Let $X$ be a smooth, projective $\mathbb{C}$-variety of dimension $n$. Fix a closed point $x \in X$ and an embedding of $X$ in $\mathbb{P}^m$ for some integer $m$. For a given $d$, denote by $\sigma\_d : \mathbb{P}^m \to \mathbb{P}^{N\_d}$ the $d$-tuple embedding. My question is: for $d \gg 0$, does there exist a linear subspace $L\_x \subset \mathbb{P}^{N\_d}$ (depending on $x$) of dimension $N\_d-n-2$, not intersecting $\sigma\_d(X)$ such that for the linear projection from $L\_x$ (sometimes called *projection with centre* $L$):
$$\pi\_{L\_x} : \sigma\_d(X) \to \mathbb{P}^{n+1}$$ we have $\pi\_{L\_x}^{-1}(\pi\_{L\_x}(\sigma\_d(x)))=\sigma\_d(x)$ i.e., the preimage of $\sigma\_d(x)$ is only $\sigma\_d(x)$ **for the chosen closed point $x$**? Any hint/reference is most welcome.


**EDIT** Note that, the choice of $L\_x$ depends on the choice of $x$.



A question on linear projection of a smooth projective variety
It is definitely not possible to expect the existence of a linear projection $\pi\_L$ such that the equality
$$
\pi\_L^{-1}(\pi\_L(\sigma\_d(x)))=\sigma\_d(x)\tag{\*}
$$
holds scheme-theoretically for each point $x \in X$. Indeed, this equality means that $\pi\_L$ defines an isomorphism of $X$ onto its image in $\mathbb{P}^{n+1}$, which is thus a smooth hypersurface. But for $n \ge 3$ a smooth hypersurface in $\mathbb{P}^{n+1}$ has Picard group isomorphic to $\mathbb{Z}$, while the Picard group of $X$ could be arbitrary.


On the other hand, for a given point $x \in X$ it is easy to find a subspace $L$ such that $(\*)$ holds. Indeed, the closure of the union of lines joining $x$ with other points of $X$ has dimension at most $n + 1$, and any $L$ disjoint from this union works.


