Let $X$ and $Y$ be smooth algebraic varieties over $\mathbb{C}$. Let $f$ be a continuous map from complex points of $X$ to $Y$. Are there Zariski opens $U$ and $V$ inside $X\times \mathbb{A}^1$ and $Y\times \mathbb{A}^1$ respectively such that $U$ contains $X\times\{0\}$ and $X\times \{1\}$ (similarly $V$ contains $Y\times \{0\}$ and $Y\times \{1\}$) with the property that a continuous map $g$ from $U$ to $V$ can be assigned to $f$ in a way that $g|_{X\times \{0\}}= f$ and $g|_{X\times \{1\}}$ is a regular morphism?
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1$\begingroup$ The restriction of g doesn't have target Y but V, so it can't be f. $\endgroup$– Fernando MuroAug 14, 2021 at 7:30
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2$\begingroup$ @FernandoMuro Probably it is meant that the restriction of $g$ maps $X\times 0$ into $Y\times 0$ sitting in $V$. $\endgroup$– Evans GambitAug 14, 2021 at 7:38
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$\begingroup$ I think $V$ is not needed: aren't we asking for some $g:U\to Y$ restricting to $f$ on $X\times \{0\}$ and to a morphism on $X\times \{1\}$? $\endgroup$– Laurent Moret-BaillyAug 14, 2021 at 16:45
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$\begingroup$ @LaurentMoret-Bailly You are right $V$ is not necessary. $\endgroup$– user127776Aug 14, 2021 at 16:54
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I think that the answer is negative.
For instance, let $X=Y$ be a smooth curve of genus $g\geq 3$ with $\operatorname{Aut}(X)=\{\mathrm{id} \}$, and take as $f \colon X \to X$ an isotopically non-trivial diffeomorphism.
answered Aug 14, 2021 at 8:57
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$\begingroup$ Does the situation of the problem imply that a diffeomorphism on one edge has to go an automorphism (regular map) on the other edge? $\endgroup$ Aug 14, 2021 at 16:00
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1$\begingroup$ See J. Harris, Algebraic Geometry, Theorem 14.9 and Corollary 14.10, for a proof of the fact that smooth bijection of smooth projective varieties is an isomorphism. $\endgroup$ Aug 14, 2021 at 16:54
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2$\begingroup$ A simple concrete counterexample could be $X=Y=\mathbb{P}^1$, $f=$ complex conjugation. This induces $\times(-1)$ on $H_2(\mathbb{P}^1,\mathbb{Z})$, so it cannot be homotopic to a morphism, right? $\endgroup$ Aug 14, 2021 at 16:55
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1$\begingroup$ @LaurentMoret-Bailly: the same argument can be applied to every anti-biholomorphism (for instance, complex conjugation if $X$ is defined over $\mathbb{R}$) of a variety of odd dimension, right? In fact, in odd dimension anti-biholomorphisms are orientation-reversing. $\endgroup$ Aug 14, 2021 at 16:57
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1$\begingroup$ @FrancescoPolizzi: Yes, assuming $X$ proper, I guess. $\endgroup$ Aug 14, 2021 at 17:50