Why is $K(U)$ =$K(X)$ for $U\subset X$ where both $U$ and $X$ are quasi projective? I am self studying algebraic geometry from these notes: https://www.dpmms.cam.ac.uk/~cb496/ag2007-final.pdf
Here is the definition of regular functions on quasi projective sets:
Let $X$ be a quasi-projective algebraic set. A function $\phi: X \rightarrow k$ is called a regular function if for every $x \in X$, there is a neighborhood $U$ of $x$, and homogeneous polynomials $F, G$ of the same degree such that on $U, \phi$ and $F / G$ are equal, in particular, $G$ has no zero on $U$. The set of regular functions on $X$ is denoted by $k[X]$ which is a $k$-algebra.
If $X$ is irreducible, a rational function $\pi: X \rightarrow k$ on $X$ is the equivalence class of a regular function on some open subset of $X$ in the sense that if $\phi_U$ and $\phi_V$ are regular functions on the open subsets $U$ and $V$ respectively, then $\phi_U$ is equivalent to $\phi_V$ if $\left.\phi_U\right|_{U \cap V}=\left.\phi_V\right|_{U \cap V}$. A rational function then is uniquely determined by some $F / G$ where $G$ is not identically zero on $X$. The set of rational functions on $X$ is denoted by $k(X)$ which is a field and is called the function field of $X$.
Note that for a quasi-projective algebraic variety $X$, unlike the affine case, $k(X)$ is not necessarily the fraction field of $k[X]$.
Exercise: Let $U \neq \emptyset$ be an open subset of a quasi-projective variety $X$. Prove that $k(U)=k(X)$.
So I have two questions: what does it mean for a rational function to be "uniquely determined by some $F / G$ where $G$ is not identically zero on $X$" and why is that the case?
And for the exercise, what does it mean to say that $k(U)=k(X)$?. I would assume we mean they are isomorphic, but saying they are literally equal feels strange because functions in the first set have open sets of $U$ as their domain and functions in the second set have open sets of $X$ so functions can't be equal if they don't even have the same domain so what does that mean?
 A: As stated in the definition, a rational function is an equivalence class of regular functions on some open set. To make this more concrete, we can write
$$
k(X)=\{(s,U)\mid U\subseteq X\text{ non-empty open set, }s\in k[U]\}/\sim
$$
where $(s,U)\sim (t,V)$ if and only if $s|_{U\cap V}=t|_{U\cap V}$ (note that any two non-empty open sets have non-trivial intersection). Write $[s,U]$ for the equivalence class of $s\in k[U]$. Then addition of $k(X)$ is defined as
$$
[s,U]+[t,V]:=[s|_{U\cap V}+t|_{U\cap V},U\cap V]
$$
and similar for multiplication. You can check that this makes $k(X)$ a field.
Now for your first question, if $\phi\in k(X)$ is a rational function, this means it is equal to $[s,U]$ for some open set $U$ and a regular function $s:U\to k$. By shrinking $U$ we may suppose that $s$ is of the form $F/G$ with $G$ non-zero on $U$. Hence, every $\phi$ can be written as the equivalence class of some $F/G$ (but note that there could be $F'/G'$ defined on another open set, but equivalent to $F/G$).
For your second question, this is basically immediate, once you understand what the elements of $k(U)$ resp. $k(X)$ really are. We have simple maps in both directions:
$$
k(U)\to k(X)\\
[s,V]\mapsto [s,V]
$$
and
$$
k(X)\to k(U)\\
[s,V]\mapsto [s|_{U\cap V}, U\cap V].
$$
You can check that these are mutually inverse field morphisms.
On your confusion about the domains: the elements of $k(X)$ aren't really functions in the strict sense, but an equivalence class of functions, with changing domains. And equality is replaced by equivalence, and two functions are said to be equivalent if and only if their restriction to a common domain is equal.
