Component-wise Zariski continuity implies Zariski continuity?

Question

I have a question while reading the answer here: Continuity of rational functions between affine algebraic sets

Original question:: Let $\varphi:X\rightarrow Y$ be a rational map between affine varieties defined on some open subset $U\subseteq X$. Show that $\varphi:U\rightarrow Y$ is continuous.

In Olórin's answer, it seems that he prove the continuity in the following steps:

1. Show that every polynomial function $f:U\rightarrow \mathbb{A}^1$ is continuous.

2. Show that every rational function defined on $U$, $\frac{f}{g}:U\rightarrow \mathbb{A}^1$, is continous.

3. Show that $\varphi:U\rightarrow \mathbb{A}^n$ is continuous, which implies that $\varphi:U\rightarrow Y$ is continuous.

From step 2 to step 3, it seems that he use the property that component-wise Zariski continuity implies Zariski continuity. It holds for product topology, but since Zariski topology on $\mathbb{A}^n$ is finer then product topology, I am not sure if it is right. Can we prove or disprove this?

Thanks in advance.

Edit: I find a counterexample in the post Function with continuous components, Zariski topology It seems that Olórin's answer need some further explanation.

Answer 1

As in the counterexample you mentioned, component-wise Zariski continuity doesn't imply Zariski continuity. However, the continuous map in that counterexample is not a morphism between varieties.

In the category of varieties, the product exists (See, for example, [Hartshorne, Exercise I.3.15.(c)]). Therefore, if you have two morphisms $f\colon X\rightarrow Y$ and $g\colon X\rightarrow Z$, then by the universal property of the product, there is a morphism $(f,g)\colon X\rightarrow Y\times Z$ which is continuous.

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Edit: Let $\varphi=(\varphi_1,\dots,\varphi_n)$ for some $\varphi_1,\dots,\varphi_n\in \mathscr{O}(U)$. For any closed subset $Z$ of $\mathbb{A}^n$, $Z$ is of the form $V(g_1,\dots,g_m)$ for some $g_1,\dots,g_m\in \mathscr{O}(\mathbb{A}^n)$. We can verify $g_i\circ\varphi=g_i(\varphi_1,\dots,\varphi_n)\in \mathscr{O}(U)$, hence $\varphi^{-1}(Z)=\{x\in U|g_i\circ \varphi(x)=0 \text{ for all }i=1,\dots m\}$ is closed in $U$. So $\varphi$ is continuous.