Difficulty calculating with rational functions

I'm having trouble calculating the image of a point $P=[0:0:1]$ under a polynomial map:$$f:V \to \mathbb{P}^1 : [x:y:z] \mapsto [y:x]$$ where $$V = \left\{ [x:y:z] \in \mathbb{P}^2 ~\mid~ zy^2 = zx^2 + x^3 \right\}.$$

What is $f([0:0:1])$?

Notice that the given polynomials don't work at this point (but they do at every other point of $V$) since $[0:0]$ is not a valid expression for a point in $\mathbb{P}^1$. I found a different expression, $$f([x:y:z]) = [zx+x^2:zy],$$ but this expression is not defined at $[0:0:1]$, $[0:1:0]$, or $[-1:0:1]$, so it is an objectively worse expression. How do I find a third expression?

Similar problem where things work:

Let $\mathbb{P}^n$ be projective space over an algebraically closed field $k$, so that $\mathbb{P}^n$ consists of all $(n+1)$-tuples $[x_0 : x_1 : \cdots : x_n ]$ with $(x_0,x_1,\ldots,x_n) \neq (0,0,\ldots,0)$ with two tuples $[x_0 : x_1 : \cdots : x_n ] = [y_0 : y_1 : \cdots : y_n ]$ considered equivalent if there is some nonzero $\lambda$ in $k$ with $\lambda x_i = y_i$ for $i=0,1,\ldots,n$.

Let $$V = \left\{ [x:y:z] \in \mathbb{P}^2 ~\mid~ x^2 + y^2 = z^2, (x,y,z) \neq(0,0,0) \right\}$$ be the (projective) circle, and let $$W = \left\{ [ x:y ] \in \mathbb{P}^1 ~\mid~ (x,y) \neq 0 \right\}$$ be the projective line, and let $$f : V \to W : [x:y:z] \mapsto \begin{cases} [ y-z : x ] & \text{ unless } [x:y:z] = [0:1:1] \\\ [ x : y+z ] & \text{ unless } [x:y:z] = [0:-1:1] \\ \end{cases}.$$

Note that this rule for $f$ is globally valid and the function is well-defined. In other words, the two definitions agree on their overlap, and every point of $V$ is covered by one of the definitions.

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You can stop looking for a third expression for $f$: that rational function is not defined at $P$.
Note carefully that I'm not saying that $f$ has a pole at $P$: I am saying that it has a point of indeterminacy at $P$ and that $f$ can't be extended to a morphism $V\to \mathbb P^1$.

In contrast, a rational map from a smooth projective curve to a projective variety is actually a morphism i.e. it has no such indeterminacy: this explains your success with the circle .
However the point $P$ on $V$ is singular : this explains your failure [or rather the failure of $f$ :-)] with the strophoid $V$ .

And actually all this is visible with the naked eye in your example: in the affine part $z=1$ of the curve, the rational map $f$ sends the point $Q=(x,y)=[x:y:1]$ to the slope $f(Q)=y/x$ of the chord $\overline{PQ}$.
In the limit when $Q$ tends to $P$ that slope tends to $+1$ or $-1$ according to the branch of the strophoid on which $Q$ lies.
However the Zariski topology is too coarse to distinguish between these two branches and that is the geometric reason why $f$ is not defined at $P$.

Edit
The last section above shows us the way to correct this failure of $f$.
We normalize the curve $V$ so that $P$ is replaced by two points $P_1, P_{-1}$ and $f$ will send these points respectively to $1$ and $-1$. Technically the normalization is $$n:\mathbb P^1\to V:[u:v]\mapsto [u^2v-v^3:u^3-uv^2:v^3]$$ and the composition of $n$ and $f$ is the everywhere defined morphism $$\tilde f=f\circ n:\mathbb P^1\to \mathbb P^1:[u:v]\mapsto [ u^3-uv^2:u^2v-v^3]=[u:v]$$
In other words $\tilde f$ is the identity!
Check that the points above $P=[0:0:1]$ are $P_1=[1:1], P_{-1}=[-1:1]$ and that $\tilde f(P_1)=[1:1]=\text {slope }(1/1), \;\tilde f(P_{-1})=[-1:1]=\text {slope } (-1/1)$

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I suspected that the singularity is the problem, but I was missing the intuition for why the map cannot be extended. Thanks! –  Andrew Aug 7 '12 at 21:02
Ah, so using the projective line for the range allows me to handle poles nicely, but there are other problems ("branches" maybe?) and my function is not single-valued at P. Is there some easy way to find a "Riemann surface" for f, where it would be single-valued? Maybe $V$ is actually the projection of $\tilde V \subset \mathbb{P}^3$ and $\tilde f$ is well-defined on $\tilde V$? –  Jack Schmidt Aug 7 '12 at 21:02
I think $\tilde V$ must be the blowup of $V$ at the nodal point. Then there will be exactly $2$ points in the exceptional locus, and we can define the map there. –  Andrew Aug 7 '12 at 21:09
@Andrew: Can you write down such a thing using explicit polynomials? It would be ideal if $\tilde V$ just used an extra variable $t$ that we ignored to get $V$, but anything explicit and simple enough for me to calculate with is fine. –  Jack Schmidt Aug 7 '12 at 21:18
@Georges: Thanks, the slope explanation is very clear, and I suspect was the point of the exercise. (I'm still trying to make f into a single-valued function, like $f(z)=\sqrt{z}$ and its two branches.) –  Jack Schmidt Aug 7 '12 at 21:20
In order for $f=(f_0:\cdots:f_n)$ to define a rational map from some $X$ to $\mathbb P^n,$ we must have not all $f_i=0.$ This is part of the definition of a rational map, since otherwise we encounter the issue of mapping to $(0:\cdots:0),$ which makes no sense. So in your second example, you are exactly modifying the expression in order to define a rational map at every point of $V.$ In the first, you should say $f:[x:y:z]\to[y:x]$ unless if $x=y=0.$ As it stands, we have not assigned a value to $f((0:0:1)),$ and in order to do so, we must find an alternative expression for $f$ in a different chart (coinciding with the first expression on an overlap), as you did in the second example.
Over the chart $z=1$ that we've been using, the blowup $\tilde V$ of $V$ can be computed as the strict transform of $V$ under $\mathrm{Bl}_{(0,0)}\mathbb A^2\subseteq\mathbb A^2\times\mathbb P^1\to\mathbb A^2.$ Away from $(0,0),$ the point $(x,y)$ corresponds to $(x,y;x:y),$ so if e.g. $x\neq 0$ we get $(x,y;1:y/x)$ as the unique point in the fibre above $(x,y).$ (As you noticed, this is the graph of your function!) But above the origin are two points, $(0,0;1:1)$ and $(0,0;1:-1)$ corresponding to the branches of your parametrization at $(0,0).$ There is a natural map $\tilde f:\tilde V\to\mathbb P^1,$ the restriction of the projection $\mathbb A^2\times\mathbb P^1\to\mathbb P^1,$ which lifts the original $f$ (and $\tilde f:\tilde V\to\mathbb P^1$ is essentially the identity, as with the normalization).
Isn't this the rational parametrization of a node by $\mathbb P^1$? I'm not quite convinced such an expression exists. –  Andrew Aug 7 '12 at 20:53
(from Georges's answer's thread) Weird, that's almost toooo easy: that's just the graph! I've got three nice answers now (a weird $f(x)=\pm\sqrt{1+x}$ answer too), so plenty to explore. If you want to add your blow-up to your answer, I'll upvote it too. –  Jack Schmidt Aug 7 '12 at 21:47