By a $k$-variety, we will mean a separated scheme of finte type over a field $k$. Let $k$ be of characteristic 0. Given a smooth quasi-projective $k$-variety $X$, there is a projective $k$-variety $\bar{X}$ containing $X$. Since char$(k)=0$, we can take a resolution of singularities $\pi:\widetilde{X}\to\bar{X}$ where $\widetilde{X}$ is a smooth projective $k$-variety. $\pi$ will be an isomorphism on the smooth locus of $\bar{X}$ which contains $X$, so that $j=\pi^{-1}|_X:X\to\widetilde{X}$ is an open immersion.

Given two smooth compactifications $j_1:X\to\widetilde{X_1}$ and $j_2:X\to\widetilde{X_2}$, can one assume (after suitable modifications) that there is a smooth morphism $g:\widetilde{X_2}\to\widetilde{X_1}$ such that $g\circ j_2=j_1$?

It is a standard fact that given two smooth compactifications $j_1,j_2$ coming from resolutions of singularities $\pi_i:\widetilde{X_i}\to\bar{X}$, $i=1,2$, we can consider the resolution of singularities $\widetilde{X_3}$ of the Zariski closure of the image of $(j_1,j_2):X\to\widetilde{X_1}\times\widetilde{X_2}$ to obtain projective birational morphisms $g_i:\widetilde{X_3}\to\widetilde{X_i}$, $i=1,2$, such that $\pi_1\circ g_1=\pi_2\circ g_2$. This implies $g_i\circ j_i=j_3$, $i=1,2$. Thus, we may assume that there is a morphism $g:\widetilde{X_2}\to\widetilde{X_1}$ such that $g\circ j_2=j_1$. However, I don't see how $g$ can be smooth. Since $g$ is projective and birational and $\widetilde{X_i}$ is smooth projective, maybe the Zariski Main Theorem implies that $g$ is an open immersion, hence smooth?

This question is related to Bondarko's comment in the question Non-uniqueness of smooth compactification.

By a $k$-variety, we will mean a separated scheme of finte type over a field $k$. Let $k$ be of characteristic 0. Given a smooth quasi-projective $k$-variety $X$, there is a projective $k$-variety $\bar{X}$ containing $X$. Since char$(k)=0$, we can take a resolution of singularities $\pi:\widetilde{X}\to\bar{X}$ where $\widetilde{X}$ is a smooth projective $k$-variety. $\pi$ will be an isomorphism on the smooth locus of $\bar{X}$ which contains $X$, so that $j=\pi^{-1}|_X:X\to\widetilde{X}$ is an open immersion.

Given two smooth compactifications $j_1:X\to\widetilde{X_1}$ and $j_2:X\to\widetilde{X_2}$, can one assume (after suitable modifications) that there is a smooth morphism $g:\widetilde{X_2}\to\widetilde{X_1}$ such that $g\circ j_2=j_1$?

It is a standard fact that given two smooth compactifications $j_1,j_2$ coming from resolutions of singularities $\pi_i:\widetilde{X_i}\to\bar{X}$, $i=1,2$, we can consider the resolution of singularities $\widetilde{X_3}$ of the Zariski closure of the image of $(j_1,j_2):X\to\widetilde{X_1}\times\widetilde{X_2}$ to obtain projective birational morphisms $g_i:\widetilde{X_3}\to\widetilde{X_i}$, $i=1,2$, such that $\pi_1\circ g_1=\pi_2\circ g_2$. This implies $g_i\circ j_i=j_3$, $i=1,2$. Thus, we may assume that there is a morphism $g:\widetilde{X_2}\to\widetilde{X_1}$ such that $g\circ j_2=j_1$. However, I don't see how $g$ can be smooth. Since $g$ is projective and birational and $\widetilde{X_i}$ is smooth projective, maybe the Zariski Main Theorem implies that $g$ is an open immersion, hence smooth?

This question is related to Bondarko's comment in the question Non-uniqueness of smooth compactification.