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?



Are continuous maps generically homotopic to a regular map?
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.


