Lemma 54.16.1. Let $X$ be a Noetherian scheme. Let $E \subset X$ be an exceptional curve of the first kind. If a contraction $X \to X'$ of $E$ exists, then it has the following universal property: for every morphism $\varphi : X \to Y$ such that $\varphi (E)$ is a point, there is a unique factorization $X \to X' \to Y$ of $\varphi$.

Proof. Let $b : X \to X'$ be a contraction of $E$. As a topological space $X'$ is the quotient of $X$ by the relation identifying all points of $E$ to one point. Namely, $b$ is proper (Divisors, Lemma 31.32.13 and Morphisms, Lemma 29.43.5) and surjective, hence defines a submersive map of topological spaces (Topology, Lemma 5.6.5). On the other hand, the canonical map $\mathcal{O}_{X'} \to b_*\mathcal{O}_ X$ is an isomorphism. Namely, this is clear over the complement of the image point $x \in X'$ of $E$ and on stalks at $x$ the map is an isomorphism by part (4) of Lemma 54.3.4. Thus the pair $(X', \mathcal{O}_{X'})$ is constructed from $X$ by taking the quotient as a topological space and endowing this with $b_*\mathcal{O}_ X$ as structure sheaf.

Given $\varphi$ we can let $\varphi ' : X' \to Y$ be the unique map of topological spaces such that $\varphi = \varphi ' \circ b$. Then the map

$\varphi ^\sharp : \varphi ^{-1}\mathcal{O}_ Y = b^{-1}((\varphi ')^{-1}\mathcal{O}_ Y) \to \mathcal{O}_ X$

$(\varphi ')^\sharp : (\varphi ')^{-1}\mathcal{O}_ Y \to b_*\mathcal{O}_ X = \mathcal{O}_{X'}$

Then $(\varphi ', (\varphi ')^\sharp )$ is a morphism of ringed spaces from $X'$ to $Y$ such that we get the desired factorization. Since $\varphi$ is a morphism of locally ringed spaces, it follows that $\varphi '$ is too. Namely, the only thing to check is that the map $\mathcal{O}_{Y, y} \to \mathcal{O}_{X', x}$ is local, where $y \in Y$ is the image of $E$ under $\varphi$. This is true because an element $f \in \mathfrak m_ y$ pulls back to a function on $X$ which is zero in every point of $E$ hence the pull back of $f$ to $X'$ is a function defined on a neighbourhood of $x$ in $X'$ with the same property. Then it is clear that this function must vanish at $x$ as desired. $\square$

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