hcju/stacksqa huggingface dataset:
By Lemma \ref{lemma-exterior-product-associative-dim-1} we have $(\alpha \times \beta) \times \gamma = \alpha \times (\beta \times \gamma)$ in $\CH_*(Y \times_S Z \times_S W)$. Consider the closed immersions $$ \Delta_{12} : X \times_S X \longrightarrow X \times_S X \times_S X, \quad (x, x') \mapsto (x, x, x') $$ and $$ \Delta_{23} : X \times_S X \longrightarrow X \times_S X \times_S X, \quad (x, x') \mapsto (x, x', x') $$ Denote $\Delta_{12}^!$ and $\Delta_{23}^!$ the corresponding bivariant classes; observe that $\Delta_{12}^!$ is the restriction (Remark \ref{remark-restriction-bivariant}) of $\Delta^!$ to $X \times_S X \times_S X$ by the map $\text{pr}_{12}$ and that $\Delta_{23}^!$ is the restriction of $\Delta^!$ to $X \times_S X \times_S X$ by the map $\text{pr}_{23}$. Thus clearly the restriction of $\Delta_{12}^!$ by $\Delta_{23}$ is $\Delta^!$ and the restriction of $\Delta_{23}^!$ by $\Delta_{12}$ is $\Delta^!$ too. Thus by Lemma \ref{lemma-gysin-commutes} we have $$ \Delta^! \circ \Delta_{12}^! = \Delta^! \circ \Delta_{23}^!  $$ Now we can prove the lemma by the following sequence of equalities: \begin{align*} (\alpha \cdot \beta) \cdot \gamma & = \Delta^!(\Delta^!(\alpha \times \beta) \times \gamma) \\ & = \Delta^!(\Delta_{12}^!((\alpha \times \beta) \times \gamma)) \\ & = \Delta^!(\Delta_{23}^!((\alpha \times \beta) \times \gamma)) \\ & = \Delta^!(\Delta_{23}^!(\alpha \times (\beta \times \gamma)) \\ & = \Delta^!(\alpha \times \Delta^!(\beta \times \gamma)) \\ & = \alpha \cdot (\beta \cdot \gamma) \end{align*} All equalities are clear from the above except perhaps for the second and penultimate one. The equation $\Delta_{23}^!(\alpha \times (\beta \times \gamma)) = \alpha \times \Delta^!(\beta \times \gamma)$ holds by Remark \ref{remark-commuting-exterior}. Similarly for the second equation.
Denote $g : X \to S$ the structure morphism. The map is the composition of the isomorphisms $$ A^p(X) \to A^{p - d + 1}(X \to S) \to \CH_{d - p}(X) $$ The first is the isomorphism $c \mapsto c \circ g^*$ of Proposition \ref{proposition-compute-bivariant} and the second is the isomorphism $c \mapsto c \cap [S]_1$ of Lemma \ref{lemma-chow-cohomology-towards-base-dim-1}. From the proof of Lemma \ref{lemma-chow-cohomology-towards-base-dim-1} we see that the inverse to the second arrow sends $\alpha \in \CH_{d - p}(X)$ to the bivariant class $c_\alpha$ which sends $\beta \in \CH_*(Y)$ for $Y$ locally of finite type over $k$ to $\alpha \times \beta$ in $\CH_*(X \times_k Y)$. From the proof of Proposition \ref{proposition-compute-bivariant} we see the inverse to the first arrow in turn sends $c_\alpha$ to the bivariant class which sends $\beta \in \CH_*(Y)$ for $Y \to X$ locally of finite type to $\Delta^!(\alpha \times \beta) = \alpha \cdot \beta$. From this the final result of the lemma follows.
