Remark 91.13.4 (Functoriality of the Lichtenbaum-Schlessinger complex). Consider a commutative square

of ring maps. Choose a factorization

with $P$ a polynomial algebra over $A$ and $P'$ a polynomial algebra over $A'$. Choose generators $f_ t$, $t \in T$ for $\mathop{\mathrm{Ker}}(P \to B)$. For $t \in T$ denote $f'_ t$ the image of $f_ t$ in $P'$. Choose $f'_ s \in P'$ such that the elements $f'_ t$ for $t \in T' = T \amalg S$ generate the kernel of $P' \to B'$. Set $F = \bigoplus _{t \in T} P$ and $F' = \bigoplus _{t' \in T'} P'$. Let $Rel = \mathop{\mathrm{Ker}}(F \to P)$ and $Rel' = \mathop{\mathrm{Ker}}(F' \to P')$ where the maps are given by multiplication by $f_ t$, resp. $f'_ t$ on the coordinates. Finally, set $TrivRel$, resp. $TrivRel'$ equal to the submodule of $Rel$, resp. $TrivRel$ generated by the elements $(\ldots , f_{t'}, 0, \ldots , 0, -f_ t, 0, \ldots )$ for $t, t' \in T$, resp. $T'$. Having made these choices we obtain a canonical commutative diagram

Moreover, tracing through the choices made in the proof of Lemma 91.13.3 the reader sees that one obtains a commutative diagram

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