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A simplicial map $g:K\to L$ is a simplicial approximation to $f:|K|\to |L|$ whenfor every vertex $v$, where $\mathrm{st}$ denotes the open star in a simplicial complex.

Fix $x$ in a simplex with vertices $v_0,\dots ,v_q$, retaining only vertices with positive barycentric coordinate at $x$. Then $x\in {\mathrm{st}}_K(v_i)$ for every $i$, so $f(x)$ lies in every ${\mathrm{st}}_L(g(v_i))$. The vertices $g(v_i)$ and the vertices carrying the barycentric coordinates of $f(x)$ consequently span a common simplex of $L$. Both $f(x)$ and $|g|(x)$ lie in its convex realization. The straight-line homotopy from a simplicial approximationis therefore well-defined and continuous, with endpoints $f$ and $|g|$.

The open star in a simplicial complex of a vertex $v$ is the union of the relative interiors of the simplices containing $v$. In barycentric coordinates it isLet $L$ be the subcomplex consisting of all simplices not containing $v$. The complement of the star is exactly $|L|$: the positive-coordinate support of a point misses $v$ precisely when it is a simplex of $L$. By the allowed closed-subcomplex theorem, $|L|$ is closed in $|K|$. Hence the star is open.

This is the open-star convention required by the question. The closed star, which also includes the opposite faces, is a different subset and need not be open.

The topological realization of a simplicial complex with infinitely many vertices is not compact. Its vertex open star in a simplicial complex form an open cover, but a vertex belongs only to its own star. Thus every finite collection of vertex stars omits all the other vertices. The argument uses the weak topology of a simplicial realization, in which a vertex star is open because its complement is the realization of the subcomplex of simplices missing that vertex. No local finiteness assumption is required.