The projection $X' \to X$ is flat as a base change of the flat morphism $S' \to S$ (Morphisms, Lemma \ref{morphisms-lemma-base-change-flat}). Hence every generic point $x'$ of an irreducible component of $X'$ maps to the generic point $x \in X$ (because generalizations lift along $X' \to X$ by Morphisms, Lemma \ref{morphisms-lemma-generalizations-lift-flat}). Let $s \in S$ be the image of $x$. Recall that the scheme $S'_s = S' \times_S s$ has the same underlying topological space as $g^{-1}(\{s\})$ (Schemes, Lemma \ref{schemes-lemma-fibre-topological}). We may view $x'$ as a point of the scheme $S'_s \times_s x$ which comes equipped with a monomorphism $S'_s \times_s x \to S' \times_S X$. Of course, $x'$ is a generic point of an irreducible component of $S'_s \times_s x$ as well. Using the flatness of $\Spec(\kappa(x)) \to \Spec(\kappa(s)) = s$ and arguing as above, we see that $x'$ maps to a generic point $s'$ of an irreducible component of $g^{-1}(\{s\})$. Hence $\delta'(s') = \delta(s) + c$ by assumption. We have $\dim_x(X_s) = \dim_{x'}(X_{s'})$ by Morphisms, Lemma \ref{morphisms-lemma-dimension-fibre-after-base-change}. Since $x$ is a generic point of an irreducible component $X_s$ (this is an irreducible scheme but we don't need this) and $x'$ is a generic point of an irreducible component of $X'_{s'}$ we conclude that $\text{trdeg}_{\kappa(s)}(\kappa(x)) =  \text{trdeg}_{\kappa(s')}(\kappa(x'))$ by Morphisms, Lemma \ref{morphisms-lemma-dimension-fibre-at-a-point}. Then $$ \delta_{X'/S'}(x') = \delta(s') + \text{trdeg}_{\kappa(s')}(\kappa(x')) = \delta(s) + c + \text{trdeg}_{\kappa(s)}(\kappa(x)) = \delta_{X/S}(x) + c $$ This proves what we want by Definition \ref{definition-delta-dimension}.
The proof is exactly the same is the proof of Lemma \ref{lemma-pullback-coherent} and we suggest the reader skip it. \medskip\noindent The statements on dimensions follow from Lemma \ref{lemma-dimension-base-change}. Part (1) follows from part (2) by Lemma \ref{lemma-cycle-closed-coherent} and the fact that the base change of the coherent module $\mathcal{O}_Z$ is $\mathcal{O}_{Z'}$. \medskip\noindent Proof of (2). As $X$, $X'$ are locally Noetherian we may apply Cohomology of Schemes, Lemma \ref{coherent-lemma-coherent-Noetherian} to see that $\mathcal{F}$ is of finite type, hence $\mathcal{F}'$ is of finite type (Modules, Lemma \ref{modules-lemma-pullback-finite-type}), hence $\mathcal{F}'$ is coherent (Cohomology of Schemes, Lemma \ref{coherent-lemma-coherent-Noetherian} again). Thus the lemma makes sense. Let $W \subset X$ be an integral closed subscheme of $\delta$-dimension $k$, and let $W' \subset X'$ be an integral closed subscheme of $\delta'$-dimension $k + c$ mapping into $W$ under $X' \to X$. We have to show that the coefficient $n$ of $[W']$ in $g^*[\mathcal{F}]_k$ agrees with the coefficient $m$ of $[W']$ in $[\mathcal{F}']_{k + c}$. Let $\xi \in W$ and $\xi' \in W'$ be the generic points. Let $A = \mathcal{O}_{X, \xi}$, $B = \mathcal{O}_{X', \xi'}$ and set $M = \mathcal{F}_\xi$ as an $A$-module. (Note that $M$ has finite length by our dimension assumptions, but we actually do not need to verify this. See Lemma \ref{lemma-length-finite}.) We have $\mathcal{F}'_{\xi'} = B \otimes_A M$. Thus we see that $$ n = \text{length}_B(B \otimes_A M) \quad \text{and} \quad m = \text{length}_A(M) \text{length}_B(B/\mathfrak m_AB) $$ Thus the equality follows from Algebra, Lemma \ref{algebra-lemma-pullback-module}.
