Is the pre-closure of the join of two projective varieties quasi-projective? I borrow notation from this answer. Let $X,Y\subseteq\mathbb{P}^{n}$ be two irreducible varieties over an algebraically closed field $k$. Consider
$$
S^{0}_{X,Y}:=\{(x,y,z)\in X\times Y\times \mathbb{P}^{N}:x\neq y, z\in\langle x,y\rangle\},
$$
and
$$
S_{X,Y}:=\overline{S^{0}_{X,Y}}\subseteq X\times Y\times\mathbb{P}^{N}
$$
(where the line means Zariski closure). Let us consider the projections
$$
\pi_{1}:S_{X,Y}\rightarrow X\times Y,
$$
$$
\pi_{2}:S_{X,Y}\rightarrow \mathbb{P}^{N}.
$$
The join of $X,Y$ is defined to be
$$
S(X,Y):=\pi_{2}(S_{X,Y}).
$$
Question: Is $\pi_2(S_{X,Y}^0)$ a quasi-projective variety?
Motivation: If you take $X=Y$, the join is equal to the secant variety of $X$. If $X$ is the Segre variety, then $\pi_2(S_{X,X}^0)$ is equal to the set of tensors of tensor rank at most $2$, and this can be extended to the set of tensors of tensor rank at most $r$ by taking more joins. My motivation is that I want to prove that the set of tensors of tensor rank at most $r$ is quasi-porjective. I have seen it mentioned that this is the case (see Landsberg; Tensors: Geometry and Applications, top of page 119), but I have been unable to prove it.
First attempt at a solution: It is clear that $S_{X,Y}^0$ is locally closed (i.e., quasi-projective). Since projection maps are open, $\pi_2$ is an open map. Since projective varieties are complete, $\pi_2$ is a closed map. The proof would be finished if $\pi_2$ sends locally closed sets (the intersection of an open set and a closed set) to locally closed sets (I would imagine such a map would be called locally closed, but I haven't found this definition anywhere).
Another idea is to show that $\pi_2|_{S_{X,Y}}:S_{X,Y} \rightarrow S(X,Y)$ is an open map. One way I know how to do this is by proving that it is flat, which in turn holds if the fibres are equidimensional and the input and output spaces are smooth. It would be nice not to have to assume smoothness here. (I don't think the join of two smooth varieties is smooth, in general).
Attempt 2: A completely different approach is to consider the abstract (or ruled) join of $X$ and $Y$ (see Flenner; Join Varieties and Intersection Theory, section 2):
$$
J_{X,Y}=\{[x:y]: \big((x \in X \text{ or } x=0) \text{ and } (y \in Y \text{ or } y=0)\big) \text{ and } \big(x,y \text{ not both zero}\big)\}\subseteq \mathbb{P}^{2N+1}
$$
Then $\pi_2(S^0_{X,Y})=\pi (J_{X,Y} \setminus \Delta)$, where
$$\pi : \mathbb{P}^{2N+1} \dashrightarrow \mathbb{P}^N$$
is given by $\pi [x: y] = [x-y]$, and $\Delta=\{[x:x] : x \in X \cap Y\}$.
This again expresses $\pi_2(S^0_{X,Y})$ as the image of a quasi-projective variety, which is in general only constructible, but maybe $\pi$ is ``locally closed" for some reason?
...
I am looking for completions of both of these attempts, but I would settle for one of the two!
 A: No.
The pre-closure of the join of two irreducible projective varieties is NOT necessarily quasi-projective.
Let $X$ be a smooth plane conic and let $Y$ be a single point of $X$. The pre-closure of the join of $X$ and $Y$ is the union of all the lines through points $x \in X$, $y \in Y$, $x \neq y$. Of course there is only one point in $Y$. So we get the union of all the lines through $Y$, except for the line through $Y$ which is tangent to the conic $X$. However we do get the point $Y$ itself, since it lies on all the other lines.
The union is the whole plane, minus one line, plus one point of that missing line. That's not quasi-projective.
A: Suppose first $X = Y = \mathbb{P}^n$. Let $V = \{x, y, L \,|\,x, y \in L \}$ where $L$ is a line in $\mathbb{P}^n$. Write $W = X \times Y = \mathbb{P}^n \times \mathbb{P}^n$ and $\Delta$ for the diagonal copy of $\mathbb{P}^n \subseteq W$. Then $V = Bl_\Delta W$ is the blowup of $W$ along the diagonal. The universal line bundle $\mathcal{O}(-1)$ has total space parametrizing points on the line, so contains your $S^0$ as the open subset away from the diagonal where $x \neq y$. If you want to allow the "point at infinity" on your line, you need to take the associated $\mathbb{P}^1$-bundle.
To restrict to $X,Y$ not necessarily equal to $\mathbb{P}^n$, pull back (take the preimage) of $X \times Y \subseteq \mathbb{P}^n \times \mathbb{P}^n$ along $V \to \mathbb{P}^n \times \mathbb{P}^n$.
