Lemma 97.27.5. In Situation 97.27.1 if there exists a solution $(f : X' \to X, T, a)$ then there is a functorial bijection $F(V) = \mathop{\mathrm{Mor}}\nolimits _ S(V, X)$ on the category of locally Noetherian schemes $V$ over $S$.

**Proof.**
Let $V$ be a locally Noetherian scheme over $S$. Let $x : V \to X$ be a morphism over $S$. Then we get an element $(Z, u', \hat x)$ in $F(V)$ as follows

$Z \subset V$ is the inverse image of $T$ by $x$,

$u' : V \setminus Z \to U' = U$ is the restriction of $x$ to $V \setminus Z$,

$\hat x : V_{/Z} \to W$ is the composition of $x_{/Z} : V_{/Z} \to X_{/T}$ with the isomorphism $a : X_{/T} \to W$.

This triple satisfies the compatibility condition because we can take $V' = V \times _{x, X} X'$, we can take $\hat x'$ the completion of the projection $x' : V' \to X'$.

Conversely, suppose given an element $(Z, u', \hat x)$ of $F(V)$. We claim there is a unique morphism $x : V \to X$ compatible with $u'$ and $\hat x$. Namely, let $V' \to V$, $\hat x'$, and $x'$ witness the compatibility between $u'$ and $\hat x$. Then Algebraization of Formal Spaces, Proposition 87.26.1 is exactly the result we need to find a unique morphism $x : V \to X$ agreeing with $\hat x$ over $V_{/Z}$ and with $x'$ over $V'$ (and a fortiori agreeing with $u'$ over $V \setminus Z$).

We omit the verification that the two constructions above define inverse bijections between their respective domains. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)