Lemma 98.5.7. In Situation 98.5.1. 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 80.6.1. 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 98.5.5 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 80.6.6. $\square$

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