Lemma 28.18.5. Let $U \to V$ be an open immersion of quasi-affine schemes. Then

is cartesian.

Lemma 28.18.5. Let $U \to V$ be an open immersion of quasi-affine schemes. Then

\[ \xymatrix{ U \ar[d] \ar[rr]_-j & & \mathop{\mathrm{Spec}}(\Gamma (U, \mathcal{O}_ U)) \ar[d] \\ U \ar[r] & V \ar[r]^-{j'} & \mathop{\mathrm{Spec}}(\Gamma (V, \mathcal{O}_ V)) } \]

is cartesian.

**Proof.**
The diagram is commutative by Schemes, Lemma 26.6.4. Write $A = \Gamma (U, \mathcal{O}_ U)$ and $B = \Gamma (V, \mathcal{O}_ V)$. Let $g \in B$ be such that $V_ g$ is affine and contained in $U$. This means that if $f$ is the image of $g$ in $A$, then $U_ f = V_ g$. By Lemma 28.18.3 we see that $j'$ induces an isomorphism of $V_ g$ with the standard open $D(g)$ of $\mathop{\mathrm{Spec}}(B)$. Thus $V_ g \times _{\mathop{\mathrm{Spec}}(B)} \mathop{\mathrm{Spec}}(A) \to \mathop{\mathrm{Spec}}(A)$ is an isomorphism onto $D(f) \subset \mathop{\mathrm{Spec}}(A)$. By Lemma 28.18.3 again $j$ maps $U_ f$ isomorphically to $D(f)$. Thus we see that $U_ f = U_ f \times _{\mathop{\mathrm{Spec}}(B)} \mathop{\mathrm{Spec}}(A)$. Since by Lemma 28.18.4 we can cover $U$ by $V_ g = U_ f$ as above, we see that $U \to U \times _{\mathop{\mathrm{Spec}}(B)} \mathop{\mathrm{Spec}}(A)$ is an isomorphism.
$\square$

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