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Let $\phi:X\dashrightarrow Y$ be an elementary small transformation (isomorphism in codimension $1$) between normal and $\mathbb{Q}$-factorial projective varieties. Then there are small contractions $f:X\rightarrow Z$ and $g:Y\rightarrow Z$ such that $g\circ \phi = f$.

Now, let $X'\subset X$ be a normal $\mathbb{Q}$-factorial projective variety such that $f_{|X'}:X'\rightarrow Z$ is $1$ to $1$ onto $f(X')$, and let $Y'\subset Y$ be the closure of $\phi(X')$ in $Y$. Assume that $Y'$ and $X'$ have the same Picard rank.

In this situation, is $\phi_{|X'}:X'\dashrightarrow Y'$ necessarily an isomorphism?

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I think not. Here is a potential example: Let $Z$ be the cone over the Segre embedding of $\mathbb P^1\times\mathbb P^1$ and $W\to Z$ the blow-up of the vertex. Further let $X$ and $Y$ be the blow ups of the Weil divisors on $Z$ obtained as the embedded cones over one of the ruling rational curves on $\mathbb P^1\times\mathbb P^1$. (I.e., $X$ corresponds to one direction and $Y$ to the other). Then the morphism $W\to Z$ factors through both $X$ and $Y$, these morphisms can be thought of as blowing down each ruling of the exceptional divisor separately. The morphisms $X\to Z$ and $Y\to Z$ are small, $X$ and $Y$ are non-singular. Now take a curve $C\subseteq W$ not contained in the exceptional divisor of $W\to Z$, say $E$, which is isomorphic to $\mathbb P^1\times\mathbb P^1$ such that $C$ is tangent to one of the rulings, but transversal to the other and such that $C$ intersects $E$ in a single point. Now let $X'$ and $Y'$ be the images of $C$ in $X$ and $Y$ respectively. Then this satisfy the requirements, the induced map between $X'$ and $Y'$ will be 1-1 (and to their image in $Z$ as well), but it will not be an isomorphism.

Addition in response to the question in the comments: Let $D_X,D_Y\subseteq Z$ the Weil divisors that give the blow ups that lead to $X$ and $Y$, i.e., $f:X=Bl_{D_X}Z\to Z$ and $g:Y=Bl_{D_Y}Z\to Z$. These correspond to the two rulings on $\mathbb P^1\times\mathbb P^1$. Now, let $X'\subseteq X$ the proper transform of $D_Y$ on $X$. Then $Y'$ is the proper transform of $D_Y$ on $Y$, but then it is actually its preimage there, so $X'\to D_X=f(X')$ is 1-1 and in fact an isomorphism, but $Y'\to D_X$ is not even 1-1. To make sure everything is OK, note that $X'\simeq D_X$ is a cone over a conic, so it is normal, while $Y'$ is its blow up at the vertex, so it is even smooth.

Finally, if both $X'\to Z$ and $Y'\to Z$ are 1-1 and both $X'$ and $Y'$ are normal, then they are isomorphic, because they are both the normalization of $Z$ (the morphisms factor through the normalization, and then use ZMT).

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  • $\begingroup$ Thank you for the answer. In your example $Y'$ is not normal. If we assume in addition that $Y'$ is normal are then $X'$ and $Y'$ isomorphic? $\endgroup$
    – J. Ross
    Jun 11 at 9:25
  • $\begingroup$ It looks like that this still fails, even if $Y'$ is smooth. However, if in addition $Y'\to f(X')$ is 1-1, then $X'\simeq Y'$. See the addition to my answer. $\endgroup$ Jun 15 at 7:08

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