Morphism of Affine Algebraic Variety

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.