# ag.algebraic geometry – One arrow reversed version of smooth extension problem

Suppose $$Y$$ is an irreducible quasi-projective $$k$$-scheme of dimension $$geq 1$$, which may be singular or non-reduced, etc., and let $$U subset Y$$ be a nonempty open subscheme.

There are various “extension” problems from $$U$$ to $$Y$$. For instance, people often need to extend a coherent sheaf or a perfect complex, etc. over $$U$$ to one on $$Y$$. Usually the “structures” are “over” $$U$$, or $$Y$$.

While working on a problem, I somehow bumped into the following kind of funny situation, where we have “structures” with arrows “reversed” in a sense.

So, here is the precise question:

For the above open immersion $$iota: U subset Y$$, suppose there is a closed immersion $$U hookrightarrow Z$$ into a smooth $$k$$-scheme. Then can we find smooth
$$X$$ together with a closed immersion $$Y hookrightarrow X$$, such that $$X$$
restricted to $$U$$ is an open neighborhood of $$U$$ in $$Z$$?

Here, the last part means that there exist (1) an open $$X’ subset X$$ and (2) an open neighborhood $$Z’subset Z$$ of $$U$$, such that $$X’ cap Y= U$$ and we have an isomorphism of $$X’$$ with $$Z’$$ under $$U$$.

At first sight, I thought it should be an obvious consequence of some usual smooth compactifications (via Nagata compactifications + desingularizations) but then I realized the situation was not like that. I still hope that it can be resolved using some clever resolution tricks, but I am running out of steams.

In case this turns out to be false, can we weaken the requirement

“we have an isomorphism $$X’simeq Z’$$ under $$U$$

to

“we have a morphism $$X’ to Z’$$ under $$U$$“?