Extending a compact surface metric across its boundary

Write $M_{+}$ for the chosen labelled copy and $B=\mathrm{\partial }{M}_{+}$. If $M=\mathrm{\varnothing }$, then $M_{+}=\mathrm{\varnothing }$ is open in $DM$ and the unique empty tensor is its own extension. More generally, if $B$ is empty, the double is the disjoint union of its labelled copies, so $M_{+}$ is open even when it is disconnected; $g$ itself is the required extension. Otherwise $B$ is closed in the compact space $M_{+}$ and is compact. The seam charts in [F1] identify $M_{+}$ locally with a closed half-space in $DM$, and its interior $M_{+}^{\circ }$ is open in $DM$.

For each $b\in B$, [F2] gives a covariant two-tensor extension $h_b$ of $g$ on some open neighbourhood of $b$ in $DM$. Replace $h_b$ by its symmetric part $(h_b+h_b^{\mathsf{T}})\mathrm{/}2$; this remains smooth and still equals $g$ on the original half-space. At $b$ it is positive definite by [F4]. In a local frame, the positive quadratic form has a positive minimum on the Euclidean unit circle; continuity of its finitely many coefficients therefore keeps it positive definite after shrinking the neighbourhood. Thus every $b$ has an open neighbourhood $U$ with a smooth positive-definite symmetric extension $g_U$ agreeing with $g$ on $U\cap M_{+}$.

The family of all such neighbourhoods covers the compact set $B$, so choose a finite subcover $U_1,\dots ,U_m$ and corresponding extensions $g_1,\dots ,g_m$; these are only finitely many selections. Set $O={\bigcup }_{i=1}^m{U}_i$. Apply [F3] to the open cover $(U_i{)}_{i=1}^m$ of the smooth manifold $O$, obtaining a smooth partition $({\rho }_i{)}_{i=1}^m$. By [F5], each product ${\rho }_i{g}_i$ extends by zero outside $U_i$ smoothly because $\mathrm{supp}{\rho }_i\subseteq U_i$. Consequently $G={\sum }_i{\rho }_i{g}_i$ is a smooth symmetric tensor on $O$. For every nonzero tangent vector $v$ at $x\in O$, each term with positive weight satisfies $g_i(v,v)>0$, at least one weight is positive, and the weights sum to one; hence $G_x(v,v)>0$. On $O\cap M_{+}$ every $g_i$ agrees with $g$, so $G=g$ there.

The open set $N=O\cup M_{+}^{\circ }$ contains the whole copy $M_{+}$, since $B\subset O$. On $O$ use $G$, and on $M_{+}^{\circ }$ use the original metric $g$. These tensors agree on their overlap by step 2.1, so they glue to a smooth Riemannian metric on $N$ extending $g$. This proves the claim. The exact ${\mathrm{A}\mathrm{C}}_{\omega }$ uses are invoking [F1] for the smooth double and [F3] for its partition of unity; the finite subcover and finite local extension selections use only finite choice.