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Suppose that $X$ is a projective threefold with at worst conifold singularities and suppose $\omega_X$ trivial. Suppose $Y$ is a projective variety with a birational morphism $f: Y\to X$ which is an isomorphism away from the conifold points and such that $f^{-1}(p) = \mathbb{P}^1$ for each conifold point $p \in X$. Can I conclude that $Y$ is smooth? i.e. that $f:Y\to X$ is a conifold resolution?

This seems too good to be true, but I was unable to come up with a counterexample and it would be really useful (to me at least) if it were true.

Edit: In light of Sasha/Jason's counterexample I would like to impose a normality condition. The most useful version to me would be to let $\bar{Y} \to Y $ be the normalization of $Y$, and ask: is $\bar{Y} \to X$ a conifold resolution? Alternatively, is it true if I assume that $Y$ is normal to begin with?

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    $\begingroup$ I suggest that you add the hypothesis that $Y$ is normal. $\endgroup$
    – Jason Starr
    Commented Aug 9, 2019 at 9:25
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    $\begingroup$ Is the equality $f^{-1}(p) = \mathbb{P}^1$ true scheme-theoretically? $\endgroup$
    – Angelo
    Commented Aug 11, 2019 at 8:02
  • $\begingroup$ @Angelo Yes, since I need to avoid silliness like tacking an embedded point onto a smooth resolution. Alternatively (and more useful from my point of view), I could only require $f^{-1}(p)=\mathbb{P}^1$ set theoretically and then ask if $\bar{Y}_{red}$ is a conifold resolution. $\endgroup$
    – Jim Bryan
    Commented Aug 11, 2019 at 16:38

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Take a conifold resolution $\tilde{X}$ and then "impose a cusp" on its exceptional fiber (in the vertical tangent direction). In other words, in a local chart replace $Spec(A)$ by the spectrum of the subring $$ A' := \{a \in A \mid \partial(a) \in \mathfrak{m} \}, $$ where $\mathfrak{m}$ is the ideal of a point on the exceptional fiber and $\partial$ is a derivation tangent to exceptional fiber. Define $Y$ by gluing $Spec(A')$ with the rest of $\tilde{X}$. This is a counterexample to your question.

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    $\begingroup$ If $f^{-1}(p)$ has a cusp, it is not isomorphic to $\Bbb{P}^1$? $\endgroup$
    – abx
    Commented Aug 9, 2019 at 8:44
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    $\begingroup$ The point by @abx is valid, but Sasha's construction is easily corrected. For instance, for the threefold $A_1$-singularity with completed local ring $R=k[[x,y,z,w]]/\langle xw-yz \rangle$, instead of taking a small resolution $\text{Proj}\ R[S,T]/\langle Sy-Tx,Sw-Tz\rangle$, take $\text{Proj}\ R[\sigma,\tau]/\langle \sigma y^2 - \tau x^2, \sigma w^2-\tau z^2\rangle$. The easy fix is to assume that the scheme $Y$ is normal. $\endgroup$
    – Jason Starr
    Commented Aug 9, 2019 at 9:24
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    $\begingroup$ @abx: You are absolutely right. One should, instead, impose a cusp in a horizontal direction. $\endgroup$
    – Sasha
    Commented Aug 9, 2019 at 10:32

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