[proofplan]
We prove the result locally and then check that the local identifications glue. On a trivializing [open set](/page/Open%20Set) for $L^{\otimes m}$, the sections $s_j$ become holomorphic functions $f_j$, and the Kodaira map is $x \mapsto [f_0(x):\cdots:f_N(x)]$. The hyperplane bundle $\mathcal{O}_{\mathbb{P}^N}(1)$ is locally generated by a nonvanishing coordinate section, so the ratios $Z_j/Z_k$ pull back to $f_j/f_k$. This gives a local isomorphism sending pulled-back homogeneous coordinates to the original sections, and the transition functions agree with those of $L^{\otimes m}$.
[/proofplan]
[step:Write the Kodaira map in a local frame of $L^{\otimes m}$]
Let $M := L^{\otimes m}$. Since $M$ is a holomorphic line bundle, choose an open set $U \subset X$ on which $M$ admits a holomorphic frame
\begin{align*}
e: U &\to M|_U.
\end{align*}
For each $j \in \{0,\dots,N\}$, define the holomorphic coefficient function
\begin{align*}
f_j: U &\to \mathbb{C}
\end{align*}
by the identity
\begin{align*}
s_j|_U = f_j e.
\end{align*}
Base point freeness means that for every $x \in U$, at least one value $s_j(x)$ is nonzero. Since $e(x)$ is a basis of the one-dimensional [vector space](/page/Vector%20Space) $M_x$, this is equivalent to saying that not all complex numbers $f_0(x),\dots,f_N(x)$ vanish. Therefore the map
\begin{align*}
\Phi_m|_U: U &\to \mathbb{P}^N \\
x &\mapsto [f_0(x):\cdots:f_N(x)]
\end{align*}
is well-defined and holomorphic.
[guided]
Set $M := L^{\otimes m}$. The point of introducing $M$ is only to simplify notation: the theorem is about the line bundle $L^{\otimes m}$.
Choose an open set $U \subset X$ on which $M$ is trivial, and choose a holomorphic frame
\begin{align*}
e: U &\to M|_U.
\end{align*}
A frame means that $e(x)$ is a basis of the one-dimensional complex vector space $M_x$ for every $x \in U$. Hence each section $s_j$ can be written uniquely on $U$ as
\begin{align*}
s_j|_U = f_j e,
\end{align*}
where
\begin{align*}
f_j: U &\to \mathbb{C}
\end{align*}
is holomorphic.
The hypothesis that $M$ is base point free says that for every $x \in X$ there is at least one section among $s_0,\dots,s_N$ that does not vanish at $x$. On $U$, this says exactly that for every $x \in U$, at least one of the coefficients $f_0(x),\dots,f_N(x)$ is nonzero, because $e(x)$ is itself nonzero. Thus the projective point
\begin{align*}
[f_0(x):\cdots:f_N(x)] \in \mathbb{P}^N
\end{align*}
is defined for every $x \in U$. In this frame, the Kodaira map is therefore
\begin{align*}
\Phi_m|_U: U &\to \mathbb{P}^N \\
x &\mapsto [f_0(x):\cdots:f_N(x)].
\end{align*}
[/guided]
[/step]
[step:Construct the local isomorphism over the standard affine charts]
For each $k \in \{0,\dots,N\}$, let
\begin{align*}
V_k := \{[z_0:\cdots:z_N] \in \mathbb{P}^N : z_k \neq 0\}
\end{align*}
be the standard affine chart. The section $Z_k$ of $\mathcal{O}_{\mathbb{P}^N}(1)$ is nonvanishing on $V_k$, so $\Phi_m^*Z_k$ is a holomorphic frame of $\Phi_m^*\mathcal{O}_{\mathbb{P}^N}(1)$ on $\Phi_m^{-1}(V_k)$.
On $U \cap \Phi_m^{-1}(V_k)$, define a holomorphic line bundle map
\begin{align*}
\Psi_{U,k}: \Phi_m^*\mathcal{O}_{\mathbb{P}^N}(1)|_{U \cap \Phi_m^{-1}(V_k)} &\to M|_{U \cap \Phi_m^{-1}(V_k)}
\end{align*}
by prescribing its value on the frame $\Phi_m^*Z_k$:
\begin{align*}
\Psi_{U,k}(\Phi_m^*Z_k) := f_k e.
