# When is the quotient of simplicial complexes a simplicial complex?

Let $K$ be a simplicial complex and let $L$ be a subcomplex of $K$.

Questions:

• Is it possible to define an operation on (some) simplicial complexes so that $K/L$ is a simplicial complex for which $|K/L|\cong |K|/|L|$?
• Is it the case that $|K|/|L|$ is always triangulable, i.e., that there's a simplicial complex $Q$ such that $|Q|\cong |K|/|L|$?

For example, if $K=\{a,b,ab\}$ and $L=\{a,b\}$ then $K/L$ would be a vertex with a self-loop. That's not a simplicial complex, but it is a cellular complex, and its homology is isomorphic to the homology of $|K|/|L|$. (In this case, to get $K/L$ I identified all simplices in $L$ with a point, preserving "connections" between them.)

For simplicial sets rather than simplicial complexes, you can define quotients perfectly fine: if $K$ is a simplicial set and $L\subset K$ is a sub-simplicial set, there is a simplicial set $K/L$ defined by $(K/L)_n=K_n/L_n$, and the canonical map $|K|/|L|\to |K/L|$ is a homeomorphism. If $K$ and $L$ are simplicial complexes, then you can consider them as simplicial sets, and take the simplicial set $K/L$; if the nondegenerate simplices of $K/L$ happen to form a simplicial complex, then you've found a natural simplicial complex whose geometric realization is $|K|/|L|$. In general, the nondegenerate simplices of a simplicial set form a simplicial complex iff the vertices of any nondegenerate simplex are all distinct and no two nondegenerate simplices have the same vertices. In the case of $K/L$, this translates to the following (quite strong!) conditions on $L$: if two vertices of a simplex of $K$ are in $L$, then the entire simplex must be in $L$, and for any simplex of $K$ not containing any vertices in $L$, there is at most one vertex in $L$ that can be added to it to give a simplex of $K$.
If you want an operation stated purely in terms of simplicial complexes without mentioning simplicial sets, you can do this as follows. Let $K$ be a simplicial complex with vertex set $V$ and $L$ be a subcomplex of $K$ with vertex set $W$ such that if two vertices of a simplex of $K$ are in $W$, then the entire simplex is in $L$, and for any simplex of $K$ not containing any vertices in $W$, there is at most one vertex in $W$ that can be added to it to give a simplex of $K$. Then you can define a simplicial complex $K/L$ as follows: the vertex set of $K/L$ is $V/W$, and a subset $S$ of $V/W$ is a simplex of $K/L$ iff it is the image of a simplex of $K$ under the quotient map $V\to V/W$. There is then a canonical homeomorphism $|V|/|W|\cong|V/W|$.
As for your second question, the answer is yes, since the geometric realization of any simplicial set is triangulable. The proof is nontrivial; see this answer on MO for an indication of some of the ideas involved and references to more detailed proofs. In particular, while there does exist a simplicial complex $Q$ such that $|Q|\cong|K|/|L|$, there is not (as far as I know) any canonical choice of such a $Q$.
• Thanks for answering. Yes, I want an operation stated in terms of simplicial complexes only. I don't quite understand the construction, however. (1) So what does $V/W$ stand for ($V$ and $W$ are sets)? (2) What simplex are you referring to in "If two vertices of $K$ are in $W$ then the entire simplex is in $L$"? (3) What exactly do you mean by "For any simplex of $K$ not containing any vertices in $W$ there is at most one vertex in $W$ that can be added to it to give a simplex of $K$"? (I am accepting the answer as soon as I understand how and why the construction works.) Commented Dec 10, 2015 at 14:27
• (1) $V/W$ is the quotient of the set $V$ by the equivalence relation that make all the points of $W$ equivalent to each other. (2) I said "two vertices of a simplex of $K$", so you have some particular simplex of $K$ you're looking at. (3) I mean that if $S$ is the set of vertices of a simplex in $K$ and $S\cap W=\emptyset$, then there cannot be two distinct vertices $x,y\in W$ such that $S\cup\{x\}$ and $S\cup\{y\}$ are both the vertex sets of simplices of $K$. (The point is that if there were, then $S\cup\{x\}$ and $S\cup\{y\}$ would have the same vertices in the quotient.) Commented Dec 10, 2015 at 19:23
• As for why it works, this should be straightforward to see once you understand it. The point is that the hypotheses on $L$ are extremely restrictive, such that when you just glue together all the vertices of $L$ and throw away all the higher-dimensional simplices of $L$ (or equivalently, collapse them together down to a single point), you still get a simplicial complex (no simplex you had before has had two of its vertices glued together, and simplices are still uniquely determined by their vertex sets). Commented Dec 10, 2015 at 19:41
• Could you elaborate on a canonical homeomorphism $f:|V/W|\to|V|/|W|$? I understand that once you have $f$ you can extend it to a homeomorphism $g:|K/L|\to|K|/|L|$. For instance, given $f$ for a suitable geometric realization, take $x\in|K/L|$ for $x=\lambda_1v_1+\ldots+\lambda_rv_r$, where $\lambda_i$'s are barycentric coordinates. Now define $g(x)=\lambda_1f(v_1)+\ldots+\lambda_rf(v_r)$, which is a homeomorphism between $|K/L|$ and $|K|/|L|$. Commented Dec 14, 2015 at 15:03