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. 



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> 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 $g\circ j\_2=j\_1$?
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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](https://mathoverflow.net/questions/124265/non-uniqueness-of-smooth-compactification).



Uniqueness of smooth compactification upto a smooth morphism
This desire fails on a much more basic level. A smooth morphism is flat, and you can't even have a flat morphism between two smooth compactifications. 


A flat morphism has equidimensional fibers. If it is also proper and birational, then it is finite and a finite birational morphism mapping onto something normal is automatically an isomorphism. 


The main point of this is that *flat* and *birational* are almost "opposites", so they very rarely come together. 


