Extending a rational map $\psi:C\dashrightarrow\Bbb{P}^1$ into a morphism: concrete example Let all varieties be projective over $\Bbb{C}$.
It is well-known that a rational from a smooth curve to another curve extends into a morphism. My question is about a concrete example of this fact.
Take $C:y^2z=x^3-xz^2$ on $\Bbb{P}^2$ and $P=(0:0:1)$ (or some smooth cubic and a point on it). Define the rational map
\begin{align*}
\phi:\Bbb{P}^2&\dashrightarrow\Bbb{P}^1\\
(x:y:z)&\mapsto(x:y)
\end{align*}
The map is defined everywhere except at $P$. Now let $\psi:=\phi|_C:C\dashrightarrow\Bbb{P}^1$. Since $C$ is smooth, then $\psi$ extends into a morphism $\psi:C\to\Bbb{P}^1$.
My question is: how can we define $\psi(P)$?
I've tried a geometric interpretation, namely, if $L:sx+ty=0$ is any line through $P$, then $(x:y)=(-t:s)$, so the preimage $\psi^{-1}(-t:s)$ should consist (it seems to me) of $L\cap C$, which has three points counted with multiplicity, including $P$.
But by this interpretation, $P$ should be in the preimage of every point in $\Bbb{P}^1$, so it looks like I'm in the wrong path.
 A: I'd like to say a few words about the motivation for the solution and how you might reason through this if you had to do this process from scratch yourself. The key idea when extending a map from a smooth curve $C$ to projective space $\Bbb P^n$ given by a formula $$[f_0:\cdots:f_n]$$ over a point $c$ where $f_i=0$ for all $i$ is that the local ring of a curve at a regular point is a discrete valuation ring. Therefore if we let $v_i$ be the valuation of $f_i$ in $\mathcal{O}_{C,c}$, we can divide by $u^{\min v_i}$ (where $u$ has valuation 1 in $\mathcal{O}_{C,c}$) in order to get a an expression $$[g_0:\cdots:g_n]$$ for our map where the valuation of some $g_i$ is zero, hence $g_i$ does not vanish upon evaluation at $c$ and our expression is defined at $c$.
Okay, how do you actually do this? We can start by computing valuations and identifying a uniformizer. For a regular plane $C$ curve cut out by the local equation $f$ near $c$ and another curve $D$ cut out by the local equation $g$ near $c$, the intersection multiplicity of $C$ and $D$ at $c$ is $\dim \mathcal{O}_c/(f,g) = \dim \mathcal{O}_{C,c}/(g)=val_c(g)$. Therefore the equation of any line not tangent to $C$ at $c$ has valuation 1, while the equation of a tangent line will have higher valuation. In our case, our curve is given by $y^2=x^3-x$ near $c=[0:0:1]$, and we have a vertical tangent line at $[0:0:1]$, so $y$ has valuation 1 in $\mathcal{O}_{C,c}$ and we'll take $y$ as our uniformizer. From the equation of our curve, we may conclude that $x$ has valuation 2: $v(x)=v(x^3-x)=v(y^2)=2$.
To actually do our division by $y$, we need to write $x$ in terms of $y$. The equation for our curve tells us how to do that: since the equation of our curve is in the maximal ideal $(x,y)$ in the local ring, we can write it as $x\sigma=y\tau$. If we set it up so that $\tau$ is not divisible by $x$, then multiplying through by an appropriate power of $\sigma$ will let us convert $x$s to $y$s. In our case, our equation is already positioned very nicely for this: as $y^2=x^3-x$, we may take $\sigma = x^2-1$ and $\tau = y$, so $$[x:y]=[x\sigma:y\sigma]=[y\tau:y\sigma]=[\tau:\sigma]=[y:x^2-1],$$ which when evaluating at $x=y=0$ gives $[0:1]$ as expected.
A: Note that
$$\begin{align}(x:y) &= (x(x^2-z^2):y(x^2-z^2)) \\&= (y^2 z : y(x^2 -z^2)) \\ &= (yz : x^2 -z^2)\end{align}$$
on the open subset of $C$ where $x^2 \neq z^2$ and $y\neq 0$.
Since the rational $C \dashrightarrow \mathbb{P}^1, (x:y:z)\mapsto (yz:x^2 -z^2)$ is defined at $P$ and agrees with the original rational map on an open subset of $C$, we can glue these together to get an actual map $C\to \mathbb{P}^1$, which will necessarily agree with $\psi$. In particular, we find $\psi(P) = (0:1)$.
