Is there a definition of the "boundary" of a simplicial complex?

Question

Let $S$ be the abstract simplicial complex with facets $\{A,B,D\}$, $\{A,C,D\}$, $\{B,C,D\}$. Its geometric representation is homeomorphic to a disc:

In this picture, the geometric representation of the sub-complex $T$ with facets $\{A,B\}$, $\{B,C\}$, $\{C,A\}$ maps to the boundary of the disc. So we can say that "the boundary of $S$ is $T$".

My question is: is this notion of "boundary" well-defined? That is, given any abstract complex $S$ with geometric representation homeomorphic to a ball, is there a unique subcomplex $T\subseteq S$, such that the geometric representation of $T$ equals the boundary of that ball?

Answer 1

You could define it as follows:

Let $S$ be an abstract simplical complex. Then define $\partial S=\{A\in S\colon \exists!B\in S, A\subsetneq B\}$ where $\exists!$ means "There is a unique".

The idea behind this definition is as follows:

• If there is no $B\supseteq A$, then $A$ is maximal/part of the "body" of the complex, and the boundary should be part of this $A$, not $A$ itself.
• If there is more than one $B\supseteq A$, then $A$ is either "too small", i.e. there are $A\subseteq B\subseteq B'$, or $A$ is on the "boundary" of two $B$, i.e. there are $B$, $B'$ with $B\cap B'=A$.

But I do not know if this definition is standard.

You can then try to prove all properties you require.