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

35.4.20.1

\begin{equation} \label{descent-equation-equalizer-f2} \xymatrix@C=8pc{ f_*(M, \theta ) \otimes _ R S \ar[r]^{\theta \circ (1_ M \otimes \delta _0^1)} & M \otimes _{S, \delta _1^1} S_2 \ar@<1ex>[r]^{(\theta \otimes \delta _2^2) \circ (1_ M \otimes \delta ^2_0)} \ar@<-1ex>[r]_{1_{M \otimes S_2} \otimes \delta ^2_1} & M \otimes _{S, \delta _{12}^1} S_3 } \end{equation}

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

\[ \xymatrix@C=8pc{ C(M \otimes _{S, \delta _1^1} S_2) \ar@<1ex>^{C(\theta \circ (1_ M \otimes \delta _0^1))}[r] \ar@<-1ex>_{C(1_ M \otimes \delta _1^1)}[r] \ar[d] & C(M) \ar[r]\ar[d] & C(f_*(M,\theta )) \ar@{-->}[d] \\ C(M \otimes _{S,\delta _{12}^1} S_3) \ar@<1ex>^{C((\theta \otimes \delta _2^2) \circ (1_ M \otimes \delta ^2_0))}[r] \ar@<-1ex>_{C(1_{M \otimes S_2} \otimes \delta ^2_1)}[r] \ar[d] & C(M \otimes _{S, \delta _1^1} S_2 ) \ar[r]^{C(\theta \circ (1_ M \otimes \delta _0^1))} \ar[d]^{C(1_ M \otimes \delta _1^1)} & C(M) \ar[d] \ar@{=}[dl] \\ C(M \otimes _{S, \delta _1^1} S_2) \ar@<1ex>[r]^{C(\theta \circ (1_ M \otimes \delta _0^1))} \ar@<-1ex>[r]_{C(1_ M \otimes \delta _1^1)} & C(M) \ar[r] & C(f_*(M,\theta )) } \]

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

\[ \xymatrix@C=8pc{ C(M \otimes _{S,\delta _2^2 \circ \delta _1^1} S_3) \ar@<1ex>^{C((\theta \otimes \delta _2^2) \circ (1_ M \otimes \delta ^2_0))}[r] \ar@<-1ex>_{C(1_{M \otimes S_2} \otimes \delta ^2_1)}[r] & C(M \otimes _{S, \delta _1^1} S_2 ) \ar[r]^{C(\theta \circ (1_ M \otimes \delta _0^1))} & C(f_*(M,\theta ) \otimes _ R S). } \]

By Lemma 35.4.10, we conclude (35.4.20.1) must also be an equalizer. $\square$

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