the projective class \begin{align*} [f_0(z):f_1(z):\cdots:f_r(z)] \end{align*} has affine coordinates \begin{align*} \left(\frac{f_1(z)}{f_0(z)},\dots,\frac{f_r(z)}{f_0(z)}\right). \end{align*} Thus the local expression of $\Phi_V$ is exactly the holomorphic map \begin{align*} z &\mapsto \bigl(g_1(z),\dots,g_r(z)\bigr), \end{align*} where $g_j=f_j/f_0$. This ratio description is the bridge between the projective geometry of sections and the differential geometry of the map $\Phi_V$.[/guided]

[step:Identify point separation with injectivity of $\Phi_V$]For $x,y \in X$, define the evaluation maps \begin{align*} \operatorname{ev}_x: V &\to L_x, & \operatorname{ev}_y: V &\to L_y \end{align*} by $\operatorname{ev}_x(s)=s(x)$ and $\operatorname{ev}_y(s)=s(y)$. These maps are nonzero because $V$ is base point free. Choosing nonzero linear identifications $L_x\cong\mathbb{C}$ and $L_y\cong\mathbb{C}$ turns them into nonzero elements of $V^*$; changing either identification multiplies the corresponding element of $V^*$ by a nonzero scalar, so the induced projective point and the kernel in $V$ are independent of the choices. Hence $\Phi_V(x)=\Phi_V(y)$ holds exactly when \begin{align*} \ker(\operatorname{ev}_x)=\ker(\operatorname{ev}_y). \end{align*} Therefore $\Phi_V$ is injective if and only if, for every $x\neq y$, one has \begin{align*} \ker(\operatorname{ev}_x) \neq \ker(\operatorname{ev}_y), \end{align*} which is precisely the condition that $V$ separates points.[/step]

[guided]We want to translate the condition that $\Phi_V$ is injective into a statement about sections. For each point $x \in X$, the evaluation map \begin{align*} \operatorname{ev}_x: V &\to L_x \end{align*} is a nonzero [linear map](/page/Linear%20Map) because $V$ is base point free. Since $L_x$ is one-dimensional, the projective point $\Phi_V(x)$ records the line spanned by this nonzero functional. Two nonzero linear functionals on a finite-dimensional [vector space](/page/Vector%20Space) determine the same projective point if and only if they differ by multiplication by a nonzero scalar. This is equivalent to their kernels being equal: if $\lambda,\mu:V\to\mathbb{C}$ are nonzero and $\mu=c\lambda$ for some $c\in\mathbb{C}^\times$, then $\ker\lambda=\ker\mu$; conversely, if their kernels agree, both descend to nonzero linear maps from the one-dimensional quotient $V/\ker\lambda$ to $\mathbb{C}$, so they differ by a nonzero scalar. Thus \begin{align*} \Phi_V(x)=\Phi_V(y) \end{align*} holds exactly when \begin{align*} \ker(\operatorname{ev}_x)=\ker(\operatorname{ev}_y). \end{align*} Consequently $\Phi_V$ is injective exactly when distinct points have distinct evaluation kernels, which is the stated point-separation condition.[/guided]

[step:Identify tangent separation with injectivity of the differential]Fix $x \in X$ and $\xi \in T_xX$. Choose $s_0 \in V$ with $s_0(x)\neq 0$, extend it to a basis $s_0,\dots,s_r$ of $V$, and use the local notation $f_j$ and $g_j=f_j/f_0$ from the first step. The differential \begin{align*} d(\Phi_V)_x: T_xX &\to T_{\Phi_V(x)}\mathbb{P}(V^*) \end{align*} is represented in the chosen affine chart by \begin{align*} \xi &\mapsto \bigl(dg_{1,x}(\xi),\dots,dg_{r,x}(\xi)\bigr). \end{align*} For $j=1,\dots,r$, define a section \begin{align*} t_j := s_j - g_j(x)s_0 \in V. \end{align*} Then $t_j(x)=0$, and in the chosen trivialisation $t_j$ is represented by the holomorphic function \begin{align*} h_j: U &\to \mathbb{C} \\ z &\mapsto f_j(z)-g_j(x)f_0(z). \end{align*} Since $g_j=f_j/f_0$, the quotient rule at $x$ gives \begin{align*} dg_{j,x}(\xi) &= \frac{dh_{j,x}(\xi)}{f_0(x)}. \end{align*} Because $f_0(x)\neq 0$, the equality $dg_{j,x}(\xi)=0$ is equivalent to $dh_{j,x}(\xi)=0$. Thus $d(\Phi_V)_x(\xi)=0$ if and only if every section $s\in V$ satisfying $s(x)=0$ has local representative $f_s$ with $df_{s,x}(\xi)=0$. Hence $d(\Phi_V)_x$ is injective if and only if $V$ separates tangent vectors at $x$.[/step]

