 Solution

Choose holomorphic coordinates centered at $p$. Locally, the blowup of a complex manifold at a point iswith $\sigma (z,[\ell ])=z$; away from $p$ this is an isomorphism, so it glues to $X\setminus \{p\}$. The exceptional divisor is $E={\sigma }^{-1}(p)\cong {\mathbb{CP}}^{n-1}$.

The proper transform is the closure of ${\sigma }^{-1}(Y\setminus \{p\})$. If $p\notin Y$, it is isomorphic to $Y$. If $p\in Y$, the holomorphic implicit function theorem supplies coordinates in which $Y=\{z_1=0\}$. In the blowup chart with $z_i=t$ and $z_j=t{u}_j$ for $j\ne i$, every chart with $i\ne 1$ describes the proper transform by $u_1=0$, while the chart $i=1$ does not meet it. These are smooth coordinate hypersurfaces, so $\tilde{Y}$ is smooth.

For a divisor $D$, the line bundle associated to a divisor $\mathcal{O}(D)$ consists locally of meromorphic functions $f$ such that $(f)+D\ge 0$. Pulling back a local defining function for $Y$ shows that its divisor iswhere $m$ is the order of vanishing at $p$ of a local defining function for $Y$. ThereforeBecause $Y$ is smooth, $m=1$ when $p\in Y$ and $m=0$ when $p\notin Y$.

Applying the definition with $D=E$ gives directlyThe map sends a section to its local meromorphic coefficient relative to the canonical meromorphic section of $\mathcal{O}(E)$; the divisor inequality is exactly the condition that these coefficients define a holomorphic section, and the inverse construction is local multiplication by that canonical section.

Solved by gpt-5.6-sol high.