In the following I will use definitions and
constructions of simplicial complexes and $\Delta$-complexes/sets from Greg Friedman's An elementary illustrated Introduction to simplicial sets.
I'm looking for an example of a $\Delta$-complex which cannot be endowed with structure of a simplicial complex or a proof that it always works. In this case one should say this space is "triangulable".
Note that of course it's easy to construct a $\Delta$-complex which isn't a simplicial complex with respect the choices of simplices comming from the $\Delta$-complex datum: see example on page 11 in Friedman's notes. And this should be not surprising at all because $\Delta$-complexes allow much more flexibility in gluing boudaries than simplicial complex. But that's not what I'm asking about: if we subdivide the $2$-cell in tree new $2$-cells, then we can endow this $\Delta$-complex with another simplex structure which turns it into a honest simplicial complex.
And so my question is if every $\Delta$-complex "triangulable" in the sense that one can endow it with structure of a simplicial complex (ie decompose it into simplices satisfying glueing relations allowed for simplicial complex) which might have nothing to do with original $\Delta$-complex structure?
Supplemental picture:
Best Answer
Here is an exercise: The 2nd barycentric subdivision of every $\Delta$-complex is a simplicial complex.
To follow up on our discussion in the comments, let me address the example edited in to your post.
Let me denote the large blue $2$-simplex as $\Sigma$, which we think of as the domain of one of the simplices in the given $\Delta$-complex $X$, with corresponding characteristic map denoted $f : \Sigma \to X$.
Let me also denote the highest vertex of $\Sigma$ as $A$, the lower right vertex of $\Sigma$ as $E$, and all the other vertices of the 2nd barycentric subdivision along $\overline{AE}$ as $A,B,C,D,E$.
A key observation here is that while it is possible that $f(A)=f(E)$, there is no other possibility of two points on $\overline{AE}$ being mapped to the same image by $f$: the function $f$ is injective on the half-open edge $[A,E)$. This follows from the definition of a $\Delta$-complex, together with the assumption that $f$ is one of the characteristic maps of the given $\Delta$-complex.
Consider also the two red outlined 2-simplices in your diagram, each of which is a simplex of the 2nd barycentric subdivision of $\Sigma$; let me denote them as $\sigma$, which has $\overline{AB}$ as one of its edges, and $\tau$ which has $\overline{BC}$ as one of its edges. Notice that $\sigma \cap \tau$ is a common $1$-simplex face of each of $\sigma$ and $\tau$.
The thing to observe is that $f$ is injective on $\sigma$, also $f$ is injective on $\tau$, and finally $f(\sigma) \cap f(\tau) = f(\sigma \cap \tau)$. This holds because $f$ is injective on $[A,C)$.