## Hartshorne Question

I have a question from Algebraic Geometry by Robin Hartshorne. My question is on the following proof:

Theorem. Let $X$ be a variety and $Y \subseteq \mathbb{A}^n$ be an affine variety. Let $x_1, \ldots, x_n$ be the coordinate functions on $\mathbb{A}^n$. A map of sets $\psi : X \rightarrow Y$ is a morphism if and only if $x_i \circ \psi$ is a regular function on $X$ for each $i=1, \ldots, n$.

Proof
$(\Rightarrow)$ If $\psi$ is a morphism then the $x_i \circ \psi$ are regular functions. This is by the definition of a morphism.

$(\Leftarrow)$ Suppose the $x_i \circ \psi$ are regular. Let $f=f(x_1, \ldots, x_n)$ be any polynomial. Then $f \circ \psi$ is also regular on $X$. Now because the closed sets of $Y$ are defined by the vanishing of polynomial functions and because regular functions are continuous, $\psi^{-1}$ takes closed sets to closed sets. So $\psi$ is continuous. Recall that regular functions on open subsets of $Y$ are locally quotients of polynomials. Hence $g \circ \psi$ is regular where $g$ is a regular function on any open subset of $Y$. Therefore $\psi$ is a morphism.

I don't understand this sentence:

Now because the closed sets of $Y$ are defined by the vanishing of polynomial functions and because regular functions are continuous, $\psi^{-1}$ takes closed sets to closed sets.

How do we get $\psi^{-1}$ takes closed sets to closed sets? I guess I just don't understand this sentence in the proof. Thank you.