The open star criterion produces a simplicial map

If $K$ is vertex-free all assertions concern empty maps. Otherwise enumerate its finitely many vertices. For each positive denominator enumerate, lexicographically, all barycentric rational grid points on every face, allowing repetitions, to obtain a sequence $d_j$. On a face with $s$ vertices, round the first $s-1$ coordinates down to multiples of $1\mathrm{/}N$ and give the last coordinate the remaining mass; every coordinate error is at most $(s-1)\mathrm{/}N$. These points approximate every point of each simplex. Hence the sequence is dense in the finite Euclidean realization, and therefore in the weak realization. The image $C=f(\mathrm{\mid }K\mathrm{\mid })$ is compact: pull an open cover back to compact $\mathrm{\mid }K\mathrm{\mid }$ and take a finite subcover.

Suppose the union of the finite supports of $f(d_j)$ were infinite. Recursively select the least index whose image support is not contained in the finite union of previously selected supports, and call the resulting images $q_n$. Each new point introduces a fresh target vertex. In a fixed closed simplex $\tau$, at most $\mathrm{\#}\tau$ of the points $q_n$ can occur, since each such occurrence introduces a fresh vertex of $\tau$. Every subset of $Q=\{q_n\}$ thus has finite closed trace on each target simplex and is weakly closed. Therefore $Q$ is closed in compact $C$, and every subset is closed in $Q$, making $Q$ discrete. Its singleton cover contradicts compactness. All repeated choices here are least natural indices, not applications of Countable Choice.

The union $W$ of those supports is therefore finite. Let $P$ contain all faces of $L$ whose vertices lie in $W$. This is finite and weakly closed: its intersection with any simplex is a finite union of closed faces. The closed set $f^{-1}(\mathrm{\mid }P\mathrm{\mid })$ contains the dense sequence, so equals $\mathrm{\mid }K\mathrm{\mid }$. By the closed-embedding assertion, $f$ regarded as a map into $\mathrm{\mid }P\mathrm{\mid }$ is continuous for its Euclidean topology.

For any nonempty source face $\sigma$, its barycenter lies in the star of every vertex of $\sigma$. Its image therefore has every $g(v)$, $v\in \sigma$, in its support. That support is a face of $L$, so its subset $g(\sigma )$ is a face. This proves simpliciality. For an arbitrary point $x$ the same reasoning applies to each positive-coordinate vertex of $x$: every corresponding $g(v)$ belongs to $\mathrm{supp}f(x)$. Consequently $\mathrm{\mid }g\mathrm{\mid }(x)={\sum }_v{x}_v{e}_{g(v)}$ lies in that very simplex. In particular the image of $g$ lies in $\mathrm{\mid }P\mathrm{\mid }$.

The realization $\mathrm{\mid }g\mathrm{\mid }$ is affine on each of finitely many closed source simplices; these maps agree on their common faces, so it is continuous into finite Euclidean $\mathrm{\mid }P\mathrm{\mid }$. Hence $H(x,t)=(1-t)f(x)+t\mathrm{\mid }g\mathrm{\mid }(x)$ is jointly continuous into its ambient Euclidean space. The common-carrier conclusion puts its image in $\mathrm{\mid }P\mathrm{\mid }$, so it is continuous into $\mathrm{\mid }P\mathrm{\mid }$ and then $\mathrm{\mid }L\mathrm{\mid }$. At $t=0,1$ it equals $f,\mathrm{\mid }g\mathrm{\mid }$, and if $f(x)=\mathrm{\mid }g\mathrm{\mid }(x)$ it is constant in $t$. If $x\in \mathrm{\mid }A\mathrm{\mid }$ and $f(x)\in \mathrm{\mid }B\mathrm{\mid }$, its carrier is a simplex of $B$, so the entire segment remains in $\mathrm{\mid }B\mathrm{\mid }$.