Lemma 3.6 of Hartshorne's Algebraic Geometry book: Let $X$ be a variety and let $Y$ be an affine variety. A map $f: X \rightarrow Y$ is a morphism if and only if $x_{i} \circ f$ is a regular function on $X$ for each $i$ where $x_{i}$ are the coordinate functions.
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For the continuity part of $f$, can you explain exactly what he means by "Since the closed sets of Y are defined by vanishing of polynomial functions, and since regular functions are continuous" to conclude that $f^{-1}$ takes closed sets to closed sets? What do those two facts tell us that we can conclude?
Suppose $U$ is a closed set of $Y$ and $g$ is a polynomial vanishing on $U$, then $g \circ f$ is regular (since g is a polynomial) and $g \circ f$ clearly vanishes on $f^{-1}(U)$. Can I argue this way, that since the diagram below 'commutes'
\begin{array} &X & \\ \downarrow^{f}&\searrow^{r}\\ Y&\rightarrow^{g}&k \end{array} where $r = g \circ f$ is the regular function by assumption, $g$ polynomial vanishing at the closed set $U$ of $Y$.
then since r is continuous (since regular), then f must be continuous?
Explicitly, what are the functions on X vanishing on U, and I believe we need either polynomials or homogeneous polynomials depending on the variety of X. Are they just $p = (\frac{p}{q}) \cdot q = r \cdot q = (g \circ f)\cdot q$? (where r is regular function above)