Is there a direct proof that an $(n-1)$-simplex in a subdivision of the standard $n$-simplex is a face of at most two of its $n$-simplices? There are many books and articles that prove Sperner's Lemma. Almost all that I have looked up happily take the following as obvious.

If $\mathcal{S}$ is a simplicial subdivision of the standard $n$-simplex, then every $(n-1)$-simplex of $\mathcal{S}$ is a face of at most two $n$-simplices of $\mathcal{S}$.

Most of the references don't even state this explicitly. Some of them prove the lemma only under an explicit assumption that the above holds. The only one that goes into a deeper discussion is a Polish book Wstęp do topologii by Engelking and Sieklucki. (I think that "Engelking, Sieklucki Topology: a geometric approach Sigma Series in Pure Mathematics, 4. Heldermann Verlag, Berlin, 1992" might be an English translation but I'm not sure since I have no access to it.) At first, they only prove Sperner's Lemma for iterated barycentric subdivisions for which the property above can be verified directly. Later, they show (using Invariance of Domain) that this property holds for all subdivisions.
My question: is there an elementary proof of the statement above? By "elementary" I mean one that avoids things like Invariance of Domain as well as homotopical and homological arguments.
 A: It depends on what you mean by a "simplicial subdivision". I can think of two different definitions, leading to two different outcomes for your question.
In one definition, a "simplicial subdivision" is obtained from the original decomposition into simplices by repeating some kind of elementary subdivision. The iterated barycentric subdivision is like this, but there are more general constructions. In that case the proof should simply be induction.
In another defnition, a "simplicial subdivsion" simply means another simplicial structure on the same space of which the original skeleta are subcomplexes. In this situation, invariance of domain is your friend where nothing else will help, it seems to me.
The theme here is that the local topology of Euclidean space is subtler than you might think. Try proving that dimension is a topological invariant without homological arguments, for example.
A: I presume this answer is untimely, but I recently faced the exact same issue. It turns out that more is true:


*

*Every $(n-1)$-dimensional subsimplex in $\mathcal S$ that lies fully in the boundary of the “big” simplex is a face of precisely one $n$-dimensional subsimplex in $\mathcal S$.

*Every $(n-1)$-dimensional subsimplex in $\mathcal S$ that does not lie fully in the boundary of the “big” simplex is a face of precisely two $n$-dimensional subsimplices in $\mathcal S$.
An elementary proof of these claims that does not rely on algebraic topology can be found in Section 23.1 of Maschler et al. (2013). This beautifully elegant and not overly complicated proof is based on the geometric intuition gained by connecting the barycenter of the face with the omitted edge(s), and considering a sequence of points on this segment close to, but not on, the face.
A: Nice to see old friends like @Lee Mosher here! As far as I can tell, the usual definition of a simplicial subdivision in the Sperner case is a finite collection of simplices which cover $\Delta^n$ and where the intersection of any two of these simplices is either empty or a face (of any dimension $<n$) of both. The subdivisions need not be built by a direct inductive process like barycentric subdivision. But each simplex in the subdivision is embedded in an affine way, so you cannot get any tamer than that!
It follows from the intersection condition that an interior point of an $n$-simplex in your subdivision lies in only that simplex; similarly, it follows that an interior point of an $(n-1)$-simplex lies only in that simplex and any $n$-simplex of which that simplex is a face.
Now suppose you have an $n-1$ dimension simplex $S$ which is part of your subdivision of $\Delta^n$. Let $x$ be an interior point of $S$. Let $H$ be the hyperplane spanned by $S$ (its "affine hull"). $H$ divides $\mathbb{R}^n$ into two half-spaces.
Consider an open ball $B$ about $x$, small enough that it intersects only $S$ and the interior of any $n$-simplices that have $S$ as its boundary, which is possible by the remarks above. $B-H$ will consist of two connected components, each an open half-ball.
Now there are two cases, corresponding to the two cases mentioned by @triple_sec: either $x$ is on the boundary of $\Delta^n$ or $x$ is in the interior of $\Delta^n$.
If $x$ is on the boundary of $\Delta^n$. It is not hard to see that $S$ must be completely contained in one of $n-1$ dimensional faces of $\Delta^n$, and that $S$ and that face have the same hyperplane $H$ as their affine hull. One component of $B-H$ will lie in the exterior of $\Delta^n$, and hence not intersect any simplices in our subdivision. The other must lie in the interior of $\Delta^n$ and it must lie in the interior of exactly one $n$-simplex of the subdivision.
If $x$ is an interior point of $\Delta^n$, each component of $B-H$ must lie in exactly one $n$-simplex of the subdivision, and it is easy to see that these simplices must be distinct.
