During this proof we use the notation $s, t : A \to B$ instead of the notation $s^\sharp, t^\sharp$. By Lemma \ref{lemma-criterion-quotient-representable} it suffices to show that $C \to A$ is finite locally free and that the map $$ t \otimes s : A \otimes_C A \longrightarrow B $$ is an isomorphism. First, note that $j$ is a monomorphism, and also finite (since already $s$ and $t$ are finite). Hence we see that $j$ is a closed immersion by Morphisms, Lemma \ref{morphisms-lemma-finite-monomorphism-closed}. Hence $A \otimes_C A \to B$ is surjective. \medskip\noindent We will perform base change by flat ring maps $C \to C'$ as in Lemma \ref{lemma-invariants-base-change}, and we will use that formation of invariants commutes with flat base change, see part (3) of the lemma cited. We will show below that for every prime $\mathfrak p \subset C$, there exists a local flat ring map $C_{\mathfrak p} \to C_{\mathfrak p}'$ such that the result holds after a base change to $C_{\mathfrak p}'$. This implies immediately that $A \otimes_C A \to B$ is injective (use Algebra, Lemma \ref{algebra-lemma-characterize-zero-local}). It also implies that $C \to A$ is flat, by combining Algebra, Lemmas \ref{algebra-lemma-local-flat-ff}, \ref{algebra-lemma-flat-localization}, and \ref{algebra-lemma-flatness-descends}. Then since $U \to \Spec(C)$ is surjective also (Lemma \ref{lemma-points}) we conclude that $C \to A$ is faithfully flat. Then the isomorphism $B \cong A \otimes_C A$ implies that $A$ is a finitely presented $C$-module, see Algebra, Lemma \ref{algebra-lemma-descend-properties-modules}. Hence $A$ is finite locally free over $C$, see Algebra, Lemma \ref{algebra-lemma-finite-projective}. \medskip\noindent By Lemma \ref{lemma-finite-locally-free-disjoint-free} we know that $A$ is a finite product of rings $A_r$ and $B$ is a finite product of rings $B_r$ such that the groupoid scheme decomposes accordingly (see the proof of Lemma \ref{lemma-integral-over-invariants}). Then also $C$ is a product of rings $C_r$ and correspondingly $C'$ decomposes as a product. Hence we may and do assume that the ring maps $s, t : A \to B$ are finite locally free of a fixed rank $r$. \medskip\noindent The local ring maps $C_{\mathfrak p} \to C_{\mathfrak p}'$ we are going to use are any local flat ring maps such that the residue field of $C_{\mathfrak p}'$ is infinite. By Algebra, Lemma \ref{algebra-lemma-flat-local-given-residue-field} such local ring maps exist. \medskip\noindent Assume $C$ is a local ring with maximal ideal $\mathfrak m$ and infinite residue field, and assume that $s, t : A \to B$ is finite locally free of constant rank $r > 0$. Since $C \subset A$ is integral (Lemma \ref{lemma-integral-over-invariants}) all primes lying over $\mathfrak m$ are maximal, and all maximal ideals of $A$ lie over $\mathfrak m$. Similarly for $C \subset B$. Pick a maximal ideal $\mathfrak m'$ of $A$ lying over $\mathfrak m$ (exists by Lemma \ref{lemma-points}). Since $t : A \to B$ is finite locally free there exist at most finitely many maximal ideals of $B$ lying over $\mathfrak m'$. Hence we conclude (by Lemma \ref{lemma-points} again) that $A$ has finitely many maximal ideals, i.e., $A$ is semi-local. This in turn implies that $B$ is semi-local as well. OK, and now, because $t \otimes s : A \otimes_C A \to B$ is surjective, we can apply Algebra, Lemma \ref{algebra-lemma-semi-local-module-basis-in-submodule} to the ring map $C \to A$, the $A$-module $M = B$ (seen as an $A$-module via $t$) and the $C$-submodule $s(A) \subset B$. This lemma implies that there exist $x_1, \ldots, x_r \in A$ such that $M$ is free over $A$ on the basis $s(x_1), \ldots, s(x_r)$. Hence we conclude that $C \to A$ is finite free and $B \cong A \otimes_C A$ by applying Lemma \ref{lemma-basis}.