Let $K$ be a field and $X\subset K^n$. Then define $$\mathcal{O}(X):=K[X_1,...,X_n]\big/I(X)$$ where $I(X):=\{f\in K[X_1,...,X_n]: f(x)=0~~\forall x\in X\}$. I need to show that $$\phi:\mathcal{O}(X)\rightarrow Map(X,K); ~~~~~\bar f\mapsto \left(\phi(\bar f):x\mapsto f(x)\right)$$ is well defined.
My idea was the following:
Let $\bar f,\bar g\in \mathcal{O}(X)$ such that $\phi(\bar f)=\phi(\bar g)$. This means that for all $x\in X$, $f(x)=g(x)$ so $f=g$ in $K[X_1,...,X_n]$ but then $\bar f=\bar g$ in $\mathcal{O}(X)$ and hence $\phi$ is well defined.
Does this work?
No it doesn't work if $K$ is a finite field, take $K=\mathbb{F}_p$, $f=0$ and $g=x^p-x$, then $f=g$ on $\mathbb{F}_p$ but $f\neq g$. What you can do however : the map $K[x_1,\ldots,x_n]\longrightarrow {\rm Map}(X,K)$ that sends $f$ to $x\mapsto f(x)$ is a morphism of rings that vanishes on the ideal $I(X)$ by definition, so it induces a ring morphism $K[x_1,\ldots,x_n]/I(X)\longrightarrow{\rm Map}(X,K)$ and this map is exactly $\phi$.