Suppose that $\alpha \in Z_k(X)$ is a $k$-cycle which is rationally equivalent to zero. By Lemma \ref{lemma-rational-equivalence-family} there exists a locally finite family of integral closed subschemes $W_i \subset X \times \mathbf{P}^1$ of $\delta$-dimension $k$ not contained in the divisors $(X \times \mathbf{P}^1)_0$ or $(X \times \mathbf{P}^1)_\infty$ of $X \times \mathbf{P}^1$ such that $\alpha = \sum ([(W_i)_0]_k - [(W_i)_\infty]_k)$. Thus it suffices to prove for $W \subset X \times \mathbf{P}^1$ integral closed of $\delta$-dimension $k$ not contained in the divisors $(X \times \mathbf{P}^1)_0$ or $(X \times \mathbf{P}^1)_\infty$ of $X \times \mathbf{P}^1$ we have \begin{enumerate} \item the base change $W' \subset X' \times \mathbf{P}^1$ satisfies the assumptions of Lemma \ref{lemma-closed-subscheme-cross-p1} with $k$ replaced by $k + c$, and \item $g^*[W_0]_k = [(W')_0]_{k + c}$ and $g^*[W_\infty]_k = [(W')_\infty]_{k + c}$. \end{enumerate} Part (2) follows immediately from Lemma \ref{lemma-pullback-coherent-base-change} and the fact that $(W')_0$ is the base change of $W_0$ (by associativity of fibre products). For part (1), first the statement on dimensions follows from Lemma \ref{lemma-dimension-base-change}. Then let $w' \in (W')_0$ with image $w \in W_0$ and $z \in \mathbf{P}^1_S$. Denote $t \in \mathcal{O}_{\mathbf{P}^1_S, z}$ the usual equation for $0 : S \to \mathbf{P}^1_S$. Since $\mathcal{O}_{W, w} \to \mathcal{O}_{W', w'}$ is flat and since $t$ is a nonzerodivisor on $\mathcal{O}_{W, w}$ (as $W$ is integral and $W \not = W_0$) we see that also $t$ is a nonzerodivisor in $\mathcal{O}_{W', w'}$. Hence $W'$ has no associated points lying on $W'_0$.
In fact, we claim it commutes on the level of cycles. Namely, let $Z \subset X$ be an integral closed subscheme of $\delta$-dimension $k$ and denote $Z' \subset X'$ its base change. By construction we have $g^*[Z] = [Z']_{k + c}$. By Lemma \ref{lemma-pullback-coherent} we have $(f')^*g^*[Z] = [Z' \times_{X'} Y']_{k + c + r}$. Conversely, we have $f^*[Z] = [Z \times_X Y]_{k + r}$ by Definition \ref{definition-flat-pullback}. By Lemma \ref{lemma-pullback-coherent-base-change} we have $g^*f^*[Z] = [(Z \times_X Y)']_{k + r + c}$. Since $(Z \times_X Y)' = Z' \times_{X'} Y'$ by associativity of fibre product we conclude.
In fact, we claim it commutes on the level of cycles. Namely, let $Z \subset Y$ be an integral closed subscheme of $\delta$-dimension $k$ and denote $Z' \subset X'$ its base change. By construction we have $g^*[Z] = [Z']_{k + c}$. By Lemma \ref{lemma-cycle-push-sheaf} we have $(f')_*g^*[Z] = [f'_*\mathcal{O}_{Z'}]_{k + c}$. By the same lemma we have $f_*[Z] = [f_*\mathcal{O}_Z]_k$. By Lemma \ref{lemma-pullback-coherent-base-change} we have $g^*f_*[Z] = [(X' \to X)^*f_*\mathcal{O}_Z]_{k + r}$. Thus it suffices to show that $$ (X' \to X)^*f_*\mathcal{O}_Z \cong f'_*\mathcal{O}_{Z'} $$ as coherent modules on $X'$. As $X' \to X$ is flat and as $\mathcal{O}_{Z'} = (Y' \to Y)^*\mathcal{O}_Z$, this follows from flat base change, see Cohomology of Schemes, Lemma \ref{coherent-lemma-flat-base-change-cohomology}.
Let $p : L \to X$ be the line bundle associated to $\mathcal{L}$ with zero section $o : X \to L$. For $\alpha \in CH_k(X)$ we know that $\beta = c_1(\mathcal{L}) \cap \alpha$ is the unique element of $\CH_{k - 1}(X)$ such that $o_*\alpha = - p^*\beta$, see Lemmas \ref{lemma-linebundle} and \ref{lemma-linebundle-formulae}. The same characterization holds after pullback. Hence the lemma follows from Lemmas \ref{lemma-pullback-base-change-pullback} and \ref{lemma-pullback-base-change-pushforward}.
Set $P = \mathbf{P}(\mathcal{E})$. The base change $P'$ of $P$ is equal to $\mathbf{P}(\mathcal{E}')$. Since we already know that flat pullback and cupping with $c_1$ of an invertible module commute with base change (Lemmas \ref{lemma-pullback-base-change-pullback} and \ref{lemma-pullback-base-change-c1}) the lemma follows from the characterization of capping with $c_i(\mathcal{E})$ given in Lemma \ref{lemma-determine-intersections}.
