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Let $i : X \rightarrow Y$ be a closed immersion corresponding to a finite type ideal sheaf. Let $f : B \rightarrow Y$ be a morphism of schemes. Suppose that for all affine open sets $U \subseteq Y$, $f^{-1}(U) \rightarrow U$ is isomorphic to $Bl_{X \cap U}(U)$, and the isomorphism commutes with the morphisms to $U$. Does this imply that $B$ is isomorphic to $Bl_X(Y)$ with $f$ being the canonical morphism?

The reason for this questions is that in Vakil's Foundations of Algebraic Geometry, the proof of theorem 22.3.2 that $Bl_X(Y) = \underline{Proj}_Y(O_Y \oplus I \oplus I^2 \oplus ...)$ begins with "Reduce to the case of affine target $Spec(R)$ with ideal $I \subseteq R$". But I'm not sure why we're allowed to do this reduction. Otherwise we just get a local isomorphism.

In Vakil, the blowup is defined by the universal property: It's the final object in the category of morphsims $W \rightarrow Y$ such that $X \times_Y W \rightarrow W$ is an effective Cartier divisor.

I expect there to be some kind of compatibility conditions involved in the isomorphsims $f^{-1}(U) \simeq Bl_{X \cap U}(U)$. I don't understand why the local isomorphisms glue to give a global isomorphism.

I think maybe the uniqueness of the final object has something to do with it.

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Let $\{U_i\}$ be an open affine covering of $Y$. The morphism $B\to Y$ you have in hand has the following property. If $V\to U_i$ is a morphism and $V_{X_{U_i}}$ is a Cartier divisor in $V$ then there is a unique $U_i$-morphism $V\to B_{U_i}$. By the definition of open immersions, this means that if $V\to Y$ is a morphism whose (set-theoretic) image lies in $U_i$ and such that $V_X$ is a Cartier divisor in $V$, then there is a unique $Y$-morphism $V\to B$.

Now suppose that $W\to Y$ is such that $W_X$ is a Cartier divisor in $W$. Then $(W_X)_{U_i}$ is a Cartier divisor in $W_{U_i}$. Hence there is a unique $Y$-morphism $W_{U_i}\to B$. For any $j$, the induced $Y$-morphism $W_{U_i\times_Y U_j}\to W_{U_i}\to B$ is also unique, since the image of $W_{U_i\times_Y U_j}$ in $Y$ lies inside $U_i$. Hence the $Y$-morphisms $W_{U_i}\to B$ and $W_{U_j}\to B$ coincide on $W_{U_i\times U_j}\simeq W_{U_i}\times_W W_{U_j}$. Since the $W_{U_i}$ form an open covering of $W$, the various $Y$-morphisms $W_{U_i}\to B$ glue to a $Y$-morphism $W\to B$ (use "glueing morphisms" in Hartshorne, p. 88). This morphism is unique because any two distinct such morphisms would differ on one of the open subsets $W_{U_i}$ of $W$, which is impossible by assumption. So $B$ has the universal property and must be the blow-up.

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