Let $\pi:\mathcal{C} \to B$ be a (flat) family of complex projective schemes of pure dimension $1$ with fixed Hilbert polynomial, in particular, for some $n \ge 3$, $\mathcal{C} \hookrightarrow \mathbb{P}^n\_B$, the composition of this closed immersion with the natural projection $\mathbb{P}^n\_B \to B$ is $\pi$ and there exists a Hilbert polynomial $P$ such that for all $b \in B$, the fiber $\pi^{-1}(b)$ is of Hilbert polynomial $P$ in $\mathbb{P}^n\_b$. Assume that $B$ is integral. Suppose that there exists a closed point $b\_0 \in B$ such that the fiber $\pi^{-1}(b\_0)$ can be embedded into $\mathbb{P}^3$ i.e., the closed immersion of $\pi^{-1}(b\_0) \hookrightarrow \mathbb{P}^n$ factors through $\mathbb{P}^3$. Is it then possible to find an open neighbourhood $U$ of $b\_0$ such that for all $u \in U$, $\pi^{-1}(u)$ can be embedded into $\mathbb{P}^3\_u$ i.e., the closed immersion of $\pi^{-1}(u) \hookrightarrow \mathbb{P}^n\_u$ factors through $\mathbb{P}^3\_u$? If not, is there any known condition on $B$ under which this happens? 


**EDIT** By "closed immersion $i:A \hookrightarrow \mathbb{P}^n$ factors through $\mathbb{P}^3$", I mean that there is a closed immersion $i\_1:A \hookrightarrow \mathbb{P}^3$ and a LINEAR embedding $i\_2:\mathbb{P}^3 \hookrightarrow \mathbb{P}^n$ such that $i=i\_2 \circ i\_1$.



Deformation of curves and closed immersions
The answer to the first question is no. In the moduli space $\mathcal{M}\_{10}$ of curves of genus 10, the complete intersections $(3,3)$ in $\Bbb{P}^3$ form a strict subvariety $\mathcal{CI}$. Pick for $B$ a subvariety of $\mathcal{M}\_{10}$ which intersect $\mathcal{CI}$ transversally at one point $[C]$. Embed the corresponding family $\mathcal{C}\rightarrow B$ into $\mathbb{P}^{9}\_B$ by the canonical embedding (replacing $B$ by an open subset). At $[C]$ this embedding factors through $C\hookrightarrow \mathbb{P}^3$, but not at the other points of $B$.


As for the second question, no condition on $B$ will ensure what you ask. You may write down conditions on the family $\mathcal{C}\rightarrow B$, but they will be essentially tautological (existence of a line bundle on $\mathcal{C}$ giving the required embedding on each fiber.)


**EDIT :** The answer to the edited question is still no. Take $B=\mathbb{C}$, and consider the embedding $\mathbb{P}^1\times B\hookrightarrow \mathbb{P}^4\times B$ given by $([X:Y],t)\mapsto ([X^4:X^3Y: tX^2Y^2: XY^3:Y^4],t)$. The embedding of $\mathbb{P}^1\times \{0\} $ in $\mathbb{P}^4$ factors through $\mathbb{P}^3$, but not those of $\mathbb{P}^1\times \{t\} $ for $t\neq 0$.