Let $s \in S$ and let $s'' \in S''$ be a generic point of an irreducible component of $(g \circ g')^{-1}(\{s\})$. Set $s' = g'(s'')$. Clearly, $s''$ is a generic point of an irreducible component of $(g')^{-1}(\{s'\})$. Moreover, since $g'$ is flat and hence generalizations lift along $g'$ (Morphisms, Lemma \ref{morphisms-lemma-base-change-flat}) we see that also $s'$ is a generic point of an irreducible component of $g^{-1}(\{s\})$. Thus by assumption $\delta'(s') = \delta(s) + c$ and $\delta''(s'') = \delta'(s') + c'$. We conclude $\delta''(s'') = \delta(s) + c + c'$ and the first part of the statement is true. \medskip\noindent For the second part, let $Z \subset X$ be an integral closed subscheme of $\delta$-dimension $k$. Denote $Z' \subset X'$ and $Z'' \subset X''$ the base changes. By definition we have $g^*[Z] = [Z']_{k + c}$. By Lemma \ref{lemma-pullback-coherent-base-change} we have $(g')^*[Z']_{k + c} = [Z'']_{k + c + c'}$. This proves the final statement.
It is clear that the rule described in the lemma gives a $\kappa$-linear map since all of the admissible relations are satisfied by the usual symbols $e_1 \wedge \ldots \wedge e_l$. It is also clearly a surjective map. Since by Lemma \ref{lemma-dimension-at-most-one} the left hand side has dimension at most one we see that the map is an isomorphism.
The significance of taking nonzero symbols in the explicit description of the map $\gamma_{K \to L \to M}$ is simply that if $(e_1, \ldots, e_l)$ is an admissible sequence in $K$, and $(\overline{f}_1, \ldots, \overline{f}_m)$ is an admissible sequence in $M$, then it is not guaranteed that $(e_1, \ldots, e_l, f_1, \ldots, f_m)$ is an admissible sequence in $L$ (where of course $f_i \in L$ signifies a lift of $\overline{f}_i$). However, if the symbol $[e_1, \ldots, e_l]$ is nonzero in $\det_\kappa(K)$, then necessarily $K = \langle e_1, \ldots, e_k\rangle$ (see proof of Lemma \ref{lemma-dimension-at-most-one}), and in this case it is true that $(e_1, \ldots, e_k, f_1, \ldots, f_m)$ is an admissible sequence. Moreover, by the admissible relations of type (b) for $\det_\kappa(L)$ we see that the value of $[e_1, \ldots, e_k, f_1, \ldots, f_m]$ in $\det_\kappa(L)$ is independent of the choice of the lifts $f_i$ in this case also. Given this remark, it is clear that an admissible relation for $e_1, \ldots, e_k$ in $K$ translates into an admissible relation among $e_1, \ldots, e_k, f_1, \ldots, f_m$ in $L$, and similarly for an admissible relation among the $\overline{f}_1, \ldots, \overline{f}_m$. Thus $\gamma$ defines a linear map of vector spaces as claimed in the lemma. \medskip\noindent By Lemma \ref{lemma-determinant-dimension-one} we know $\det_\kappa(L)$ is generated by any single symbol $[x_1, \ldots, x_{k + m}]$ such that $(x_1, \ldots, x_{k + m})$ is an admissible sequence with $L = \langle x_1, \ldots, x_{k + m}\rangle$. Hence it is clear that the map $\gamma_{K \to L \to M}$ is surjective and hence an isomorphism. \medskip\noindent Property (1) holds because \begin{eqnarray*} & & \det\nolimits_\kappa(v)([e_1, \ldots, e_k, f_1, \ldots, f_m]) \\ & = & [v(e_1), \ldots, v(e_k), v(f_1), \ldots, v(f_m)] \\ & = & \gamma_{K' \to L' \to M'}([u(e_1), \ldots, u(e_k)] \otimes [w(f_1), \ldots, w(f_m)]). \end{eqnarray*} Property (2) means that given a symbol $[\alpha_1, \ldots, \alpha_a]$ generating $\det_\kappa(A)$, a symbol $[\gamma_1, \ldots, \gamma_c]$ generating $\det_\kappa(C)$, a symbol $[\zeta_1, \ldots, \zeta_g]$ generating $\det_\kappa(G)$, and a symbol $[\iota_1, \ldots, \iota_i]$ generating $\det_\kappa(I)$ we have \begin{eqnarray*} & & [\alpha_1, \ldots, \alpha_a, \tilde\gamma_1, \ldots, \tilde\gamma_c, \tilde\zeta_1, \ldots, \tilde\zeta_g, \tilde\iota_1, \ldots, \tilde\iota_i] \\ & = & (-1)^{cg} [\alpha_1, \ldots, \alpha_a, \tilde\zeta_1, \ldots, \tilde\zeta_g, \tilde\gamma_1, \ldots, \tilde\gamma_c, \tilde\iota_1, \ldots, \tilde\iota_i] \end{eqnarray*} (for suitable lifts $\tilde{x}$ in $E$) in $\det_\kappa(E)$. This holds because we may use the admissible relations of type (c) $cg$ times in the following order: move the $\tilde\zeta_1$ past the elements $\tilde\gamma_c, \ldots, \tilde\gamma_1$ (allowed since $\mathfrak m\tilde\zeta_1 \subset A$), then move $\tilde\zeta_2$ past the elements $\tilde\gamma_c, \ldots, \tilde\gamma_1$ (allowed since $\mathfrak m\tilde\zeta_2 \subset A + R\tilde\zeta_1$), and so on. \medskip\noindent Part (3) of the lemma is obvious. This finishes the proof.
If the length of $M$ as an $R$-module is $l$, then the length of $M$ as an $R'$-module (i.e., $M_{R'}$) is $l$ as well, see Algebra, Lemma \ref{algebra-lemma-pushdown-module}. Note that an admissible sequence $x_1, \ldots, x_l$ of $M$ over $R$ is an admissible sequence of $M$ over $R'$ as $\mathfrak m'$ maps into $\mathfrak m$. The isomorphism is obtained by mapping the symbol $[x_1, \ldots, x_l] \in \det\nolimits_{R, \kappa}(M)$ to the corresponding symbol $[x_1, \ldots, x_l] \in \det\nolimits_{R', \kappa}(M)$. It is immediate to verify that this is functorial for isomorphisms and compatible with the isomorphisms $\gamma$ of Lemma \ref{lemma-det-exact-sequences}.
Denote $f_M \in \kappa^*$ the element such that $\det\nolimits_\kappa(u_M) = f_M \text{id}_{\det\nolimits_\kappa(M)}$. Suppose that $0 \to K \to L \to M \to 0$ is a short exact sequence of finite $R$-modules. Then we see that $u_k$, $u_L$, $u_M$ give an isomorphism of short exact sequences. Hence by Lemma \ref{lemma-det-exact-sequences} (1) we conclude that $f_K f_M = f_L$. This means that by induction on length it suffices to prove the lemma in the case of length $1$ where it is trivial.
Let us prove (1). Set $\psi = 0$. Then we may, with notation as above Definition \ref{definition-periodic-determinant}, identify $K_\varphi = I_\psi = 0$, $I_\varphi = K_\psi = M$. With these identifications, the map $$ \gamma_\varphi : \kappa \otimes \det\nolimits_\kappa(M) = \det\nolimits_\kappa(K_\varphi) \otimes \det\nolimits_\kappa(I_\varphi) \longrightarrow \det\nolimits_\kappa(M) $$ is identified with $\det_\kappa(\varphi^{-1})$. On the other hand the map $\gamma_\psi$ is identified with the identity map. Hence $\gamma_\psi \circ \sigma \circ \gamma_\varphi^{-1}$ is equal to $\det_\kappa(\varphi)$ in this case. Whence the result. We omit the proof of (2).
