Let $X,\tilde{X}$ be two smooth projective varieties over $\mathbb{C}$, and let $\pi:\tilde{X}\rightarrow X$ be a projective morphism. Let us moreover assume that there exists a smooth closed subvariety $Y\subset X$, such that $\pi$ is isomorphism outside $Y$, and $\pi^{-1}(Y)\rightarrow Y$ is a projective bundle of rank = codim($Y,X$).

I want to show that $\tilde{X}$ is indeed the blowup of $X$ along $Y$. I have seen smoething along this line mentioined in a paper. Of course, there will be a map from $\tilde{X}$ to the blowup by the universal perty of blowups, but I can't show it to be an isomorphism. I know that any projective birational morphism is a blowup along some closed subscheme, but it's not clear from the proof that the closed subscheme is indeed $Y$.

Any help would be appreciated.

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    $\begingroup$ The induced morphism $\tilde{X}\to Bl_Y(X)$ is a birational morphism of smooth projective varieties with the same Picard number (equal to the Picard number of $X$ plus one). This implies that it is an isomorphism (the equality of Picard numbers implies that there are no exceptional divisors). $\endgroup$ – Olivier Benoist yesterday
  • $\begingroup$ @OlivierBenoist : Thank you for your prompt response. Could you please explain how you found the picard number of $\tilde{X}$? Also, why does same picard number imply isomorphism? Is there a reference to this? $\endgroup$ – yojusmath yesterday
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    $\begingroup$ The isomorphism statement follows from this post. $\endgroup$ – R. van Dobben de Bruyn yesterday
  • $\begingroup$ @R.vanDobbendeBruyn : In our case, isn't the exceptional subvariety for the map $\tilde{X}\rightarrow Bl_Y(X)$of codimension 1? Could you please explain how you are concluding from the post you have mentioned? $\endgroup$ – yojusmath 14 hours ago
  • $\begingroup$ If the Picard ranks are the same there cannot be a codimension $1$ exceptional locus, because the exceptional divisor would necessary be linearly independent in Néron–Severi. $\endgroup$ – R. van Dobben de Bruyn 12 hours ago

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