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This seems like a problem that should be easy if true, but I wasn't able to prove it in general or disprove it. Given a complex projective variety $X$ and a sub-variety $Y$ and a closed point $p$ on $Y$. The question is: Is it possible to move $Y$ so it does not include the point $p$? You can assume any form of positivity on $Y$. For example you can assume it is ample or a complete intersection.

By moving I mean the following: there is a continuous map $F: Y(\mathbb{C})\times [0,1] \rightarrow X(\mathbb{C})$, where $ F(-, 0)$ is the closed immersion of $Y$ in $X$. Furthermore the image of $F(-,1)$ does not include $p$. Moreover $F$ at any fixed time $t$ is an algebraic map ($F(-,t)$ is an algebraic map).

Another equivalent way of stating the problem is the following. Let's look at the connected component of the Hom scheme $\mathcal{H}om(Y,X)$ that contains the closed immersion of $Y$ into $X$. Is there another morphism in the same connected component that its image avoids $p$?

If untrue is there any general form for varieties $X$ that this holds?

I am interested in a very specific case where $X=\text{Sym}^n(\mathbb{P}^r_{\mathbb{C}})$. $Y$ is a complete intersection that contains a diagonal element $p$ i.e. points with multiplicity $n$ of the symmetric product.

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    $\begingroup$ No in general: if you blow up a point in a (smooth) surface, the exceptional curve cannot move. $\endgroup$
    – abx
    Commented May 3, 2022 at 13:18
  • $\begingroup$ In the case when $X$ is singular, we do not expect to move divisors off singular points, e.g. in a quadric cone $xy = z^2$ lines making up the cone all pass through the singular point $0$ and there is no way to move them off it -- in the sense of linear equivalence, and probably in the sense you are asking as well. $\endgroup$ Commented May 4, 2022 at 7:45

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