Lemma 35.4.20. If $f$ is universally injective, then the diagram
obtained by tensoring (35.4.19.1) over $R$ with $S$ is an equalizer.
Lemma 35.4.20. If $f$ is universally injective, then the diagram
obtained by tensoring (35.4.19.1) over $R$ with $S$ is an equalizer.
Proof. By Lemma 35.4.12 and Remark 35.4.13, the map $C(1_ N \otimes f): C(N \otimes _ R S) \to C(N)$ can be split functorially in $N$. This gives the upper vertical arrows in the commutative diagram
in which the compositions along the columns are identity morphisms. The second row is the coequalizer diagram (35.4.17.1); this produces the dashed arrow. From the top right square, we obtain auxiliary morphisms $C(f_*(M,\theta )) \to C(M)$ and $C(M) \to C(M\otimes _{S,\delta _1^1} S_2)$ which imply that the first row is a split coequalizer diagram. By Remark 35.4.11, we may tensor with $S$ inside $C$ to obtain the split coequalizer diagram
By Lemma 35.4.10, we conclude (35.4.20.1) must also be an equalizer. $\square$
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