\end{align*}
Since $x \in \Phi_m^{-1}(V_k)$ implies $f_k(x) \neq 0$, the section $f_k e$ is a holomorphic frame of $M$ on $U \cap \Phi_m^{-1}(V_k)$. Hence $\Psi_{U,k}$ is a holomorphic line bundle isomorphism on this open set.
[guided]
Fix an index $k \in \{0,\dots,N\}$. The standard affine chart
\begin{align*}
V_k := \{[z_0:\cdots:z_N] \in \mathbb{P}^N : z_k \neq 0\}
\end{align*}
is exactly the region where the homogeneous coordinate section $Z_k$ does not vanish. Therefore $Z_k$ is a frame of $\mathcal{O}_{\mathbb{P}^N}(1)$ over $V_k$, and its pullback $\Phi_m^*Z_k$ is a frame of $\Phi_m^*\mathcal{O}_{\mathbb{P}^N}(1)$ over $\Phi_m^{-1}(V_k)$.
On the intersection $U \cap \Phi_m^{-1}(V_k)$, define
\begin{align*}
\Psi_{U,k}: \Phi_m^*\mathcal{O}_{\mathbb{P}^N}(1)|_{U \cap \Phi_m^{-1}(V_k)} &\to M|_{U \cap \Phi_m^{-1}(V_k)}
\end{align*}
by declaring
\begin{align*}
\Psi_{U,k}(\Phi_m^*Z_k) := f_k e.
\end{align*}
This determines a unique holomorphic bundle map because $\Phi_m^*Z_k$ is a frame. It is an isomorphism because $f_k e$ is also a frame: if $x \in \Phi_m^{-1}(V_k)$, then the $k$-th homogeneous coordinate of $\Phi_m(x)$ is nonzero, and in the local expression $\Phi_m(x)=[f_0(x):\cdots:f_N(x)]$ this means $f_k(x)\neq 0$.
[/guided]
[/step]
[step:Check compatibility on overlaps of affine charts]
Let $j,k \in \{0,\dots,N\}$. Define the holomorphic transition function
\begin{align*}
r_{jk}: V_j \cap V_k &\to \mathbb{C} \\
[\zeta_0:\cdots:\zeta_N] &\mapsto \frac{\zeta_j}{\zeta_k}.
\end{align*}
This is well-defined because replacing the nonzero representative $(\zeta_0,\dots,\zeta_N) \in \mathbb{C}^{N+1}\setminus\{0\}$ by a scalar multiple leaves the quotient $\zeta_j/\zeta_k$ unchanged. On $V_j \cap V_k$, the homogeneous coordinate sections satisfy
\begin{align*}
Z_j = r_{jk} Z_k.
\end{align*}
Pulling back by $\Phi_m$ and using $\Phi_m|_U(x)=[f_0(x):\cdots:f_N(x)]$, we obtain on $U \cap \Phi_m^{-1}(V_j \cap V_k)$:
\begin{align*}
\Phi_m^*Z_j = \frac{f_j}{f_k}\Phi_m^*Z_k.
\end{align*}
Therefore
\begin{align*}
\Psi_{U,k}(\Phi_m^*Z_j)
&= \Psi_{U,k}\left(\frac{f_j}{f_k}\Phi_m^*Z_k\right) \\
&= \frac{f_j}{f_k} f_k e \\
&= f_j e \\
&= s_j|_U.
\end{align*}
In particular, taking $j$ as the distinguished nonvanishing coordinate gives
\begin{align*}
\Psi_{U,k}(\Phi_m^*Z_j)=\Psi_{U,j}(\Phi_m^*Z_j)
\end{align*}
on the overlap, so the maps $\Psi_{U,k}$ agree on common domains.
[guided]
We must verify that the local definitions do not depend on which nonzero homogeneous coordinate we use. Let $j,k \in \{0,\dots,N\}$. Define
\begin{align*}
r_{jk}: V_j \cap V_k &\to \mathbb{C} \\
[\zeta_0:\cdots:\zeta_N] &\mapsto \frac{\zeta_j}{\zeta_k}.
\end{align*}
This map is well-defined because if $(\zeta_0,\dots,\zeta_N)$ is replaced by $(\lambda\zeta_0,\dots,\lambda\zeta_N)$ with $\lambda \in \mathbb{C}^\times$, then the quotient remains
\begin{align*}
\frac{\lambda\zeta_j}{\lambda\zeta_k}=\frac{\zeta_j}{\zeta_k}.
