Is a square a $\Delta$-complex?

Question

Notation and definitions:

• $\Delta^n$ is the standard $n$-simplex with ordered vertices $[v_0,\ldots,v_n]$;
• $[v_0,\ldots,\hat{v_i},\ldots,v_n]$, where the hat denotes omission, is a face of $\Delta^n$ and is an $(n-1)$-simplex, and this restriction respects the ordering;
• $\mathrm{int}\Delta^n$ is $\Delta^n$ minus the union of all its faces.

The definition of a $\Delta$-complex, a generalisation of a simplicial complex, from Hatcher's Algebraic Topology is as follows:

Let $X$ be a space. A $\Delta$-complex structure on $X$ is given by a collection of continuous maps $$\{\sigma_\alpha\colon\Delta^n\to X \mid \alpha\in A\}$$ such that

1. The restriction $\sigma_\alpha\mid_{\mathrm{int}{\Delta^n}}$ is injective, and each point in $X$ is in the image of exactly one such restriction;
2. each restriction of $\sigma_\alpha$ to some face of $\Delta^n$ is one of the maps $\sigma_\beta\colon\Delta^{n-1}\to X$ (identifying the face of $\Delta^n$ with some $\Delta^{n-1}$ by the canonical linear homomorphism respecting the ordering of the vertices);
3. $A\subset X$ is open if and only if $\sigma_\alpha^{-1}(A)$ is open in $\Delta^n$ for each $\sigma_\alpha$.

My question is, why is the following space $X$ only a $\Delta$-complex when we include the orange $1$-simplex? Note that there are no $2$-simplices in the above picture, and the arrows represent the ordering of the vertices, and not side identification.

I feel like the clue is in the name: a square is most certainly not a triangle (at least, not to my knowledge), and so it makes sense that it wouldn't merit the honour of being a $\Delta$-complex. I'm sure this is something obvious, but really can't quite spot what exactly...

Bonus question: what are these $\Delta$-complexes called? That is, if I were doing a literature search for more information, what would I search for?

Answer 1

As a short answer to my question: yes, a square (with no $2$-simplices) as in the picture does have the structure of a $\Delta$-complex. Though if we 'fill it in' with a $2$-simplex, we actually need to use two, and also add in the $1$-simplex that is represented by the orange line in the picture.