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I am trying to understand how the Zariski topology is different from the usual topology on the affine space $A^n$. Let $X$ be an affine algebraic subset of $A^n$, the $n$-dimensional affine space over $k$. Let $f_i:X\rightarrow A^1$, $i=1,...,k$ be continuous functions, both $X$ and $A^1$ being equipped with the Zariski topology. Is the function $f:X\rightarrow A^k, x\mapsto (f_1(x),...,f_k(x))$ continuous when $A^k$ is equipped with the Zariski topology? I know that $f$ is continuous for the product topology, but the Zariski topology is finer the the product topology.

I appreciate any help.

Jiu
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1 Answers1

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NO, $f$ need not be continuous in the Zariski topology !
For a counterexample take $X=\mathbb A^1$ and let $f_1:\mathbb A^1\to \mathbb A^1$ be the map permuting $0$ and $1$ and leaving $x$ fixed if $x\neq 0,1$.
Like all permutations of $\mathbb A^1$ the map $f_1$ is continuous in the Zariski topology.
Now taking the identity for $f_2:\mathbb A^1\to\mathbb A^1$, let us show that the map $$f:\mathbb A^1\to \mathbb A^1\times \mathbb A^1:x\mapsto (f_1(x),f_2(x)=(f_1(x),x)$$ is not continuous for the Zariski topology.
Indeed the diagonal $\Delta=\{(x,x)\vert x\in \mathbb A^1\} \subset \mathbb A^1\times \mathbb A^1$ is closed in the product endowed with its Zariski topology but its inverse image $f^{-1}{\Delta}=\mathbb A^1\setminus \{0,1\}\subset \mathbb A^1$ is not closed if $k$ is an infinite field.

Georges Elencwajg
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