TITLE: A map from a curve $C$ to $\mathbb{P}^n$ which represents a generator of $H^2(C,\mathbb{Z})$.
QUESTION [5 upvotes]: Let $C$ be a smooth complex curve. By the standard algebraic topology (recall that $K(\mathbb{Z},2)=\mathbb{P}^\infty$), there exists a map $\phi:C\rightarrow \mathbb{P}^n$ for sufficiently large $n$ that corresponds to a generator of $H^2(C,\mathbb{Z})$. How can one construct the map $\phi$ and can it be algebraic?

REPLY [7 votes]: It cannot be algebraic, unless $C\cong \mathbb{P}^1$. What you ask is that  $\phi^*:H^2(\mathbb{P}^n,\mathbb{Z})\rightarrow H^2(C,\mathbb{Z})$ takes the class $h$ of a hyperplane section  of $\mathbb{P}^n$ to the class of a point $p$ of $C$. But if $\phi$ is algebraic, we have $\phi^*h=rd[p]$, where $r$ is the degree of the finite map $C\rightarrow \phi(C)$ and $d=\deg(\phi(C))$. If $r=d=1$, $\phi$ is an isomorphism onto a line in $\mathbb{P}^n$.