Let us reformulate the statement of the proposition. Consider the category $\mathcal{C}$ whose objects, denoted $Y/X$, are locally quasi-finite syntomic morphism $Y \to X$ of locally Noetherian schemes and whose morphisms $b/a : Y'/X' \to Y/X$ are commutative diagrams $$ \xymatrix{ Y' \ar[d] \ar[r]_b & Y \ar[d] \\ X' \ar[r]^a & X } $$ which induce an isomorphism of $Y'$ with an open subscheme of $X' \times_X Y$. The proposition means that for every object $Y/X$ of $\mathcal{C}$ we have an isomorphism $c_{Y/X} : \det(\NL_{Y/X}) \to \omega_{Y/X}$ with $c_{Y/X}(\delta(\NL_{Y/X})) = \tau_{Y/X}$ and for every morphism $b/a : Y'/X' \to Y/X$ of $\mathcal{C}$ we have $b^*c_{Y/X} = c_{Y'/X'}$ via the identifications $b^*\det(\NL_{Y/X}) = \det(\NL_{Y'/X'})$ and $b^*\omega_{Y/X} = \omega_{Y'/X'}$ described above. \medskip\noindent Given $Y/X$ in $\mathcal{C}$ and $y \in Y$ we can find an affine open $V \subset Y$ and $U \subset X$ with $f(V) \subset U$ such that there exists some isomorphism $$ \det(\NL_{Y/X})|_V \longrightarrow \omega_{Y/X}|_V $$ mapping $\delta(\NL_{Y/X})|_V$ to $\tau_{Y/X}|_V$. This follows from picking affine opens as in Lemma \ref{lemma-syntomic-quasi-finite} part (5), the affine local description of $\delta(\NL_{Y/X})$ in Remark \ref{remark-local-description-delta}, and Lemma \ref{lemma-different-quasi-finite-complete-intersection}. If the annihilator of the section $\tau_{Y/X}$ is zero, then these local maps are unique and automatically glue. Hence if the annihilator of $\tau_{Y/X}$ is zero, then there is a unique isomorphism $c_{Y/X} : \det(\NL_{Y/X}) \to \omega_{Y/X}$ with $c_{Y/X}(\delta(\NL_{Y/X})) = \tau_{Y/X}$. If $b/a : Y'/X' \to Y/X$ is a morphism of $\mathcal{C}$ and the annihilator of $\tau_{Y'/X'}$ is zero as well, then $b^*c_{Y/X}$ is the unique isomorphism $c_{Y'/X'} : \det(\NL_{Y'/X'}) \to \omega_{Y'/X'}$ with $c_{Y'/X'}(\delta(\NL_{Y'/X'})) = \tau_{Y'/X'}$. This follows formally from the fact that $b^*\delta(\NL_{Y/X}) = \delta(\NL_{Y'/X'})$ and $b^*\tau_{Y/X} = \tau_{Y'/X'}$. \medskip\noindent We can summarize the results of the previous paragraph as follows. Let $\mathcal{C}_{nice} \subset \mathcal{C}$ denote the full subcategory of $Y/X$ such that the annihilator of $\tau_{Y/X}$ is zero. Then we have solved the problem on $\mathcal{C}_{nice}$. For $Y/X$ in $\mathcal{C}_{nice}$ we continue to denote $c_{Y/X}$ the solution we've just found. \medskip\noindent Consider morphisms $$ Y_1/X_1 \xleftarrow{b_1/a_1} Y/X \xrightarrow{b_2/a_2} Y_2/X_2 $$ in $\mathcal{C}$ such that $Y_1/X_1$ and $Y_2/X_2$ are objects of $\mathcal{C}_{nice}$. {\bf Claim.} $b_1^*c_{Y_1/X_1} = b_2^*c_{Y_2/X_2}$. We will first show that the claim implies the proposition and then we will prove the claim. \medskip\noindent Let $d, n \geq 1$ and consider the locally quasi-finite syntomic morphism $Y_{n, d} \to X_{n, d}$ constructed in Example \ref{example-universal-quasi-finite-syntomic}. Then $Y_{n, d}$ is an irreducible regular scheme and the morphism $Y_{n, d} \to X_{n, d}$ is locally quasi-finite syntomic and \'etale over a dense open, see Lemma \ref{lemma-universal-quasi-finite-syntomic-etale}. Thus $\tau_{Y_{n, d}/X_{n, d}}$ is nonzero for example by Lemma \ref{lemma-different-ramification}. Now a nonzero section of an invertible module over an irreducible regular scheme has vanishing annihilator. Thus $Y_{n, d}/X_{n, d}$ is an object of $\mathcal{C}_{nice}$. \medskip\noindent Let $Y/X$ be an arbitrary object of $\mathcal{C}$. Let $y \in Y$. By Lemma \ref{lemma-locally-comes-from-universal} we can find $n, d \geq 1$ and morphisms $$ Y/X \leftarrow V/U \xrightarrow{b/a} Y_{n, d}/X_{n, d} $$ of $\mathcal{C}$ such that $V \subset Y$ and $U \subset X$ are open. Thus we can pullback the canonical morphism $c_{Y_{n, d}/X_{n, d}}$ constructed above by $b$ to $V$. The claim guarantees these local isomorphisms glue! Thus we get a well defined global isomorphism $c_{Y/X} : \det(\NL_{Y/X}) \to \omega_{Y/X}$ with $c_{Y/X}(\delta(\NL_{Y/X})) = \tau_{Y/X}$. If $b/a : Y'/X' \to Y/X$ is a morphism of $\mathcal{C}$, then the claim also implies that the similarly constructed map $c_{Y'/X'}$ is the pullback by $b$ of the locally constructed map $c_{Y/X}$. Thus it remains to prove the claim. \medskip\noindent In the rest of the proof we prove the claim. We may pick a point $y \in Y$ and prove the maps agree in an open neighbourhood of $y$. Thus we may replace $Y_1$, $Y_2$ by open neighbourhoods of the image of $y$ in $Y_1$ and $Y_2$. Thus we may assume there are morphisms $$ Y_{n_1, d_1}/X_{n_1, d_1} \leftarrow Y_1/X_1 \quad\text{and}\quad Y_2/X_2 \rightarrow Y_{n_2, d_2}/X_{n_2, d_2} $$ These are morphisms of $\mathcal{C}_{nice}$ for which we know the desired compatibilities. Thus we may replace $Y_1/X_1$ by $Y_{n_1, d_1}/X_{n_1, d_1}$ and $Y_2/X_2$ by $Y_{n_2, d_2}/X_{n_2, d_2}$. This reduces us to the case that $Y_1, X_1, Y_2, X_2$ are of finite type over $\mathbf{Z}$. (The astute reader will realize that this step wouldn't have been necessary if we'd defined $\mathcal{C}_{nice}$ to consist only of those objects $Y/X$ with $Y$ and $X$ of finite type over $\mathbf{Z}$.) \medskip\noindent Assume $Y_1, X_1, Y_2, X_2$ are of finite type over $\mathbf{Z}$. After replacing $Y, X, Y_1, X_1, Y_2, X_2$ by suitable open neighbourhoods of the image of $y$ we may assume $Y, X, Y_1, X_1, Y_2, X_2$ are affine. We may write $X = \lim X_\lambda$ as a cofiltered limit of affine schemes of finite type over $X_1 \times X_2$. For each $\lambda$ we get $$ Y_1 \times_{X_1} X_\lambda \quad\text{and}\quad X_\lambda \times_{X_2} Y_2 $$ If we take limits we obtain $$ \lim Y_1 \times_{X_1} X_\lambda = Y_1 \times_{X_1} X \supset Y \subset X \times_{X_2} Y_2 = \lim X_\lambda \times_{X_2} Y_2 $$ By Limits, Lemma \ref{limits-lemma-descend-opens} we can find a $\lambda$ and opens $V_{1, \lambda} \subset Y_1 \times_{X_1} X_\lambda$ and $V_{2, \lambda} \subset X_\lambda \times_{X_2} Y_2$ whose base change to $X$ recovers $Y$ (on both sides). After increasing $\lambda$ we may assume there is an isomorphism $V_{1, \lambda} \to V_{2, \lambda}$ whose base change to $X$ is the identity on $Y$, see Limits, Lemma \ref{limits-lemma-descend-finite-presentation}. Then we have the commutative diagram $$ \xymatrix{ & Y/X \ar[d] \ar[ld]_{b_1/a_1} \ar[rd]^{b_2/a_2} \\ Y_1/X_1 & V_{1, \lambda}/X_\lambda \ar[l] \ar[r] & Y_2/X_2 } $$ Thus it suffices to prove the claim for the lower row of the diagram and we reduce to the case discussed in the next paragraph. \medskip\noindent Assume $Y, X, Y_1, X_1, Y_2, X_2$ are affine of finite type over $\mathbf{Z}$. Write $X = \Spec(A)$, $X_i = \Spec(A_i)$. The ring map $A_1 \to A$ corresponding to $X \to X_1$ is of finite type and hence we may choose a surjection $A_1[x_1, \ldots, x_n] \to A$. Similarly, we may choose a surjection $A_2[y_1, \ldots, y_m] \to A$. Set $X'_1 = \Spec(A_1[x_1, \ldots, x_n])$ and $X'_2 = \Spec(A_2[y_1, \ldots, y_m])$. Set $Y'_1 = Y_1 \times_{X_1} X'_1$ and $Y'_2 = Y_2 \times_{X_2} X'_2$. We get the following diagram $$ Y_1/X_1 \leftarrow Y'_1/X'_1 \leftarrow Y/X \rightarrow Y'_2/X'_2 \rightarrow Y_2/X_2 $$ Since $X'_1 \to X_1$ and $X'_2 \to X_2$ are flat, the same is true for $Y'_1 \to Y_1$ and $Y'_2 \to Y_2$. It