Lemma 86.3.4. Let $(A_1, I_1) \to (A_2, I_2)$ be as in Remark 86.2.3 with $A_1$ and $A_2$ Noetherian. Let $B_1$ be in (86.2.0.2) for $(A_1, I_1)$. Let $B_2$ be the base change of $B_1$. Then there is a canonical map

\[ \mathop{N\! L}\nolimits _{B_1/A_1} \otimes _{B_2} B_1 \to \mathop{N\! L}\nolimits _{B_2/A_2} \]

which induces and isomorphism on $H^0$ and a surjection on $H^{-1}$.

**Proof.**
Choose a presentation $B_1 = A_1[x_1, \ldots , x_ r]^\wedge /J_1$. Since $A_2/I_2^ n[x_1, \ldots , x_ r] = A_1/I_1^{cn}[x_1, \ldots , x_ r] \otimes _{A_1/I_1^{cn}} A_2/I_2^ n$ we have

\[ A_2[x_1, \ldots , x_ r]^\wedge = (A_1[x_1, \ldots , x_ r]^\wedge \otimes _{A_1} A_2)^\wedge \]

where we use $I_2$-adic completion on both sides (but of course $I_1$-adic completion for $A_1[x_1, \ldots , x_ r]^\wedge $). Set $J_2 = J_1 A_2[x_1, \ldots , x_ r]^\wedge $. Arguing similarly we get the presentation

\begin{align*} B_2 & = (B_1 \otimes _{A_1} A_2)^\wedge \\ & = \mathop{\mathrm{lim}}\nolimits \frac{A_1/I_1^{cn}[x_1, \ldots , x_ r]}{J_1(A_1/I_1^{cn}[x_1, \ldots , x_ r])} \otimes _{A_1/I_1^{cn}} A_2/I_2^ n \\ & = \mathop{\mathrm{lim}}\nolimits \frac{A_2/I_2^ n[x_1, \ldots , x_ r]}{J_2(A_2/I_2^ n[x_1, \ldots , x_ r])} \\ & = A_2[x_1, \ldots , x_ r]^\wedge /J_2 \end{align*}

for $B_2$ over $A_2$. As a consequence obtain a commutative diagram

\[ \xymatrix{ \mathop{N\! L}\nolimits ^\wedge _{B_1/A_1} : \ar[d] & J_1/J_1^2 \ar[r]_-{\text{d}} \ar[d] & \bigoplus B_1\text{d}x_ i \ar[d] \\ \mathop{N\! L}\nolimits ^\wedge _{B_2/A_2} : & J_2/J_2^2 \ar[r]^-{\text{d}} & \bigoplus B_2\text{d}x_ i } \]

The induced arrow $J_1/J_1^2 \otimes _{B_1} B_2 \to J_2/J_2^2$ is surjective because $J_2$ is generated by the image of $J_1$. This determines the arrow displayed in the lemma. We omit the proof that this arrow is well defined up to homotopy (i.e., indepedent of the choice of the presentations up to homotopy). The statement about the induced map on cohomology modules follows easily from the discussion (details omitted).
$\square$

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