I am reading the following lemma from Munkres' Elements of Algebraic Topology:

Lemma 2.1 A collection $K$ of simplices is a simplicial complex if and only if the follow hold:

  1. Every face of a simplex of $K$ is in $K$.

  2. Every pair of distinct simplices of $K$ have disjoint interiors.

Now I am trying to follow the proof that if 1 and 2 above hold then $K$ is a simplicial complex. Although munkres does not define what he means by "distinct simplices", I am told by Mixedmath that this means two simplices that don't have a common vertex. The proof according to Munkres goes as follows:

Let $\sigma$ and $\tau$ be two distinct simplices of $K$ such that they have disjoint interiors. Let $\sigma'$ be the face of $\sigma$ that is spanned by those vertices $b_0,\ldots,b_m$ of $\sigma$ that lie in $\tau$. The claim now is that $\sigma \cap \tau$ is equal to $\sigma'$. Now one direction I understand the other which I don't is when he shows that

$$\sigma \cap \tau \subseteq \sigma'.$$

The line I don't understand is this:

Let $x \in \sigma \cap \tau$. Then $x \in \textrm{Int}\, s \cap \textrm{Int} \hspace{2mm} t$ for some faces $s$ of $\sigma$ and $t$ of $\tau$.

How does this follow from the assumption of (2) above?

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    $\begingroup$ Hey - that's not fair - I asked you how he defined it! It seems that distinct simplices means that they have disjoint interiors! $\endgroup$ – davidlowryduda Aug 12 '12 at 4:19
  • $\begingroup$ @mixedmath I don't get how (2) means that distinct simplices have disjoint interiors. That's just an assumption in the proof. $\endgroup$ – user38268 Aug 12 '12 at 4:24

"Distinct simplices" means that they are distinct (that is, are not identical). The line you don't understand doesn't use (2); it uses the fact that if $x \in \sigma$ then $x \in \text{int}(s)$ for a face $s$ of $\sigma$ (which is actually unique). $s$ is precisely the minimal face of $\sigma$ (under inclusion) containing $x$ (if $x$ is not contained in the interior of $s$ then it is in the boundary of $s$ which is contained in a strict face of $s$).

  • $\begingroup$ Thanks for your answer. I am trying to understand what you wrote above in the case that $\sigma$ is a triangle. In this context, the faces of $\sigma$ are the three edges yes? How would this automatically mean that if $x \in \sigma$ then it lies on any one of the edges? $\endgroup$ – user38268 Aug 12 '12 at 4:42
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    $\begingroup$ @BenjaLim: a triangle has $8$ faces: the whole thing, the three edges, the three vertices, and the empty face. $\endgroup$ – Qiaochu Yuan Aug 12 '12 at 4:44
  • $\begingroup$ In that case, wouldn't it be immediate that if $x \in \sigma$ then $x \in s$ for some $s$? Why the need to mention the interior? $\endgroup$ – user38268 Aug 12 '12 at 4:46
  • $\begingroup$ @BenjaLim: because... the interior of $s$ is strictly contained in $s$? I don't understand the question. $x \in s$ is a strictly weaker statement than $x \in \text{int}(s)$. $\endgroup$ – Qiaochu Yuan Aug 12 '12 at 4:47
  • $\begingroup$ I guess you are right. If my point $s$ is not on the inside of the triangle, then it will be on one of the three edges. If it is a vertex, then $x$ is in the interior of that vertex which is the vertex itself. If it is on an edge and not a vertex, then it is in the interior of the edge. Is this right? $\endgroup$ – user38268 Aug 12 '12 at 4:49

I don't have the proof or the book in front of me, but I bet it looks like this:

First, let's suppose $K$ is a simplicial complex. Then $K$ contains the faces of its simplices, and we want to show that every point in $K$ belongs to the interior of a unique simplex of $K$. We know that if $x$ is in $K$, then it belongs to the interior of a face $\sigma$ of some simplex in $K$, as every point in a simplex belongs to the interior of some face. And $\sigma$ is a face in $K$, and thus $x$ belongs to the interior of at least one simplex of $K$.

Suppose that $x$ was in the interior of two distinct simplices $\sigma$ and $\tau$. Then $x$ belongs to the intersection of $\sigma$ and $\tau$, which is a face, as the intersection of two simplices in a simplicial complex is always a face. Thus $x$ is in some common face $\sigma \cap \tau$ of $\sigma$ and $\tau$. This is a problem, as then this common face is a proper face of one or the other of the simplices $\sigma$ and $\tau$. It can't be in both (because $\sigma \neq \tau$, and $x$ is in the interior of both $\sigma$ and $\tau$). Thus the simplex $\sigma$ of $K$ that contains $x$ is unique.

In the other direction, showing that it's a simplicial complex, is very simple. $K$ contains all the faces of its simplices, so it only remains to check that if $\sigma$ and $\tau$ are any two simplices with non-empty intersection, then $\sigma \cap \tau$ is a common face of $\sigma$ and $\tau$. So let $x \in \sigma \cap \tau$. We now know that $x$ is in the interior of a unique simplex $\omega$ of $K$. And any point of $\sigma$ or $\tau$ belongs to the interior of a unique face of that simplex, and all faces of $\sigma$ and $\tau$ belong to $K$. Thus $\omega$ is a common face of $\sigma$ and $\tau$, and it is in fact the face we want. Its uniqueness comes from the uniqueness of $\omega$.


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