Defintiion: Let $U$ be an open susbset of an affine variety $X \subseteq \Bbb A^n$. A map $\varphi:U \rightarrow k$ is called regular at the point $P \in U$, if there is a neighborhood $V$ of $P$ in $U$ such that there are polynomials $f,g \in k[x_1, \ldots, x_n]$ with $g(Q) \not= 0$, and $\varphi(Q) = \frac{f(Q)}{g(Q)}$ for all $Q \in V$. It is called regular on $U$ if it is regular at every point in $U$. Denote the set of such functions as $O_X(U)$.
Claim: Suppsoe $\varphi_1, \varphi_2 \in O_X(V)$ coniciding $U \subseteq V$ where $V$ is an irreducible affine variety. Then $\varphi_1= \varphi_2$.
In remark 3.7 pg 25 a proof is given.
We first prove $\bar{U} =V$. I understand this part.
Then author claim $\varphi_1=\varphi_2$ on $\bar{U}$. Why is this? If $\varphi_1, \varphi_2$ are continuous then $U \subseteq (\varphi_1- \varphi_2) ^{-1}(0)$.
But I interpreted the original map as set theoretic map since for a general $k$ there is no topology. Or does the note imply otherwise?
We require Lemma 3.6 from the document you hyperlinked.
Let $\varphi_1$ and $\varphi_2$ be regular functions on an irreducible affine variety $V$ which agree on some nonempty open subset $U$ of $V$. Let $$S:=\{x\in V\mid \varphi_1(x)=\varphi_2(x)\}=\{x\in V\mid (\varphi_1-\varphi_2)(x)=0\}.$$ Therefore, $$U\subseteq S.$$ Since $\varphi_1-\varphi_2$ is a regular function on $V$, by Lemma 3.6, $S$ is closed in $V$. Thus, $$V=\overline{U}\subseteq S\subseteq V.$$ Hence, $$S=V.$$ That is, $\varphi_1=\varphi_2$.