The Euclidean trace quotient is the fixed-atlas H1/2 boundary space

On the common smooth core, the identity between the fixed-atlas and Euclidean $W^{1,2}$ norms is bounded in both directions. Indeed, each fixed chart representative is obtained by multiplying by a fixed smooth cutoff and composing with a smooth flattening chart, so [F3] bounds the finite fixed-atlas sum by the Euclidean norm. Conversely, the finite partition of unity writes $u$ as the sum of its localized chart pieces; pull each piece back and use [F3] to bound the Euclidean $W^{1,2}$ norm by the fixed-atlas sum. The constants are finite because the atlas and partition are finite. By the density and completeness in [F3], the identity extends to bounded inverse maps on the completions. Since the two norms are equivalent on $C_c^{\mathrm{\infty }}(\mathrm{\Omega })$, their closures give the same zero-boundary subspace.

For $u\in H_{\mathrm{f}\mathrm{i}\mathrm{x}}^1$, take its localized boundary-chart representatives in the fixed-atlas definition. Apply [F4] to each representative (reflecting $t<0$ to $t>0$ when required), and sum the finitely many resulting local traces. The fixed-atlas norm [F1] gives a bounded map $T:H_{\mathrm{f}\mathrm{i}\mathrm{x}}^1\to H_{\mathrm{f}\mathrm{i}\mathrm{x}}^{1\mathrm{/}2}$. For smooth $u$ this sum is the classical boundary restriction because the partition equals one near the boundary and the half-space traces agree with classical restriction. As an $L^2$-valued map, this trace agrees with the Euclidean trace [F5]: the maps are bounded into $L^2$ by [F2] and [F5], and the Euclidean and fixed $H^1$ norms are equivalent by step 1.1, so equality on the dense smooth core extends to all of $H^1$.

If $Tu=0$ in $H_{\mathrm{f}\mathrm{i}\mathrm{x}}^{1\mathrm{/}2}$, it is zero as an $L^2$ class by [F2]; the $L^2$ trace agreement of step 2.1 and [F6] imply $u\in H_{0,\mathrm{E}}^1$. Step 1.1 identifies this with $H_{0,\mathrm{f}\mathrm{i}\mathrm{x}}^1$. Conversely, if $u\in H_{0,\mathrm{f}\mathrm{i}\mathrm{x}}^1$, approximate it in the fixed $H^1$ norm by interior compactly supported smooth functions. Each has zero classical trace, so continuity of $T$ gives $Tu=0$. Therefore $\ker T=H_{0,\mathrm{f}\mathrm{i}\mathrm{x}}^1$.

By [F7] and the boundary-norm equivalence [F2], the Euclidean $L^2$ trace has a bounded right inverse $R:H_{\mathrm{f}\mathrm{i}\mathrm{x}}^{1\mathrm{/}2}(\mathrm{\partial }\mathrm{\Omega })\to H_{\mathrm{E}}^1(\mathrm{\Omega })$. The norm equivalence of step 1.1 makes $R$ bounded into $H_{\mathrm{f}\mathrm{i}\mathrm{x}}^1$ as well. For every boundary class $f$, the Euclidean trace of $Rf$ is $f$; by step 2.1 the fixed trace agrees with it in $L^2$, hence as an $H_{\mathrm{f}\mathrm{i}\mathrm{x}}^{1\mathrm{/}2}$ class. So $T$ is onto. The kernel identity of step 3.1 induces a bounded bijection $\bar{T}:H_{\mathrm{f}\mathrm{i}\mathrm{x}}^1\mathrm{/}{H}_{0,\mathrm{f}\mathrm{i}\mathrm{x}}^1\to H_{\mathrm{f}\mathrm{i}\mathrm{x}}^{1\mathrm{/}2}$. For any representative $u+v$ with $v\in H_{0,\mathrm{f}\mathrm{i}\mathrm{x}}^1$, step 2.1 gives $\mathrm{\parallel }Tu{\mathrm{\parallel }}_{H^{1\mathrm{/}2}}\le C\mathrm{\parallel }u+v{\mathrm{\parallel }}_{H_{\mathrm{f}\mathrm{i}\mathrm{x}}^1}$; taking the infimum shows $\bar{T}$ is bounded. Also $\mathrm{\parallel }{\bar{T}}^{-1}f\mathrm{\parallel }\le \mathrm{\parallel }Rf{\mathrm{\parallel }}_{H_{\mathrm{f}\mathrm{i}\mathrm{x}}^1}\le C_R\mathrm{\parallel }f{\mathrm{\parallel }}_{H^{1\mathrm{/}2}}$, so its inverse is bounded. Composing with the quotient identity isomorphism of step 1.1 proves the asserted Euclidean quotient identification.