\end{align*}
On the projective affine overlap $V_j \cap V_k$, both $Z_j$ and $Z_k$ are frames of $\mathcal{O}_{\mathbb{P}^N}(1)$, and their transition relation is
\begin{align*}
Z_j = r_{jk} Z_k.
\end{align*}
This is the standard transition rule for the hyperplane bundle: the function $r_{jk}$ is the holomorphic affine coordinate quotient on $V_j \cap V_k$.
Pull this identity back along $\Phi_m$. Since in the frame $e$ we have
\begin{align*}
\Phi_m(x)=[f_0(x):\cdots:f_N(x)],
\end{align*}
the pulled-back transition function is
\begin{align*}
(r_{jk}\circ \Phi_m)(x)=\frac{f_j(x)}{f_k(x)}.
\end{align*}
Hence on $U \cap \Phi_m^{-1}(V_j \cap V_k)$,
\begin{align*}
\Phi_m^*Z_j = \frac{f_j}{f_k}\Phi_m^*Z_k.
\end{align*}
Applying the map $\Psi_{U,k}$ gives
\begin{align*}
\Psi_{U,k}(\Phi_m^*Z_j)
&= \Psi_{U,k}\left(\frac{f_j}{f_k}\Phi_m^*Z_k\right) \\
&= \frac{f_j}{f_k}\Psi_{U,k}(\Phi_m^*Z_k) \\
&= \frac{f_j}{f_k} f_k e \\
&= f_j e \\
&= s_j|_U.
\end{align*}
The same computation with $\Psi_{U,j}$ gives $\Psi_{U,j}(\Phi_m^*Z_j)=s_j|_U$. Since $\Phi_m^*Z_j$ is a frame on the overlap, the two bundle maps agree there. Thus the local isomorphisms are compatible across affine chart overlaps.
[/guided]
[/step]
[step:Check compatibility under changes of frame for $L^{\otimes m}$]
Let $U'$ be another trivializing open set for $M$, with holomorphic frame
\begin{align*}
e': U' &\to M|_{U'}.
\end{align*}
On $U \cap U'$, there is a nowhere-vanishing [holomorphic function](/page/Holomorphic%20Function)
\begin{align*}
g: U \cap U' &\to \mathbb{C}^\times
\end{align*}
such that $e'=g e$. Writing
\begin{align*}
s_j|_{U'} = f'_j e',
\end{align*}
we have on $U \cap U'$:
\begin{align*}
f'_j g e = f'_j e' = s_j = f_j e,
\end{align*}
and hence $f'_j g=f_j$ for every $j$.
For $x \in U \cap U' \cap \Phi_m^{-1}(V_k)$,
\begin{align*}
\Psi_{U',k}(\Phi_m^*Z_k)
= f'_k e'
= f'_k g e
= f_k e
= \Psi_{U,k}(\Phi_m^*Z_k).
\end{align*}
Since $\Phi_m^*Z_k$ is a frame on this set, $\Psi_{U',k}=\Psi_{U,k}$ there. Thus the local isomorphisms are also compatible under changes of trivialization of $M$.
[/step]
[step:Glue the local isomorphisms and identify the coordinate sections]
The sets $U \cap \Phi_m^{-1}(V_k)$, as $U$ ranges over trivializing open sets for $M$ and $k$ ranges over $\{0,\dots,N\}$, cover $X$. The compatibility checks above show that the local holomorphic line bundle isomorphisms $\Psi_{U,k}$ glue to a global holomorphic line bundle isomorphism
\begin{align*}
\Psi: \Phi_m^*\mathcal{O}_{\mathbb{P}^N}(1) &\to M.
\end{align*}
Since $M=L^{\otimes m}$, this gives
\begin{align*}
\Phi_m^*\mathcal{O}_{\mathbb{P}^N}(1) \cong L^{\otimes m}.
\end{align*}
Finally, for each $j \in \{0,\dots,N\}$ and for every local trivializing open set $U$, the overlap computation gives
\begin{align*}
\Psi(\Phi_m^*Z_j)|_U = s_j|_U.
\end{align*}
Because these identities hold on a cover of $X$, they glue to the global identity
\begin{align*}
\Psi(\Phi_m^*Z_j)=s_j.
\end{align*}
Thus the homogeneous coordinate sections pull back to $s_0,\dots,s_N$ under the isomorphism, completing the proof.
[/step]
