**Proof.**
Let $E = B_1$ viewed as a set. Consider the surjection $A_1[E] \to B_1$ with kernel $J$ used to define the naive cotangent complex by the formula

\[ \mathop{N\! L}\nolimits _{B_1/A_1} = (J/J^2 \to \Omega _{A_1[E]/A_1} \otimes _{A_1[E]} B_1) \]

in Algebra, Section 10.134. Since $\Omega _{A_1[E]/A_1} \otimes B_1$ is a free $B_1$-module we have

\[ \mathop{\mathrm{Ext}}\nolimits ^1_{B_1}(\mathop{N\! L}\nolimits _{B_1/A_1}, N_2) = \frac{\mathop{\mathrm{Hom}}\nolimits _{B_1}(J/J^2, N_2)}{\mathop{\mathrm{Hom}}\nolimits _{B_1}(\Omega _{A_1[E]/A_1} \otimes B_1, N_2)} \]

We will construct an obstruction in the module on the right. Let $J' = \mathop{\mathrm{Ker}}(A'_1[E] \to B_1)$. Note that there is a surjection $J' \to J$ whose kernel is $I_1A_1[E]$. For every $e \in E$ denote $x_ e \in A_1[E]$ the corresponding variable. Choose a lift $y_ e \in B'_1$ of the image of $x_ e$ in $B_1$ and a lift $z_ e \in B'_2$ of the image of $x_ e$ in $B_2$. These choices determine $A'_1$-algebra maps

\[ A'_1[E] \to B'_1 \quad \text{and}\quad A'_1[E] \to B'_2 \]

The first of these gives a map $J' \to N_1$, $f' \mapsto f'(y_ e)$ and the second gives a map $J' \to N_2$, $f' \mapsto f'(z_ e)$. A calculation shows that these maps annihilate $(J')^2$. Because the left square of the diagram (involving $c_1$ and $c_2$) commutes we see that these maps agree on $I_1A_1[E]$ as maps into $N_2$. Observe that $B'_1$ is the pushout of $J' \to A'_1[E]$ and $J' \to N_1$. Thus, if the maps $J' \to N_1 \to N_2$ and $J' \to N_2$ agree, then we obtain a map $B'_1 \to B'_2$ fitting into the diagram. Thus we let the obstruction be the class of the map

\[ J/J^2 \to N_2,\quad f \mapsto f'(z_ e) - \nu (f'(y_ e)) \]

where $\nu : N_1 \to N_2$ is the given map and where $f' \in J'$ is a lift of $f$. This is well defined by our remarks above. Note that we have the freedom to modify our choices of $z_ e$ into $z_ e + \delta _{2, e}$ and $y_ e$ into $y_ e + \delta _{1, e}$ for some $\delta _{i, e} \in N_ i$. This will modify the map above into

\[ f \mapsto f'(z_ e + \delta _{2, e}) - \nu (f'(y_ e + \delta _{1, e})) = f'(z_ e) - \nu (f'(z_ e)) + \sum (\delta _{2, e} - \nu (\delta _{1, e}))\frac{\partial f}{\partial x_ e} \]

This means exactly that we are modifying the map $J/J^2 \to N_2$ by the composition $J/J^2 \to \Omega _{A_1[E]/A_1} \otimes B_1 \to N_2$ where the second map sends $\text{d}x_ e$ to $\delta _{2, e} - \nu (\delta _{1, e})$. Thus our obstruction is well defined and is zero if and only if a lift exists.

Part (2) comes from the observation that given two maps $\varphi , \psi : B'_1 \to B'_2$ fitting into the diagram, then $\varphi - \psi $ factors through a map $D : B_1 \to N_2$ which is an $A_1$-derivation:

\begin{align*} D(fg) & = \varphi (f'g') - \psi (f'g') \\ & = \varphi (f')\varphi (g') - \psi (f')\psi (g') \\ & = (\varphi (f') - \psi (f'))\varphi (g') + \psi (f')(\varphi (g') - \psi (g')) \\ & = gD(f) + fD(g) \end{align*}

Thus $D$ corresponds to a unique $B_1$-linear map $\Omega _{B_1/A_1} \to N_2$. Conversely, given such a linear map we get a derivation $D$ and given a ring map $\psi : B'_1 \to B'_2$ fitting into the diagram the map $\psi + D$ is another ring map fitting into the diagram.
$\square$

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