# Restriction of small transformations

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?

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).
• 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? Jun 11 at 9:25
• 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. Jun 15 at 7:08