Quasi affine scheme universal embedding

Question

Suppose $X$ is quasi affine scheme, it is well known that it open embeds in $Y=\mathrm{Spec}(\mathcal{O}(X))$. In fact, by adjunction, any map into an affine scheme factors uniquely through $Y$. Is it true that for any open embedding $X \to Z$, $Z$ affine, the induced map $Y \to Z$ is open embedding?

Answer 1

No, this is not true. Consider $X=\Bbb A^2\setminus (0,0)$, so that $Y=\Bbb A^2$. Let $Z$ be the union of the $xy$-plane and the $z$-axis inside $\Bbb A^3$. Then $X\hookrightarrow Z$ by $(x,y)\mapsto (x,y,0)$ is an open embedding, but the induced map $Y\to Z$ does not have open image.