Proof of (1). It follows from (\ref{equation-sequence-functorial}) that (1)(a) and (1)(b) are equivalent. Also, if $A \to B$ is surjective, then (1)(a) and (1)(b) hold. Assume (1)(a). Since the kernel of $\text{d}_A$ is the image of $H_1(L_{k/\Lambda})$ which also maps to $\mathfrak m_B/\mathfrak m_B^2$ we conclude that $\mathfrak m_B/\mathfrak m_B^2 \to \mathfrak m_A/\mathfrak m_A^2$ is surjective. Hence $B \to A$ is surjective by Lemma \ref{lemma-surjective}. This finishes the proof of (1). \medskip\noindent Proof of (2). The equivalence of (2)(b) and (2)(c) is immediate from (\ref{equation-sequence-functorial}). \medskip\noindent Assume (2)(b). Let $g : C \to B$ be a ring map in $\mathcal{C}_\Lambda$ such that $f \circ g$ is surjective. We conclude that $\mathfrak m_C/\mathfrak m_C^2 \to \mathfrak m_A/\mathfrak m_A^2$ is surjective by Lemma \ref{lemma-surjective}. Hence $\Im(\text{d}_C) \to \Im(\text{d}_A)$ is surjective and by the assumption we see that $\Im(\text{d}_C) \to \Im(\text{d}_B)$ is surjective. It follows that $C \to B$ is surjective by (1). \medskip\noindent Assume (2)(a). Then $f$ is surjective and we see that $\Omega_{B/\Lambda} \otimes_B k \to \Omega_{A/\Lambda} \otimes_A k$ is surjective. Let $K$ be the kernel. Note that $K = \text{d}_B(\Ker(\mathfrak m_B/\mathfrak m_B^2 \to \mathfrak m_A/\mathfrak m_A^2))$ by (\ref{equation-sequence-functorial}). Choose a splitting $$ \Omega_{B/\Lambda} \otimes_B k = \Omega_{A/\Lambda} \otimes_A k \oplus K $$ of $k$-vector space. The map $\text{d} : B \to \Omega_{B/\Lambda}$ induces via the projection onto $K$ a map $D : B \to K$. Set $C = \{b \in B \mid D(b) = 0\}$. The Leibniz rule shows that this is a $\Lambda$-subalgebra of $B$. Let $\overline{x} \in k$. Choose $x \in B$ mapping to $\overline{x}$. If $D(x) \not = 0$, then we can find an element $y \in \mathfrak m_B$ such that $D(y) = D(x)$. Hence $x - y \in C$ is an element which maps to $\overline{x}$. Thus $C \to k$ is surjective and $C$ is an object of $\mathcal{C}_\Lambda$. Similarly, pick $\omega \in \Im(\text{d}_A)$. We can find $x \in \mathfrak m_B$ such that $\text{d}_B(x)$ maps to $\omega$ by (1). If $D(x) \not = 0$, then we can find an element $y \in \mathfrak m_B$ which maps to zero in $\mathfrak m_A/\mathfrak m_A^2$ such that $D(y) = D(x)$. Hence $z = x - y$ is an element of $\mathfrak m_C$ whose image $\text{d}_C(z) \in \Omega_{C/k} \otimes_C k$ maps to $\omega$. Hence $\Im(\text{d}_C) \to \Im(\text{d}_A)$ is surjective. We conclude that $C \to A$ is surjective by (1). Hence $C \to B$ is surjective by assumption. Hence $D = 0$, i.e., $K = 0$, i.e., (2)(c) holds. This finishes the proof of (2). \medskip\noindent Proof of (3). If $k'/k$ is separable, then $H_1(L_{k/\Lambda}) = \mathfrak m_\Lambda/\mathfrak m_\Lambda^2 \otimes_{k'} k$, see Lemma \ref{lemma-H1-separable-case}. Hence $\Im(\text{d}_A) = \mathfrak m_A/(\mathfrak m_\Lambda A + \mathfrak m_A^2)$ and similarly for $B$. Thus (3) follows from (2). \medskip\noindent Proof of (4). A section $s$ of $f$ is not surjective (by definition a small extension has nontrivial kernel), hence $f$ is not essentially surjective. Conversely, assume $f$ is a small extension but not an essential surjection. Choose a ring map $C \to B$ in $\mathcal{C}_\Lambda$ which is not surjective, such that $C \to A$ is surjective. Let $C' \subset B$ be the image of $C \to B$. Then $C' \not = B$ but $C'$ surjects onto $A$. Since $f : B \to A$ is a small extension, $\text{length}_C(B) = \text{length}_C(A) + 1$. Thus $\text{length}_C(C') \leq \text{length}_C(A)$ since $C'$ is a proper subring of $B$. But $C' \to A$ is surjective, so in fact we must have $\text{length}_C(C') = \text{length}_C(A)$ and $C' \to A$ is an isomorphism which gives us our section.