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Regular map between quasiprojective varieties are continuous From this, we know that regular maps between quasi-projective varieties are continuous.

My question- If $f:X \to Y$ is continuous, and for every $x \in X$ and for every regular function $\varphi$ in a neighbourhood of $f(x)$ the function $f^*\varphi$ is regular in neighbourhood of $x$ then $f$ is regular.

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I think that your question has been put in the wrong way!

Example. Let $\mathbb{K}=\overline{\mathbb{Z}_p}$ be the algebraic closure of the field of integer numbers modulo $p\in\mathbb{P}$, and let $X=Y=\mathbb{A}^1_{\mathbb{K}}$; considering the following maps of $X$ in itself: \begin{gather} O:x\in X\to 0\in X,\\ F:x\in X\to x^p-x\in X; \end{gather} one has that as maps $O=F$, but as regular maps $O\neq F$ because $O^{*}$ is the zero endomorphism of $\mathbb{K}[t]$ and $F^{*}$ is the Frobenius endomorphism of $\mathbb{K}[t]$. $\triangle$

From all this, your question can be put in this right way: let $f:X\to Y$ be a continuous map between quasi-projective varieties and, for any $y\in Y$ there exists an open subset $U_y$ of $Y$, $f^{*}_{U_y}:\varphi\in\mathcal{O}_Y(U_y)\to(\varphi\circ f)\in\mathcal{O}_X(f^{-1}(U_y))$ is a well-defined function; where $\mathcal{O}_Y(U_y)$ is the ring of regular functions of $Y$ on $U_y$, and similar statement holds for $\mathcal{O}_X(f^{-1}(U_y))$.

Does $\left(f,\left\{f^{*}_{U_y}\right\}_{y\in Y}\right)$ define a regular morphism?

Easily one proves that $f^{*}_{U_y}$ is a morphism of rings for any $y\in Y$; for any $V\subseteq U_y$ open subset, one has that $r^{U_y}_V:\varphi\in\mathcal{O}_Y(U_y)\to\varphi_{|V}\in\mathcal{O}_Y(V)$ is a well-defined morphism of rings, and $r^{f^{-1}(U_y)}_{f^{-1}(V)}\circ f^{*}_{U_y}=f^{*}_V\circ r^{U_y}_V$.

By Shafarevich - Basic Algebraic Geometry 1, II ed. (1994) Springer-Verlag, lemma I.4.2.2, you can assume that $V_y\subseteq U_y$ is an affine open neighbourhood of $y$ in $Y$ and $W_y\subseteq f^{-1}(V_y)$ is an affine open subset; so $f_{|W_y}$ is a regular map of affine open subsets, by definition $f$ is a regular morphism at any point of $X$, that is $f$ is a regular map.

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  • $\begingroup$ for me field K is algebraically closed $\endgroup$
    – user345777
    May 24 '18 at 13:31
  • $\begingroup$ Ok, I can change $\mathbb{Z}_p$ with its algebraic closure $\overline{\mathbb{Z}_p}$ and the example holds the same. $\endgroup$ May 24 '18 at 14:46
  • $\begingroup$ I am actually the beginner in the subject, I don't understand the language of schemes yet (so,i can't say if we are on the same page) i am reading the first chapter of Shafarevich basic algebraic geometry, and in that he has left above question as exercise. $\endgroup$
    – user345777
    May 25 '18 at 22:21
  • $\begingroup$ I'm addicted by schemes X-D ... but the reasoning holds. I check on Shafarevich's book as he introduces the regular maps of quasi-projective varieties, and I shall put my answer in "Shafarevich like" way. I promise. ;) $\endgroup$ May 26 '18 at 9:57
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    $\begingroup$ the comment he makes is on page no 38. $\endgroup$
    – user345777
    May 28 '18 at 19:37

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