Let us abbreviate $I_{\varphi, i} = \Im(\varphi_i)$, $K_{\varphi, i} = \Ker(\varphi_i)$, $I_{\psi, i} = \Im(\psi_i)$, and $K_{\psi, i} = \Ker(\psi_i)$. Observe that we have a commutative square $$ \xymatrix{ & 0 \ar[d] & 0 \ar[d] & 0 \ar[d] & \\ 0 \ar[r] & K_{\varphi, 1} \ar[r] \ar[d] & K_{\varphi, 2} \ar[r] \ar[d] & K_{\varphi, 3} \ar[r] \ar[d] & 0 \\ 0 \ar[r] & M_1 \ar[r] \ar[d] & M_2 \ar[r] \ar[d] & M_3 \ar[r] \ar[d] & 0 \\ 0 \ar[r] & I_{\varphi, 1} \ar[r] \ar[d] & I_{\varphi, 2} \ar[r] \ar[d] & I_{\varphi, 3} \ar[r] \ar[d] & 0 \\ & 0  & 0  & 0  & } $$ of finite length $R$-modules with exact rows and columns. The top row is exact since it can be identified with the sequence $I_{\psi, 1} \to I_{\psi, 2} \to I_{\psi, 3} \to 0$ of images, and similarly for the bottom row. There is a similar diagram involving the modules $I_{\psi, i}$ and $K_{\psi, i}$. By definition $\det_\kappa(M_2, \varphi_2, \psi_2)$ corresponds, up to a sign, to the composition of the left vertical maps in the following diagram $$ \xymatrix{ \det_\kappa(M_1) \otimes \det_\kappa(M_3) \ar[r]^\gamma \ar[d]^{\gamma^{-1} \otimes \gamma^{-1}} & \det_\kappa(M_2) \ar[d]^{\gamma^{-1}} \\ \det\nolimits_\kappa(K_{\varphi, 1}) \otimes \det\nolimits_\kappa(I_{\varphi, 1}) \otimes \det\nolimits_\kappa(K_{\varphi, 3}) \otimes \det\nolimits_\kappa(I_{\varphi, 3}) \ar[d]^{\sigma \otimes \sigma} \ar[r]^-{\gamma \otimes \gamma} & \det\nolimits_\kappa(K_{\varphi, 2}) \otimes \det\nolimits_\kappa(I_{\varphi, 2}) \ar[d]^\sigma \\ \det\nolimits_\kappa(K_{\psi, 1}) \otimes \det\nolimits_\kappa(I_{\psi, 1}) \otimes \det\nolimits_\kappa(K_{\psi, 3}) \otimes \det\nolimits_\kappa(I_{\psi, 3}) \ar[d]^{\gamma \otimes \gamma} \ar[r]^-{\gamma \otimes \gamma} & \det\nolimits_\kappa(K_{\psi, 2}) \otimes \det\nolimits_\kappa(I_{\psi, 2}) \ar[d]^\gamma \\ \det_\kappa(M_1) \otimes \det_\kappa(M_3) \ar[r]^\gamma & \det_\kappa(M_2) } $$ The top and bottom squares are commutative up to sign by applying Lemma \ref{lemma-det-exact-sequences} (2). The middle square is trivially commutative (we are just switching factors). Hence we see that $\det\nolimits_\kappa(M_2, \varphi_2, \psi_2) = \epsilon \det\nolimits_\kappa(M_1, \varphi_1, \psi_1) \det\nolimits_\kappa(M_3, \varphi_3, \psi_3) $ for some sign $\epsilon$. And the sign can be worked out, namely the outer rectangle in the diagram above commutes up to \begin{eqnarray*} \epsilon & = & (-1)^{\text{length}(I_{\varphi, 1})\text{length}(K_{\varphi, 3}) + \text{length}(I_{\psi, 1})\text{length}(K_{\psi, 3})} \\ & = & (-1)^{\text{length}(I_{\varphi, 1})\text{length}(I_{\psi, 3}) + \text{length}(I_{\psi, 1})\text{length}(I_{\varphi, 3})} \end{eqnarray*} (proof omitted). It follows easily from this that the signs work out as well.
Let $\mathfrak q$ be a minimal prime of the support of $M$. Then $M_{\mathfrak q}$ is a finite length $A_{\mathfrak q}$-module, see Algebra, Lemma \ref{algebra-lemma-support-point}. Hence both $\varphi$ and $\psi$ induce isomorphisms $M_{\mathfrak q} \to M_{\mathfrak q}$. Thus the support of $M/\varphi\psi M$ is $\{\mathfrak m_A\}$ and hence it has finite length (see lemma cited above). Finally, the kernel of $\varphi$ on $M/\varphi\psi M$ is clearly $\psi M/\varphi\psi M$, and hence the kernel of $\varphi$ is the image of $\psi$ on $M/\varphi\psi M$. Similarly the other way since $M/\varphi\psi M = M/\psi\varphi M$ by assumption.
