On Pg. $9$ of these notes, Lackenby introduces Abstract Simplicial Complexes. The topological realization of a simplicial complex $K$, denoted $|K|$, is given by the following recipe:
The author claims that:
For each $\sigma\in \Sigma$, there is a homeomorphic copy of $\triangle_\sigma$ in $|K|$. Also, $|K|$ is a disjoint union of the interiors of the simplices.
How do I prove this result? I am struggling to find an explicit homeomorphism possibly due to the not-so-explicit construction of $|K|$. What even is the topology on $|K|$? It is not clear.
Intuitively, it is immediate from the construction itself that for every $\sigma$ we choose a copy $\triangle_\sigma$. However, the identification via face-inclusion does not seem explicit.
Please let me know if any notation is not clear, although the attached notes are self-contained.
Best Answer
The construction of $\lvert K \rvert$ is in fact somewhat vague. But we can make it precise:
If $\sigma \in K$ has $n+1$ elements, we say that $\sigma$ has dimension $\dim \sigma = n$.
For each $\sigma \in K$ let $\Delta_\sigma = \Delta^{\dim \sigma}$ be the standard simplex having the dimension of $\sigma$. Choose a bijection $b_\sigma : \sigma \to V(\Delta_\sigma)$. This is meant by labelling the vertices with the elements of $\sigma$.
Form the disjoint union of standard simplices $\Delta_\sigma$: $$D = \bigcup_{\sigma \in K} \Delta_\sigma \times \{\sigma\}$$
Define an equivalence relation on $D$ by identifying $\Delta_\sigma$ with a geometric face of $\Delta_\tau$ whenever $\sigma \subset \tau$. More precisely, define a vertex map $$f_{\sigma,\tau} : V(\Delta_\sigma) \stackrel{b_\sigma^{-1}}{\to} \sigma \hookrightarrow \tau \stackrel{b_\tau}{\to} V(\Delta_\tau)$$ and let $\phi_{\sigma,\tau} : \Delta_\sigma \to \Delta_\tau$ denote ist affine extension. This map embeds $\Delta_\sigma$ as a geometric face of $\Delta_\tau$.
The above equivalence relation $\sim$ identifies each $(x,\sigma) \in D$ with $(\phi_{\sigma,\tau}(x),\tau) \in D$. Note that two distinct points of $\Delta_\sigma \times \{\sigma\}$ are never equivalent.
Define $\lvert K \rvert = D/\sim$ and give it the quotient topology induced by the quotient map $p : D \to \lvert K \rvert$.
It is now clear that the map $$i_\sigma : \Delta_\sigma \to \lvert K \rvert, x \to p(x,\sigma)$$ is a continuous bijection onto its image. It is also easy to see that $i_\sigma$ is a closed map, thus an embedding.
The interior of $\Delta'_\sigma = i_\sigma(\Delta_\sigma)$ is defined as the image under $i_\sigma$ of the interior of $\Delta_\sigma$. Note that this concept is different from the topological interior of $\Delta'_\sigma$ in $\lvert K \rvert$; in fact, the topological interior of $\Delta'_\sigma $ is in general not open in $\lvert K \rvert$. So perhaps it would be better to call it the simplicial interior of $\Delta'_\sigma$.