follows easily that the annihilators of $\tau_{Y'_1/X'_1}$ and $\tau_{Y'_2/X'_2}$ are zero. Hence $Y'_1/X'_1$ and $Y'_2/X'_2$ are in $\mathcal{C}_{nice}$. Thus the outer morphisms in the displayed diagram are morphisms of $\mathcal{C}_{nice}$ for which we know the desired compatibilities. Thus it suffices to prove the claim for $Y'_1/X'_1 \leftarrow Y/X \rightarrow Y'_2/X'_2$. This reduces us to the case discussed in the next paragraph. \medskip\noindent Assume $Y, X, Y_1, X_1, Y_2, X_2$ are affine of finite type over $\mathbf{Z}$ and $X \to X_1$ and $X \to X_2$ are closed immersions. Consider the open embeddings $Y_1 \times_{X_1} X \supset Y \subset X \times_{X_2} Y_2$. There is an open neighbourhood $V \subset Y$ of $y$ which is a standard open of both $Y_1 \times_{X_1} X$ and $X \times_{X_2} Y_2$. This follows from Schemes, Lemma \ref{schemes-lemma-standard-open-two-affines} applied to the scheme obtained by glueing $Y_1 \times_{X_1} X$ and $X \times_{X_2} Y_2$ along $Y$; details omitted. Since $X \times_{X_2} Y_2$ is a closed subscheme of $Y_2$ we can find a standard open $V_2 \subset Y_2$ such that $V_2 \times_{X_2} X = V$. Similarly, we can find a standard open $V_1 \subset Y_1$ such that $V_1 \times_{X_1} X = V$. After replacing $Y, Y_1, Y_2$ by $V, V_1, V_2$ we reduce to the case discussed in the next paragraph. \medskip\noindent Assume $Y, X, Y_1, X_1, Y_2, X_2$ are affine of finite type over $\mathbf{Z}$ and $X \to X_1$ and $X \to X_2$ are closed immersions and $Y_1 \times_{X_1} X = Y = X \times_{X_2} Y_2$. Write $X = \Spec(A)$, $X_i = \Spec(A_i)$, $Y = \Spec(B)$, $Y_i = \Spec(B_i)$. Then we can consider the affine schemes $$ X' = \Spec(A_1 \times_A A_2) = \Spec(A') \quad\text{and}\quad Y' = \Spec(B_1 \times_B B_2) = \Spec(B') $$ Observe that $X' = X_1 \amalg_X X_2$ and $Y' = Y_1 \amalg_Y Y_2$, see More on Morphisms, Lemma \ref{more-morphisms-lemma-basic-example-pushout}. By More on Algebra, Lemma \ref{more-algebra-lemma-fibre-product-finite-type} the rings $A'$ and $B'$ are of finite type over $\mathbf{Z}$. By More on Algebra, Lemma \ref{more-algebra-lemma-module-over-fibre-product} we have $B' \otimes_A A_1 = B_1$ and $B' \times_A A_2 = B_2$. In particular a fibre of $Y' \to X'$ over a point of $X' = X_1 \amalg_X X_2$ is always equal to either a fibre of $Y_1 \to X_1$ or a fibre of $Y_2 \to X_2$. By More on Algebra, Lemma \ref{more-algebra-lemma-flat-module-over-fibre-product} the ring map $A' \to B'$ is flat. Thus by Lemma \ref{lemma-syntomic-quasi-finite} part (3) we conclude that $Y'/X'$ is an object of $\mathcal{C}$. Consider now the commutative diagram $$ \xymatrix{ & Y/X \ar[ld]_{b_1/a_1} \ar[rd]^{b_2/a_2} \\ Y_1/X_1 \ar[rd] & & Y_2/X_2 \ar[ld] \\ & Y'/X' } $$ Now we would be done if $Y'/X'$ is an object of $\mathcal{C}_{nice}$. Namely, then pulling back $c_{Y'/X'}$ around the two sides of the square, we would obtain the desired conclusion. Now, in fact, it is true that $Y'/X'$ is an object of $\mathcal{C}_{nice}$\footnote{Namely, the structure sheaf $\mathcal{O}_{Y'}$ is a subsheaf of $(Y_1 \to Y')_*\mathcal{O}_{Y_1} \times (Y_2 \to Y')_*\mathcal{O}_{Y_2}$.}. But it is amusing to note that we don't even need this. Namely, the arguments above show that, after possibly shrinking all of the schemes $X, Y, X_1, Y_1, X_2, Y_2, X', Y'$ we can find some $n, d \geq 1$, and extend the diagram like so: $$ \xymatrix{ & Y/X \ar[ld]_{b_1/a_1} \ar[rd]^{b_2/a_2} \\ Y_1/X_1 \ar[rd] & & Y_2/X_2 \ar[ld] \\ & Y'/X' \ar[d] \\ & Y_{n, d}/X_{n, d} } $$ and then we can use the already given argument by pulling back from $c_{Y_{n, d}/X_{n, d}}$. This finishes the proof.