A question about the function field of the cuspidal cubic I'm trying to understand function fields of affine varieties. In the book I'm reading, a function field of a variety $V = V(I)$ was defined to be the set
$$
K_V = \left\{\frac{f}{g}, f,g\in \mathcal{O}(V) : g \notin I\right\}
$$
where $\frac{f_1}{f_2} \sim \frac{g_1}{g_2} \iff f_1 g_2 - f_2 g_1 \in I$.
Now to understand this better I'm trying to the following exercise:

First of all I'm struggling to understand of the domain of the function $\phi^*$, why is it the function field of $V$, $K_V$? The map $\phi$ has codomain $V$ right, but I don't quite what is $F$ here. Also to show that $\phi^*$ is a homomorphism I need to show that $\phi^*(FG) = \phi^*(F)\phi^*(G)$, but not really sure how to proceed here.
 A: An element $F$ of the function field $K_V$ is equivalent to a rational map $\hat{F}: V \dashrightarrow \mathbb{P}^1$. We can pre-compose this with $\phi$ to get a map
$$
\mathbb{A}^1 \overset{\phi}{\to} V \overset{\hat{F}}{\dashrightarrow} \mathbb{P}^1 \, .
$$
So we get a rational map (and in fact, a morphism) $\mathbb{A}^1 \dashrightarrow \mathbb{P}^1$, which as above, is the same thing as an element of the function field of $\mathbb{A}^1$, which is $k(t)$, the field of rational functions in one variable. Thus the morphism $\phi: \mathbb{A}^1 \to V$ induces a $k$-algebra homomorphism by pre-composition:
\begin{align*}
\phi^*: K_V &\to k(t)\\
F &\mapsto F \circ \phi
\end{align*}
In this way, we get a contravariant functor from the category of varieties over $k$ to the category of function fields over $k$.
For instance, in your example where $V: y^2 = x^3$ is the cuspidal cubic and $\phi: \mathbb{A}^1 \to V, t \mapsto (t^2, t^3)$, then
$$
\phi^*(f) = f(\phi(t)) = f(t^2,t^3)
$$
for all $f \in K_V$.
