First, note that the lemma holds in case $l = 1$. Namely, in this case $x_1$ is a basis of $M$ over $R/\mathfrak q$ and $y_1$ is a basis of $N$ over $R/\mathfrak q$ and we have $\varphi(x_1) = fy_1$ for some $f \in R$. Thus $\varphi$ is injective if and only if $f \not \in \mathfrak q$. Moreover, $\Coker(\varphi) = R/(f, \mathfrak q)$ and hence the lemma holds by definition of $\text{ord}_{R/q}(f)$ (see Algebra, Definition \ref{algebra-definition-ord}). \medskip\noindent In fact, suppose more generally that $\varphi(x_i) = f_iy_i$ for some $f_i \in R$, $f_i \not \in \mathfrak q$. Then the induced maps $$ \langle x_1, \ldots, x_i\rangle / \langle x_1, \ldots, x_{i - 1}\rangle \longrightarrow \langle y_1, \ldots, y_i\rangle / \langle y_1, \ldots, y_{i - 1}\rangle $$ are all injective and have cokernels isomorphic to $R/(f_i, \mathfrak q)$. Hence we see that $$ \text{length}_R(\Coker(\varphi)) = \sum \text{ord}_{R/\mathfrak q}(f_i). $$ On the other hand it is clear that $$ [\varphi(x_1), \ldots, \varphi(x_l)] = f_1 \ldots f_l [y_1, \ldots, y_l] $$ in this case from the admissible relation (b) for symbols. Hence we see the result holds in this case also. \medskip\noindent We prove the general case by induction on $l$. Assume $l > 1$. Let $i \in \{1, \ldots, l\}$ be minimal such that $\varphi(x_1) \in \langle y_1, \ldots, y_i\rangle$. We will argue by induction on $i$. If $i = 1$, then we get a commutative diagram $$ \xymatrix{ 0 \ar[r] & \langle x_1 \rangle \ar[r] \ar[d] & \langle x_1, \ldots, x_l \rangle \ar[r] \ar[d] & \langle x_1, \ldots, x_l \rangle / \langle x_1 \rangle \ar[r] \ar[d] & 0 \\ 0 \ar[r] & \langle y_1 \rangle \ar[r] & \langle y_1, \ldots, y_l \rangle \ar[r] & \langle y_1, \ldots, y_l \rangle / \langle y_1 \rangle \ar[r] & 0 } $$ and the lemma follows from the snake lemma and induction on $l$. Assume now that $i > 1$. Write $\varphi(x_1) = a_1 y_1 + \ldots + a_{i - 1} y_{i - 1} + a y_i$ with $a_j, a \in R$ and $a \not \in \mathfrak q$ (since otherwise $i$ was not minimal). Set $$ x'_j = \left\{ \begin{matrix} x_j & \text{if} & j = 1 \\ ax_j & \text{if} & j \geq 2 \end{matrix} \right. \quad\text{and}\quad y'_j = \left\{ \begin{matrix} y_j & \text{if} & j < i \\ ay_j & \text{if} & j \geq i \end{matrix} \right. $$ Let $M' = \langle x'_1, \ldots, x'_l \rangle$ and $N' = \langle y'_1, \ldots, y'_l \rangle$. Since $\varphi(x'_1) = a_1 y'_1 + \ldots + a_{i - 1} y'_{i - 1} + y'_i$ by construction and since for $j > 1$ we have $\varphi(x'_j) = a\varphi(x_i) \in \langle y'_1, \ldots, y'_l\rangle$ we get a commutative diagram of $R$-modules and maps $$ \xymatrix{ M' \ar[d] \ar[r]_{\varphi'} & N' \ar[d] \\ M \ar[r]^\varphi & N } $$ By the result of the second paragraph of the proof we know that $\text{length}_R(M/M') = (l - 1)\text{ord}_{R/\mathfrak q}(a)$ and similarly $\text{length}_R(M/M') = (l - i + 1)\text{ord}_{R/\mathfrak q}(a)$. By a diagram chase this implies that $$ \text{length}_R(\Coker(\varphi')) = \text{length}_R(\Coker(\varphi)) + i\ \text{ord}_{R/\mathfrak q}(a). $$ On the other hand, it is clear that writing $$ [\varphi(x_1), \ldots, \varphi(x_l)] = f [y_1, \ldots, y_l], \quad [\varphi'(x'_1), \ldots, \varphi(x'_l)] = f' [y'_1, \ldots, y'_l] $$ we have $f' = a^if$. Hence it suffices to prove the lemma for the case that $\varphi(x_1) = a_1y_1 + \ldots a_{i - 1}y_{i - 1} + y_i$, i.e., in the case that $a = 1$. Next, recall that $$ [y_1, \ldots, y_l] = [y_1, \ldots, y_{i - 1}, a_1y_1 + \ldots a_{i - 1}y_{i - 1} + y_i, y_{i + 1}, \ldots, y_l] $$ by the admissible relations for symbols. The sequence $y_1, \ldots, y_{i - 1}, a_1y_1 + \ldots + a_{i - 1}y_{i - 1} + y_i, y_{i + 1}, \ldots, y_l$ satisfies the conditions (3), (4) of the lemma also. Hence, we may actually assume that $\varphi(x_1) = y_i$. In this case, note that we have $\mathfrak q x_1 = 0$ which implies also $\mathfrak q y_i = 0$. We have $$ [y_1, \ldots, y_l] = - [y_1, \ldots, y_{i - 2}, y_i, y_{i - 1}, y_{i + 1}, \ldots, y_l] $$ by the third of the admissible relations defining $\det_{\kappa(\mathfrak q)}(N_{\mathfrak q})$. Hence we may replace $y_1, \ldots, y_l$ by the sequence $y'_1, \ldots, y'_l = y_1, \ldots, y_{i - 2}, y_i, y_{i - 1}, y_{i + 1}, \ldots, y_l$ (which also satisfies conditions (3) and (4) of the lemma). Clearly this decreases the invariant $i$ by $1$ and we win by induction on $i$.