*(Originally posted at m.se without answers.)*

Let $T$ be a set of triangles in an abstract simplicial complex, with orientation of the triangles chosen such that

$$\partial \left( \sum \limits_{t \in T} t \right) = 0$$

where the boundary operator is as usual, and we consider coefficients in $\mathbb{Z}$. Let $S$ be the support of $T$, i.e. the set of nodes in the complex that appear in some $t \in T$.

Does there necessarily exist a $3$-dimensional geometric realization of $T$, in the following sense:

Is there a polyhedron $P$ with vertices $N$ and all triangular faces $F$, as well as a map $\phi : N \to S$ such that $\phi(F) = T$?

(where $\phi(F) = \bigcup \limits_{(a, b, c) \in F} (\phi(a), \phi(b), \phi(c))$)

I mean "polyhedron" here in the classical sense: a 3d solid with finitely many faces, all of which are "flat," such that only adjacent faces intersect and they do so only at a boundary point or a line. The polyhedron can be non-convex (although I suspect that this question is equivalent if we restrict to the convex case).