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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.

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  • $\begingroup$ Let $\bar{X}$ be a smooth compactification of $X$ with (non-empty) boundary $D$. Now, blow-up a point in $D$. Write $\bar{X}'\to \bar{X}$ for this blow-up. Then $\bar{X}'$ is another compactification of $X$. This is your $g$, right? This morphism $g$ is not smooth. So what am I missing in your question? $\endgroup$ – Ariyan Javanpeykar Apr 13 '16 at 19:48
  • $\begingroup$ As you point out, a morphism between smooth compactifications is not in general smooth. My question can also be framed as : given 2 smooth compactifications, does there exist a third with smooth morphisms to the first two? $\endgroup$ – user86186 Apr 14 '16 at 4:42
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    $\begingroup$ The answer is no, I think. Take $X$ to be a smooth projective minimal surface (of positive Kodaira dimension to be safe). Let $D$ be a smooth irreducible curve in $X$, and let $U =X\setminus D$. Then $X$ is a smooth compactification of $U$. Any other smooth compactification of $U$, $X'$ say, will map uniquely to $X$, and this morphism is smooth if and only if it is an isomorphism. Is that convincing? $\endgroup$ – Ariyan Javanpeykar Apr 14 '16 at 6:57
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    $\begingroup$ So, it seems that your question is not a "right" one. Are you interested in other results in this direction (say, in motivic or cohomological ones)? $\endgroup$ – Mikhail Bondarko Apr 14 '16 at 15:41
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    $\begingroup$ I have several papers on this subject (arxiv.org/find/math/1/au:+Bondarko_M/0/1/0/all/0/1) including the survey arxiv.org/abs/0903.0091 that is unfortunately far from being perfect. If you don't like my style and/or Voevodsky motives then you can also read the (somewhat "classical") paper Gillet H., Soulé C. Descent, motives and K-theory// J. reine und angew. Math. 478, 1996, 127–176. $\endgroup$ – Mikhail Bondarko Apr 15 '16 at 13:04
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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.

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