How can the definition of triangulation be applied to quotient spaces?

Question

I've been taught the following definitions in my topology class:

1. An $n$-simplex is the convex hull of a set of $n+1$ affinely independent points (i.e. it is a subset of $\mathbb{R}^k$, $k \geq n$).

2. A simplicial complex is a set $\mathcal{T}$ of simplices that follow the two rules here. I understand that these are sometimes called geometric simplicial complexes, and I understand that there is also such a thing as an abstract simplicial complex, but since my class is using geometric simplices, I'd like to understand it through the geometric lens.

3. There is a topological space $|\mathcal{T}|$ that is the union of all the simplices in the set, given the subspace topology (my professor calls this the support, although I think the more accurate term is the "underlying space"?).

4. A triangulation of a space $X$ is a homeomorphism $f : |\mathcal{T}| \to X$.

With these definitions, my professor taught us that if, for example, we were to triangulate the cylinder, we could create a simplicial complex whose underlying space is the unit square $[0, 1]^2$, then identify the edges in the usual way. However, he explained that not all complexes are valid triangulations. For example, the naive method of taking a diagonal from top left to top right does not work, but if you first split it into three rectangles from left to right, then add a diagonal to each rectangle, it is valid (assuming you're identifying the left and right edges).

He explained this by showing that the first definition would violate the axioms of a simplicial complex, while the second does not. This makes sense to me conceptually, especially after reading this answer. What I'm still struggling with is how those rules can be applied in this case based on the definition, for two reasons:

1. How can we define the homeomorphism required? If $|\mathcal{T}| = [0, 1]^2$, my understanding is that the unit square and the cylinder are not homeomorphic, so how can we show a homeomorphism? I don't see any way to convert to the quotient space from $|\mathcal{T}|$
2. Even if we do define a homeomorphism, is it not true that both of the above simplicial complexes have the same underlying space $|\mathcal{T}|$? How can one have a homeomorphism while the other does not, if this operator just takes the union of all the simplices and gives it the subspace topology?

I'm assuming I've either misrepresented or misunderstood one of the definitions here, but I've been unable to find an alternate explanation anywhere online (I'm sure one exists, but I haven't been able to figure out what to look up to find it). If anyone could explain to me how this works, either using the definitions above, or explaining which of the definitions are wrong, I would appreciate it!

Answer 1

If you want to understand the cylinder as a geometric simplicial complex then you are going to have to drop the idea of strictly using $[0,1]^2$ --- it is not a cylinder itself; and its quotient space by identifying the left side to the right side is not a subset of $\mathbb R^n$ for any $n$.

On the other hand, here is something to try.

Consider a simplicial structure on $[0,1]^2$ as you have described: first split it into three rectangles from left to right, then add a diagonal to each rectangle. This puts a simplicial structure on $[0,1]^2$ with six 2-simplices. You seem to understand already that if you now identify the left side to the right side, the resulting quotient space will be homeomorphic to the cylinder.

Next, think about the standard embedding $\mathbb R^2 \hookrightarrow \mathbb R^3$ defined by $(x,y) \to (x,y,0)$, and think about the image of $[0,1]^2$ under this embedding.

Now here's the challenge: Is there some way to bend or warp this image to get a quotient map $f : [0,1]^2 \to X \subset \mathbb R^3$ such that $X$ is a simplicial complex, $f$ is a simplicial map, and $f$ identifies the left and right sides of $[0,1]^2$ in the manner needed, and $f$ makes no other identifications? If so, you will have constructed a simplicial complex $X \subset \mathbb R^3$ homeomorphic to the cylinder.

You can even try this with a piece of square cardboard, scored along two vertical segments and three diagonal segments in the pattern of the simplicial structure on $[0,1]^2$, and then you can try bending along the score lines.