Let $P_\bullet \to B$ be a resolution of $B$ over $A$ (Remark \ref{remark-resolution}). We will identify $L_{B/A}$ with $\Omega_{P_\bullet/A} \otimes B$. To construct the map we make some choices. \medskip\noindent Choose an $A$-algebra map $\psi : P_0 \to P$ compatible with the given maps $P_0 \to B$ and $P \to B$. \medskip\noindent Write $P_1 = A[S]$ for some set $S$. For $s \in S$ we may write $$ \psi(d_0(s) - d_1(s)) = \sum p_{s, t} f_t $$ for some $p_{s, t} \in P$. Think of $F = \bigoplus_{t \in T} P$ as a $(P_1, P_1)$-bimodule via the maps $(\psi \circ d_0, \psi \circ d_1)$. By Lemma \ref{lemma-polynomial-ring-unique} we obtain a unique $A$-biderivation $\lambda : P_1 \to F$ mapping $s$ to the vector with coordinates $p_{s, t}$. By construction the composition $$ P_1 \longrightarrow F \longrightarrow P $$ sends $f \in P_1$ to $\psi(d_0(f) - d_1(f))$ because the map $f \mapsto \psi(d_0(f) - d_1(f))$ is an $A$-biderivation agreeing with the composition on generators. \medskip\noindent For $g \in P_2$ we claim that $\lambda(d_0(g) - d_1(g) + d_2(g))$ is an element of $Rel$. Namely, by the last remark of the previous paragraph the image of $\lambda(d_0(g) - d_1(g) + d_2(g))$ in $P$ is $$ \psi((d_0 - d_1)(d_0(g) - d_1(g) + d_2(g))) $$ which is zero by Simplicial, Section \ref{simplicial-section-complexes}). \medskip\noindent The choice of $\psi$ determines a map $$ \text{d}\psi \otimes 1 : \Omega_{P_0/A} \otimes B \longrightarrow \Omega_{P/A} \otimes B $$ Composing $\lambda$ with the map $F \to F \otimes B$ gives a usual $A$-derivation as the two $P_1$-module structures on $F \otimes B$ agree. Thus $\lambda$ determines a map $$ \overline{\lambda} : \Omega_{P_1/A} \otimes B \longrightarrow F \otimes B $$ Finally, We obtain a $B$-linear map $$ q : \Omega_{P_2/A} \otimes B \longrightarrow Rel/TrivRel $$ by mapping $\text{d}g$ to the class of $\lambda(d_0(g) - d_1(g) + d_2(g))$ in the quotient. \medskip\noindent The diagram $$ \xymatrix{ \Omega_{P_3/A} \otimes B \ar[r] \ar[d] & \Omega_{P_2/A} \otimes B \ar[r] \ar[d]_q & \Omega_{P_1/A} \otimes B \ar[r] \ar[d]_{\overline{\lambda}} & \Omega_{P_0/A} \otimes B \ar[d]_{\text{d}\psi \otimes 1} \\ 0 \ar[r] & Rel/TrivRel \ar[r] & F \otimes B \ar[r] & \Omega_{P/A} \otimes B } $$ commutes (calculation omitted) and we obtain the map of the lemma. By Remark \ref{remark-explicit-comparison-map} and Lemma \ref{lemma-relation-with-naive-cotangent-complex} we see that this map induces isomorphisms $H_1(L_{B/A}) \to H_1(L)$ and $H_0(L_{B/A}) \to H_0(L)$. \medskip\noindent It remains to see that our map $L_{B/A} \to L$ induces an isomorphism $H_2(L_{B/A}) \to H_2(L)$. Choose a resolution of $B$ over $A$ with $P_0 = P = A[u_i]$ and then $P_1$ and $P_2$ as in Example \ref{example-resolution-length-two}. In Remark \ref{remark-elucidate-degree-two} we have constructed an exact sequence $$ \wedge^2_B(J_0/J_0^2) \to \text{Tor}_2^{P_0}(B, B) \to H^{-2}(L_{B/A}) \to 0 $$ where $P_0 = P$ and $J_0 = \Ker(P \to B) = I$. Calculating the Tor group using the short exact sequences $0 \to I \to P \to B \to 0$ and $0 \to Rel \to F \to I \to 0$ we find that $\text{Tor}_2^P(B, B) = \Ker(Rel \otimes B \to F \otimes B)$. The image of the map $\wedge^2_B(I/I^2) \to \text{Tor}_2^P(B, B)$ under this identification is exactly the image of $TrivRel \otimes B$. Thus we see that $H_2(L_{B/A}) \cong H_2(L)$. \medskip\noindent Finally, we have to check that our map $L_{B/A} \to L$ actually induces this isomorphism. We will use the notation and results discussed in Example \ref{example-resolution-length-two} and Remarks \ref{remark-elucidate-degree-two} and \ref{remark-surjection} without further mention. Pick an element $\xi$ of $\text{Tor}_2^{P_0}(B, B) = \Ker(I \otimes_P I \to I^2)$. Write $\xi = \sum h_{t', t}f_{t'} \otimes f_t$ for some $h_{t', t} \in P$. Tracing through the exact sequences above we find that $\xi$ corresponds to the image in $Rel \otimes B$ of the element $r \in Rel \subset F = \bigoplus_{t \in T} P$ with $t$th coordinate $r_t = \sum_{t' \in T} h_{t', t}f_{t'}$. On the other hand, $\xi$ corresponds to the element of $H_2(L_{B/A}) = H_2(\Omega)$ which is the image via $\text{d} : H_2(\mathcal{J}/\mathcal{J}^2) \to H_2(\Omega)$ of the boundary of $\xi$ under the $2$-extension $$ 0 \to \text{Tor}_2^\mathcal{O}(\underline{B}, \underline{B}) \to  \mathcal{J} \otimes_\mathcal{O} \mathcal{J} \to \mathcal{J} \to \mathcal{J}/\mathcal{J}^2 \to 0 $$ We compute the successive transgressions of our element. First we have $$ \xi = (d_0 - d_1)(- \sum s_0(h_{t', t} f_{t'}) \otimes x_t) $$ and next we have $$ \sum s_0(h_{t', t} f_{t'}) x_t = d_0(v_r) - d_1(v_r) + d_2(v_r) $$ by our choice of the variables $v$ in Example \ref{example-resolution-length-two}. We may choose our map $\lambda$ above such that $\lambda(u_i) = 0$ and $\lambda(x_t) = - e_t$ where $e_t \in F$ denotes the basis vector corresponding to $t \in T$. Hence the construction of our map $q$ above sends $\text{d}v_r$ to $$ \lambda(\sum s_0(h_{t', t} f_{t'}) x_t) = \sum\nolimits_t \left(\sum\nolimits_{t'} h_{t', t}f_{t'}\right) e_t $$ matching the image of $\xi$ in $Rel \otimes B$ (the two minus signs we found above cancel out). This agreement finishes the proof.