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:


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*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.)
 A: 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 $Q$.
