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Tag 08X7

Chapter 34: Descent > Section 34.4: Descent for universally injective morphisms

Lemma 34.4.20. If $f$ is universally injective, then the diagram \begin{equation} \tag{34.4.20.1} \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 (34.4.19.1) over $R$ with $S$ is an equalizer.

Proof. By Lemma 34.4.12 and Remark 34.4.13, 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 (34.4.17.1); 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 34.4.11, 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 34.4.10, we conclude (34.4.20.1) must also be an equalizer. $\square$

    The code snippet corresponding to this tag is a part of the file descent.tex and is located in lines 1207–1223 (see updates for more information).

    \begin{lemma}
    \label{lemma-descent-lemma}
    If $f$ is universally injective, then the diagram
    \begin{equation}
    \label{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 (\ref{equation-equalizer-f}) over $R$ with $S$ is an 
    equalizer.
    \end{lemma}
    
    \begin{proof}
    By
    Lemma \ref{lemma-split-surjection} and
    Remark \ref{remark-functorial-splitting},
    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
    (\ref{equation-coequalizer-CM}); 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 \ref{remark-adjunction}, 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 \ref{lemma-C-is-faithful}, we conclude
    (\ref{equation-equalizer-f2}) must also be an equalizer.
    \end{proof}

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