[proofplan] We construct the isomorphism locally from defining equations for the divisor. On a coordinate neighbourhood where $Y$ is cut out by a [holomorphic function](/page/Holomorphic%20Function) $f$, the differential $df$ annihilates tangent vectors to $Y$, so it descends to a linear functional on the normal line. Using the local frame of $\mathcal{O}_X(Y)$ in which the canonical section is represented by $f$, this functional identifies the normal line with $\mathcal{O}_X(Y)|_Y$. The only point to check is compatibility on overlaps; this follows because both $df$ along $Y$ and the local frame of $\mathcal{O}_X(Y)$ transform by the same nowhere-zero factor. [/proofplan]

[step:Choose local defining equations and frames for the divisor line bundle] Let $(U_i)_{i \in I}$ be an open cover of $X$ such that, for each $i \in I$, there is a holomorphic function $f_i: U_i \to \mathbb{C}$ with \begin{align*} Y \cap U_i = \{p \in U_i : f_i(p) = 0\}, \end{align*} and such that $df_i|_p \neq 0$ for every $p \in Y \cap U_i$. This is possible because $Y$ is a smooth effective divisor. For each $i \in I$, let \begin{align*} e_i: U_i \to \mathcal{O}_X(Y)|_{U_i} \end{align*} be the local holomorphic frame determined by the convention that the canonical section $s_Y$ is represented on $U_i$ by \begin{align*} s_Y|_{U_i} = f_i e_i. \end{align*} On an overlap $U_i \cap U_j$, the two local defining functions define the same effective Cartier divisor, so there is a nowhere-zero holomorphic function \begin{align*} u_{ij}: U_i \cap U_j \to \mathbb{C}^\times \end{align*} such that \begin{align*} f_i = u_{ij} f_j. \end{align*} The equality $f_i e_i = f_j e_j$ for the canonical section gives \begin{align*} e_j = u_{ij} e_i \end{align*} on $U_i \cap U_j$. [/step]

[step:Use the differential of a defining equation to identify the local normal line]Fix $i \in I$ and $p \in Y \cap U_i$. Let \begin{align*} q_p: T_pX \to N_{Y/X,p} := T_pX / T_pY \end{align*} denote the quotient map. Since $Y \cap U_i$ is the zero set of $f_i$, every vector $v \in T_pY$ satisfies \begin{align*} df_i|_p(v) = 0. \end{align*} Therefore the [linear map](/page/Linear%20Map) \begin{align*} df_i|_p: T_pX \to \mathbb{C} \end{align*} factors uniquely through the quotient $q_p$. Define \begin{align*} \Phi_i|_p: N_{Y/X,p} \to \mathcal{O}_X(Y)|_p \end{align*} by \begin{align*} \Phi_i|_p(q_p(v)) = df_i|_p(v) e_i(p) \end{align*} for $v \in T_pX$. This is well-defined because replacing $v$ by $v+w$ with $w \in T_pY$ does not change $df_i|_p(v)$. Since $df_i|_p \neq 0$ and $Y$ has codimension $1$, the induced map $N_{Y/X,p} \to \mathbb{C}$ is a nonzero linear map between one-dimensional complex vector spaces. Hence $\Phi_i|_p$ is a complex-linear isomorphism for every $p \in Y \cap U_i$. As $df_i$ and $e_i$ vary holomorphically, these fibrewise maps assemble into a holomorphic line bundle isomorphism \begin{align*} \Phi_i: N_{Y/X}|_{Y \cap U_i} \to \mathcal{O}_X(Y)|_{Y \cap U_i}. \end{align*}[/step]

[guided]The reason the differential appears is that the tangent space to a smooth hypersurface is exactly the kernel of the differential of a local defining equation. Fix $p \in Y \cap U_i$. The defining equation is a holomorphic map $f_i: U_i \to \mathbb{C}$ with $Y \cap U_i = f_i^{-1}(0)$ and $df_i|_p \neq 0$. Since $f_i$ is identically zero on $Y \cap U_i$, differentiating the restriction of $f_i$ to $Y \cap U_i$ gives \begin{align*} df_i|_p(v) = 0 \end{align*} for every $v \in T_pY$. Thus $df_i|_p$ cannot distinguish two tangent vectors in $T_pX$ that differ by a vector tangent to $Y$. If \begin{align*} q_p: T_pX \to T_pX / T_pY \end{align*} is the quotient map, this means $df_i|_p$ descends to a well-defined linear map on the quotient. We define the local bundle map by \begin{align*} \Phi_i|_p(q_p(v)) = df_i|_p(v) e_i(p), \end{align*} where $e_i(p)$ is the value at $p$ of the local frame of $\mathcal{O}_X(Y)$ satisfying $s_Y|_{U_i} = f_i e_i$. We must check that this is an isomorphism. The normal space $N_{Y/X,p}=T_pX/T_pY$ is one-dimensional because $Y$ is a smooth divisor, hence a smooth hypersurface. The induced linear functional on this one-dimensional quotient is nonzero because $df_i|_p \neq 0$. A nonzero linear map between one-dimensional complex vector spaces is an isomorphism. Therefore $\Phi_i|_p$ is an isomorphism for every $p \in Y \cap U_i$. Since the formula uses only the holomorphic differential $df_i$ and the holomorphic frame $e_i$, these pointwise maps vary holomorphically with $p$, giving a holomorphic line bundle isomorphism \begin{align*} \Phi_i: N_{Y/X}|_{Y \cap U_i} \to \mathcal{O}_X(Y)|_{Y \cap U_i}. \end{align*}[/guided]

[step:Check that the local identifications agree on overlaps] Let $p \in Y \cap U_i \cap U_j$ and let $v \in T_pX$. On $U_i \cap U_j$ we have $f_i = u_{ij} f_j$, where $u_{ij}$ is nowhere zero and holomorphic. Differentiating at $p$ gives \begin{align*} df_i|_p(v) = d(u_{ij} f_j)|_p(v). \end{align*} By the product rule, \begin{align*} d(u_{ij} f_j)|_p(v) = u_{ij}(p) df_j|_p(v) + f_j(p) du_{ij}|_p(v). \end{align*} Since $p \in Y$ and $f_j$ vanishes on $Y$, we have $f_j(p)=0$. Hence \begin{align*} df_i|_p(v) = u_{ij}(p) df_j|_p(v). \end{align*} Using also $e_j = u_{ij} e_i$, we obtain \begin{align*} df_i|_p(v) e_i(p) = u_{ij}(p) df_j|_p(v) e_i(p) = df_j|_p(v) e_j(p). \end{align*} Therefore \begin{align*} \Phi_i|_p(q_p(v)) = \Phi_j|_p(q_p(v)). \end{align*} Thus the local maps $\Phi_i$ and $\Phi_j$ agree on $Y \cap U_i \cap U_j$. [/step]

[step:Glue the local maps to obtain the canonical global isomorphism] Because the local holomorphic isomorphisms \begin{align*} \Phi_i: N_{Y/X}|_{Y \cap U_i} \to \mathcal{O}_X(Y)|_{Y \cap U_i} \end{align*} agree on all pairwise overlaps, they glue to a global holomorphic line bundle isomorphism \begin{align*} \Phi: N_{Y/X} \to \mathcal{O}_X(Y)|_Y. \end{align*} The construction depends only on the divisor $Y$, its canonical section $s_Y$, and the local defining equations used to represent that section; the overlap computation shows that changing the local defining equation does not change the glued map. Hence the isomorphism is canonical. This proves \begin{align*} N_{Y/X} \cong \mathcal{O}_X(Y)|_Y. \end{align*} [/step]