# If it quacks like a conifold resolution and it waddles like a conifold resolution, $\ldots$

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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• I suggest that you add the hypothesis that $Y$ is normal. – Jason Starr Aug 9 at 9:25
• Is the equality $f^{-1}(p) = \mathbb{P}^1$ true scheme-theoretically? – Angelo Aug 11 at 8:02
• @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. – Jim Bryan Aug 11 at 16:38

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.
• If $f^{-1}(p)$ has a cusp, it is not isomorphic to $\Bbb{P}^1$? – abx Aug 9 at 8:44
• 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. – Jason Starr Aug 9 at 9:24