If $a \in A^*$, then the equality follows from the equality $\text{length}(M/bM) = \text{length}(M'/bM')$ and Lemma \ref{lemma-symbol-when-one-is-a-unit}. Similarly if $b$ is a unit the lemma holds as well (by the symmetry of Lemma \ref{lemma-multiplicativity-symbol}). Hence we may assume that $a, b \in \mathfrak m_A$. This in particular implies that $\mathfrak m$ is not an associated prime of $M$, and hence $\alpha : M \to M'$ is injective. This permits us to think of $M$ as a submodule of $M'$. By assumption $M'/M$ is a finite $A$-module with support $\{\mathfrak m_A\}$ and hence has finite length. Note that for any third module $M''$ with $M \subset M'' \subset M'$ the maps $M \to M''$ and $M'' \to M'$ satisfy the assumptions of the lemma as well. This reduces us, by induction on the length of $M'/M$, to the case where $\text{length}_A(M'/M) = 1$. Finally, in this case consider the map $$ \overline{\alpha} : M/abM \longrightarrow M'/abM'. $$ By construction the cokernel $Q$ of $\overline{\alpha}$ has length $1$. Since $a, b \in \mathfrak m_A$, they act trivially on $Q$. It also follows that the kernel $K$ of $\overline{\alpha}$ has length $1$ and hence also $a$, $b$ act trivially on $K$. Hence we may apply Lemma \ref{lemma-tricky}. Thus it suffices to see that the two maps $\alpha_i : Q \to K$ are the same. In fact, both maps are equal to the map $q = x' \bmod \Im(\overline{\alpha}) \mapsto abx' \in K$. We omit the verification.
Choose a filtration by $A$-submodules $$ 0 = M_0 \subset M_1 \subset \ldots \subset M_n = M $$ such that each quotient $M_j/M_{j - 1}$ is isomorphic to $A/\mathfrak p_j$ for some prime ideal $\mathfrak p_j$ of $A$. See Algebra, Lemma \ref{algebra-lemma-filter-Noetherian-module}. For each $j$ we have either $\mathfrak p_j = \mathfrak q_i$ for some $i$, or $\mathfrak p_j = \mathfrak m_A$. Moreover, for a fixed $i$, the number of $j$ such that $\mathfrak p_j = \mathfrak q_i$ is equal to $\text{length}_{A_{\mathfrak q_i}}(M_{\mathfrak q_i})$ by Algebra, Lemma \ref{algebra-lemma-filter-minimal-primes-in-support}. Hence $d_{M_j}(a, b)$ is defined for each $j$ and $$ d_{M_j}(a, b) = \left\{ \begin{matrix} d_{M_{j - 1}}(a, b) d_{A/\mathfrak q_i}(a, b) & \text{if} & \mathfrak p_j = \mathfrak q_i \\ d_{M_{j - 1}}(a, b) & \text{if} & \mathfrak p_j = \mathfrak m_A \end{matrix} \right. $$ by Lemma \ref{lemma-symbol-short-exact-sequence} in the first instance and Lemma \ref{lemma-symbol-compare-modules} in the second. Hence the lemma.
By multiplicativity it suffices to prove this when $x, y \in A$. Let $t \in A$ be a uniformizer. Write $x = t^bu$ and $y = t^bv$ for some $a, b \geq 0$ and $u, v \in A^*$. Set $l = a + b$. Then $t^{l - 1}, \ldots, t^b$ is an admissible sequence in $(x)/(xy)$ and $t^{l - 1}, \ldots, t^a$ is an admissible sequence in $(y)/(xy)$. Hence by Remark \ref{remark-more-elementary} we see that $d_A(x, y)$ is characterized by the equation $$ [t^{l - 1}, \ldots, t^b, v^{-1}t^{b - 1}, \ldots, v^{-1}] = (-1)^{ab} d_A(x, y) [t^{l - 1}, \ldots, t^a, u^{-1}t^{a - 1}, \ldots, u^{-1}]. $$ Hence by the admissible relations for the symbols $[x_1, \ldots, x_l]$ we see that $$ d_A(x, y) = (-1)^{ab} u^a/v^b \bmod \mathfrak m_A $$ as desired.