Let's say I have a topological space $X$ that's a polyhedron, then there exists a simplicial complex $K$ and a homeomorphism $h : |K| \to X$. If we have a subspace $A \subseteq X$, is there a subcomplex $L$ of $K$ such that $h(|L|) = A$?
It's easy to see that $h^{-1}(A)$ is a subspace of $|K|$ (which is the geometric realization of the simplicial complex $K$), but I don't have an idea if there is a subcomplex of $K$ whose geometric realization is $h^{-1}(A)$.
If it's the case that we can't find a subcomplex of $K$ from subspaces of $X$, under what conditions can we determine if a subspace of $X$ has a corresponding subcomplex of $K$?
Let us call $h : \lvert K \rvert \to X$ a triangulation of the polyhedron $X$. Polyhedra have many triangulations, and I suggest to consider the following more general question:
If we have a subspace $A \subset X$, is there a triangulation $h : \lvert K \rvert \to X$ and a subcomplex $L \subset K$ such that $h(\lvert L \rvert) = A$?
The answer is "no". The minimal requirement is that $A$ is a polyhedron, hence any $A \subset X$ which cannot be triangulated (e.g. a copy of the Cantor set) provides a counterexample. But even if $A$ is a polyhedron, the answer is in general "no". As an example (with a non-trvial proof!) take the Alexander horned sphere $A$ sitting in $S^3$ (see https://en.wikipedia.org/wiki/Alexander_horned_sphere).
I do not know general conditions assuring that $A$ is triangulated by a subcomplex, and I doubt that such conditions exist, but perhaps somebody else can help.