How to prove that following set is closed

Let

$$ f=(f_1,\ldots,f_n):\mathbb A^m(k)\longrightarrow \mathbb A^n(k) $$

be a polynomial map. This means that each coordinate function

$$ f_i\in k[T_1,\ldots,T_m] $$

is a polynomial, and

$$ f(x_1,\ldots,x_m) = \bigl(f_1(x_1,\ldots,x_m),\ldots,f_n(x_1,\ldots,x_m)\bigr). $$

Associated with $f$ is a $k$-algebra homomorphism

$$ \Gamma(f): k[T'_1,\ldots,T'_n] \longrightarrow k[T_1,\ldots,T_m]. $$

It is defined by

$$ \Gamma(f)(T'_i)=f_i. $$

Therefore, for any polynomial

$$ g\in k[T'_1,\ldots,T'_n], $$

we have

$$ \Gamma(f)(g)=g(f_1,\ldots,f_n). $$

In other words, $\Gamma(f)(g)$ is obtained by substituting $f_i$ for the variable $T'_i$.

The zero set $V(\mathfrak a)$

Let

$$ \mathfrak a\subseteq k[T'_1,\ldots,T'_n] $$

be an ideal. Its zero set is

$$ V(\mathfrak a) = \left\{ y\in\mathbb A^n(k): g(y)=0\text{ for every }g\in\mathfrak a \right\}. $$

Thus a point $y$ belongs to $V(\mathfrak a)$ exactly when every polynomial in $\mathfrak a$ vanishes at $y$.

We now prove that

$$ f^{-1}(V(\mathfrak a)) = V(\Gamma(f)(\mathfrak a)). $$

Take a point

$$ x\in\mathbb A^m(k). $$

Then

$$ x\in f^{-1}(V(\mathfrak a)) $$

means, by the definition of inverse image, that

$$ f(x)\in V(\mathfrak a). $$

By the definition of $V(\mathfrak a)$, this is equivalent to saying that

$$ g(f(x))=0 $$

for every $g\in\mathfrak a$.

But

$$ g(f(x)) = g\bigl(f_1(x),\ldots,f_n(x)\bigr). $$

On the other hand,

$$ \Gamma(f)(g)=g(f_1,\ldots,f_n), $$

so evaluating this polynomial at $x$ gives

$$ \Gamma(f)(g)(x) = g\bigl(f_1(x),\ldots,f_n(x)\bigr) = g(f(x)). $$

Therefore,

$$ g(f(x))=0 $$

for every $g\in\mathfrak a$ if and only if

$$ \Gamma(f)(g)(x)=0 $$

for every $g\in\mathfrak a$.

This is exactly the condition that $x$ belong to the common zero set of the polynomials $\Gamma(f)(g)$, where $g$ ranges over $\mathfrak a$. Hence

$$ x\in f^{-1}(V(\mathfrak a)) \iff x\in V(\Gamma(f)(\mathfrak a)). $$

Since this holds for every $x\in\mathbb A^m(k)$, we obtain

$$ \boxed{ f^{-1}(V(\mathfrak a)) = V(\Gamma(f)(\mathfrak a)). } $$

Strictly speaking, the set

$$ \Gamma(f)(\mathfrak a) = \{\Gamma(f)(g):g\in\mathfrak a\} $$

need not be an ideal of $k[T_1,\ldots,T_m]$.

Therefore, one may write the right-hand side more precisely as

$$ V\left( \left\langle \Gamma(f)(g):g\in\mathfrak a \right\rangle \right), $$

where the angle brackets mean the ideal generated by the polynomials $\Gamma(f)(g)$.

This does not change the zero set, because a point at which all the generators vanish is also a point at which every polynomial combination of the generators vanishes.

The Zariski topology on $\mathbb A^m(k)$ is defined by declaring sets of the form

$$ V(S) = \left\{ x\in\mathbb A^m(k): h(x)=0\text{ for every }h\in S \right\} $$

to be closed, where $S$ is any collection of polynomials in

$$ k[T_1,\ldots,T_m]. $$

Equivalently, the closed sets are exactly the sets

$$ V(\mathfrak b), $$

where $\mathfrak b$ is an ideal of $k[T_1,\ldots,T_m]$.

Now every polynomial $\Gamma(f)(g)$ lies in

$$ k[T_1,\ldots,T_m]. $$

Therefore,

$$ V(\Gamma(f)(\mathfrak a)) $$

is the common zero set of a collection of polynomials in the variables $T_1,\ldots,T_m$. Hence it is Zariski closed by definition.

We have shewn that the inverse image of every closed set is closed:

$$ f^{-1}(V(\mathfrak a)) = V(\Gamma(f)(\mathfrak a)). $$

Therefore every morphism of affine algebraic sets is continuous in the Zariski topology.