[guided]The local formula for $\Phi_V$ reduces the differential to the derivatives of the ratios $g_j=f_j/f_0$. We must connect those derivatives with sections that vanish at $x$, because tangent-vector separation is stated using such sections. Choose $s_0\in V$ with $s_0(x)\neq 0$ and extend it to a basis $s_0,\dots,s_r$. In the affine chart determined by $s_0$, the map $\Phi_V$ is represented by \begin{align*} z &\mapsto \bigl(g_1(z),\dots,g_r(z)\bigr), \end{align*} where \begin{align*} g_j:U&\to\mathbb{C},\\ z&\mapsto \frac{f_j(z)}{f_0(z)}. \end{align*} Therefore $d(\Phi_V)_x(\xi)=0$ precisely when \begin{align*} dg_{j,x}(\xi)=0 \end{align*} for every $j=1,\dots,r$. Now define \begin{align*} t_j := s_j-g_j(x)s_0 \in V. \end{align*} This section vanishes at $x$, because in the trivialisation its representative is \begin{align*} h_j(z)=f_j(z)-g_j(x)f_0(z), \end{align*} and \begin{align*} h_j(x)=f_j(x)-\frac{f_j(x)}{f_0(x)}f_0(x)=0. \end{align*} Since $g_j=f_j/f_0$, differentiating at $x$ gives \begin{align*} dg_{j,x}(\xi) &= d\left(\frac{f_j}{f_0}\right)_x(\xi) \\ &= \frac{df_{j,x}(\xi)f_0(x)-f_j(x)df_{0,x}(\xi)}{f_0(x)^2} \\ &= \frac{d(f_j-g_j(x)f_0)_x(\xi)}{f_0(x)} \\ &= \frac{dh_{j,x}(\xi)}{f_0(x)}. \end{align*} Because $f_0(x)\neq 0$, this vanishes exactly when $dh_{j,x}(\xi)=0$. Every section $s\in V$ with $s(x)=0$ can be written as a linear combination of the sections $t_1,\dots,t_r$. Indeed, writing $s=\sum_{j=0}^r a_js_j$, the condition $s(x)=0$ implies \begin{align*} a_0f_0(x)+\sum_{j=1}^r a_jf_j(x)=0, \end{align*} so \begin{align*} a_0=-\sum_{j=1}^r a_j\frac{f_j(x)}{f_0(x)}=-\sum_{j=1}^r a_jg_j(x), \end{align*} and hence \begin{align*} s=\sum_{j=1}^r a_j(s_j-g_j(x)s_0)=\sum_{j=1}^r a_jt_j. \end{align*} Therefore the derivatives of all vanishing sections in the direction $\xi$ vanish if and only if all $dg_{j,x}(\xi)$ vanish. This proves that $d(\Phi_V)_x$ is injective exactly when $V$ separates tangent vectors at $x$.[/guided]

[step:Deduce the forward implication from the definition of embedding] Assume $\Phi_V$ is a holomorphic embedding. Then $\Phi_V$ is injective, so by the previous point-separation equivalence, $V$ separates points. Also $d(\Phi_V)_x$ is injective for every $x\in X$, so by the tangent-separation equivalence, $V$ separates tangent vectors at every point of $X$. Thus a holomorphic embedding implies both separation conditions. [/step]

[step:Use the separation conditions to prove that $\Phi_V$ is an embedding]Assume now that $V$ separates points and tangent vectors. By the point-separation equivalence, $\Phi_V$ is injective. By the tangent-separation equivalence, $d(\Phi_V)_x$ is injective for every $x\in X$, so $\Phi_V$ is a holomorphic immersion. It remains to upgrade an injective holomorphic immersion to a holomorphic embedding. Since $X$ is compact and $\mathbb{P}(V^*)$ is Hausdorff, the continuous injective map \begin{align*} \Phi_V: X &\to \Phi_V(X) \end{align*} is a homeomorphism onto its image. Also, because $d(\Phi_V)_x$ is injective at every $x$, the rank of $d(\Phi_V)_x$ is constantly $\dim_{\mathbb{C}}X$ on $X$. Thus the holomorphic constant-rank local normal form for immersions applies around each $x\in X$ and gives holomorphic coordinates in which $\Phi_V$ is locally the standard inclusion of a complex coordinate subspace. Therefore $\Phi_V(X)$ is locally a complex submanifold of $\mathbb{P}(V^*)$, and $\Phi_V$ is locally biholomorphic from $X$ onto this image. Combining the homeomorphism onto the image with the local holomorphic immersion normal form, $\Phi_V$ is a holomorphic embedding. This proves the reverse implication and completes the proof.[/step]

[guided]We now assume the two separation conditions and prove that $\Phi_V$ is an embedding. First, point separation says that distinct points have distinct evaluation kernels. By the equivalence already proved, this means $\Phi_V$ is injective. Second, tangent-vector separation says that for each nonzero $\xi\in T_xX$, some section vanishing at $x$ has nonzero derivative in the direction $\xi$. By the differential computation above, this means \begin{align*} d(\Phi_V)_x(\xi)\neq 0 \end{align*} for every nonzero $\xi\in T_xX$. Hence \begin{align*} d(\Phi_V)_x:T_xX\to T_{\Phi_V(x)}\mathbb{P}(V^*) \end{align*} is injective for every $x\in X$, so $\Phi_V$ is a holomorphic immersion. An embedding requires more than injectivity and injectivity of the differential: the map must also identify $X$ with its image as a complex submanifold. The global topological part comes from compactness. Since $X$ is compact and projective space is Hausdorff, the continuous injective map \begin{align*} \Phi_V:X\to \Phi_V(X) \end{align*} is a homeomorphism onto its image. The local holomorphic part comes from the holomorphic constant-rank normal form for immersions (citing a result not yet in the wiki: Holomorphic Constant Rank Theorem). Let $n:=\dim_{\mathbb{C}}X$. Its hypotheses are satisfied because $\Phi_V$ is holomorphic and $d(\Phi_V)_x$ has constant rank equal to $n$ near each point. Therefore, for each $x\in X$, there are holomorphic coordinates near $x$ and near $\Phi_V(x)$ in which $\Phi_V$ is represented as \begin{align*} (z_1,\dots,z_n)\mapsto (z_1,\dots,z_n,0,\dots,0). \end{align*} In particular, the image is locally a complex submanifold and $\Phi_V$ is locally biholomorphic onto that image. The global homeomorphism property and the local holomorphic normal form together say exactly that $\Phi_V$ is a holomorphic embedding. This proves the converse implication.[/guided]