This is a pretty bad book to learn algebraic geometry from if you've never seen it before. I'm trying to verify the following assertion. Let $k$ be an algebraically closed field, and $X, Y$ be affine algebraic varieties. If $\psi: k[Y] \rightarrow k[X]$ is a $k$-algebra homomorphism, one defines a map $$\phi: X \rightarrow Y$$ as follows. For $x \in X$, let $F_x: k[X] \rightarrow k$ be the homomorphism given by $F_x(f) = f(x)$. Then $F_x \circ \psi$ is a homomorphism $k[Y] \rightarrow k$, so its kernel is a maximal ideal, necessarily equal to (for exactly one $y \in Y$) $\mathfrak M_y = \{ g \in k[Y] : g(y) = 0\}$. We then set $y = \phi(x)$.
The claim is that $\phi$ is a continuous function.
I haven't gotten too far with this. A closed set in $Y$ looks like $\mathcal V_Y(I)$ for $I$ a radical ideal of $k[Y]$, where $\mathcal V_Y(I)$ is the set of $y \in Y$ for which $h(y) = 0$ for all $h \in I$. To say that $\phi(x) = y$ just means that $\psi(g)(x) = g(y)$ for all $g \in k[Y]$.
So the preimage $\phi^{-1} \mathcal V_Y(I)$ (which I want to show is closed) is the set of all $x \in X$ with the following property: there exists a $y \in Y$ with the property that $h(y) = 0$ for all $h \in I$, and that $g(y) = \psi(g)(x)$ for all $g \in k[Y]$. In particular $\psi(g)(x) = 0$ whenever $g \in I$.
How should I go about this? Should I try to show the complement is open?