Lemma 99.5.7. In Situation 99.5.1. Let
\[ \xymatrix{ Z \ar[r] \ar[d] & Z' \ar[d] \\ Y \ar[r] & Y' } \]
be a pushout in the category of schemes over $S$ where $Z \to Z'$ is a thickening and $Z \to Y$ is affine, see More on Morphisms, Lemma 37.14.3. Then the functor on fibre categories
\[ \mathcal{C}\! \mathit{oh}_{X/B, Y'} \longrightarrow \mathcal{C}\! \mathit{oh}_{X/B, Y} \times _{\mathcal{C}\! \mathit{oh}_{X/B, Z}} \mathcal{C}\! \mathit{oh}_{X/B, Z'} \]
is an equivalence.
Proof.
Observe that the corresponding map
\[ B(Y') \longrightarrow B(Y) \times _{B(Z)} B(Z') \]
is a bijection, see Pushouts of Spaces, Lemma 81.6.1. Thus using the commutative diagram
\[ \xymatrix{ \mathcal{C}\! \mathit{oh}_{X/B, Y'} \ar[r] \ar[d] & \mathcal{C}\! \mathit{oh}_{X/B, Y} \times _{\mathcal{C}\! \mathit{oh}_{X/B, Z}} \mathcal{C}\! \mathit{oh}_{X/B, Z'} \ar[d] \\ B(Y') \ar[r] & B(Y) \times _{B(Z)} B(Z') } \]
we see that we may assume that $Y'$ is a scheme over $B'$. By Remark 99.5.5 we may replace $B$ by $Y'$ and $X$ by $X \times _ B Y'$. Thus we may assume $B = Y'$. In this case the statement follows from Pushouts of Spaces, Lemma 81.6.6.
$\square$
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