Contact nodal coordinates are independently attainable

Fix a contact vertex $a_i$. On a nondegenerate triangle $K$ with vertices $b_0,b_1,b_2$, the matrix with columns $b_1-b_0$ and $b_2-b_0$ is invertible by [F1]. For $x\in K$, write $(s,t)$ for its image under the inverse affine coordinate map $x=b_0+s(b_1-b_0)+t(b_2-b_0)$, and define ${\lambda }_0=1-s-t$, ${\lambda }_1=s$, ${\lambda }_2=t$. These affine functions take the Kronecker values at the three vertices. Define ${\varphi }_i{\mathrm{\mid }}_K={\lambda }_j$ if $a_i=b_j$, and ${\varphi }_i{\mathrm{\mid }}_K=0$ if $a_i$ is not a vertex of $K$. This finite cellwise construction gives a piecewise-affine scalar function with value $1$ at $a_i$ and $0$ at every other mesh vertex.

On any common facet, the two affine restrictions of ${\varphi }_i$ agree at every facet vertex: the value is $1$ exactly at $a_i$ if that vertex lies on the facet, and is $0$ at the other vertices. Affine functions on an edge are determined by their two endpoint values, so the restrictions agree on every common edge; at a shared vertex they both have the same prescribed nodal value. Thus ${\varphi }_i$ is globally continuous and belongs to the conforming piecewise-affine space [F4]. By [F5], ${\varphi }_i\in H^1(\mathrm{\Omega })$.

No contact vertex lies on ${\mathrm{\Gamma }}_D$ by [F6]. Since ${\mathrm{\Gamma }}_D$ is a union of boundary edges, ${\varphi }_i$ vanishes at both endpoints of each such edge and, being affine on the edge, vanishes on the entire segment. The vector field $w_i:={\varphi }_{i}n$ is continuous on $\bar{\mathrm{\Omega }}$ and belongs to $H^1(\mathrm{\Omega };{\mathbb{R}}^2)$ because ${\varphi }_i\in H^1(\mathrm{\Omega })$ by step 2.1. Its ordinary boundary restriction vanishes on ${\mathrm{\Gamma }}_D$, so the vector-valued trace agreement in [F3] gives $T_D{w}_i=0$; hence $w_i\in V_h$ by [F2]. At each contact vertex, $B_j(w_i)={\varphi }_i(a_j)(n\cdot n)={\delta }_{ij}$ by [F6]. For any $y=(y_0,\dots ,y_N)\in {\mathbb{R}}^{N+1}$, the finite sum $w={\sum }_i{y}_i{w}_i$ lies in $V_h$ and satisfies $B_j(w)=y_j$ for every $j$. Hence $B$ is onto.