Morphism of Affine Algebraic Variety Is it true that a morphism of affine algebraic varieties is continuous in Zariski topology? How should I proceed? thank you
 A: Let $K$ be a field.
Let $\mathbb{A}^n$ and $\mathbb{A}^m$ be affine spaces over $K$.
Let $X$ be a closed subset of $\mathbb{A}^n$ and $Y$ be a closed subset of $\mathbb{A}^m$.
Let $F_1,\dots,F_m \in K[X_1,\dots,X_n]$.
Let $f:X \rightarrow Y$ be a morphism defined by $f(x) = (F_1(x),\dots,F_m(x))$. We prove that $f$ is continuous.
Let $T$ be a closed subset of $Y$.
It suffices to prove that $f^{-1}(T)$ is closed in $X$.
Since T is a closed subset of $\mathbb{A}^m$, there exist polynomials $G_1,\dots,G_r \in K[Y_1,\dots,Y_m]$ such that $T$ is the set of common zeros of $G_1,\dots,G_r$.
Let $H_i = G_i(F_1(X_1,\dots,X_n),\dots,F_m(X_1,\dots,X_n))$ for $i = 1,\dots,r$.
Let $S$ be the intersection of $X$ and the set of common zeros of $H_1,\dots,H_r$.
If $f(x) \in T$, then $H_1(x) = \cdots = H_r(x) = 0$. Hence $x \in S$.
Conversely if $x \in S$, then $H_1(x) = \cdots = H_r(x) = 0$. Hence $f(x) \in T$.
Hence $f^{-1}(T) = S$.
This completes the proof.
A: The reverse image of the $\cap$ of an arbitrary family of sets is equal to the $\cap$ of their respective reverse images: article.
Let $R$ be a commutative ring with $1$.  Let $\mathfrak{a} \subset R[X_1, \dots, X_n]$ be an ideal.
Then $V(\mathfrak{a}) = \bigcap_{g \in \mathfrak{a}} V(g)$.  Thus it sufficies to prove continuity at each $V(g)$ by the link above.
But $f^{-1}(V(g)) = \{x \in R^n : f(x) \in V(g)\}$, but $f(x) \in V(g) \iff g(f(x)) = 0 \iff x \in V(g\circ f) \subset R^{n}$.  
Clearly composing with $f$ puts $g \circ f \in R[X_1, \dots, X_n]$, thus taking $V(g\circ f)$ makes sense.
Thus morphisms of these "algebraic sets" over a ring are continuous.  Replace $R$ with $k$ and you have your proof.  Note that $V(g)$ was taken in $R^m$ (the "codomain") and $V(g\circ f)$ was taken in $R^n$, but notationally we've used "$V$" for both.
