Single $\Delta$-complex structure on $S^3$ Is it possible to identify pairs of faces of $\Delta^3$ to produce a $\Delta$-complex structure on $S^3$ having a single simplex?
I'm thinking to identify left with right and behind with bottom. But I cannot visualize what I can obtain and how can I make sure I get $S^3$?
 A: Identify $[012] \sim [013]$ and $[123]\sim[023]$. The first identification $[013] \sim [012]$ happens by stretching one face around to meet the other. Then we are left with a double-cone, and completing the next identification $[123] \sim [023]$ is the same identifying the faces of the cone as mirror images, or via reflection. (I found this easiest to visualize by drawing two pictures from the perspective of each tip of the cone.)
We now just need to see that $S^3$ is homotopic to $D^3/\sim$ with $x \sim r(x)$ for $x \in \partial D^3$, and $r$ being reflection through some plane passing through an equator.
To see it, instead imagine the 3-cell of $D^3$ as being the space outside of the boundary sphere. Then, to identify the boundary sphere via the reflection map is just to crush it to a disc. We may now contract this disc to a point, and what we have is $S^3$ since we have started with $D^3$ and glued precisely all of the boundary points together, i.e. $D^3/\partial D^3 \cong S^3$.
We can be a little more rigorous and see that the space is even homeomorphic to $S^3$. Consider $D^3$ inside the one-point compactification of $\mathbb{R}^3$, which is homeomorphic to $S^3$. Then we embed $D^3 \to \widehat{\mathbb{R}^3}$ via $x \mapsto \dfrac{x}{||x||^2}$, mapping $0$ to the point at infinity. What we have then is $S^3$ with an open ball we need to remove. Consider a Euclidean neighborhood around this ball, then removing it and crushing its boundary to a disc, we still have a Euclidean neighborhood. And this is our space, $S^3$.
As a check, I have computed the homology in the three cases, and the only time we get the correct homology for $S^3$ is with the above identification.
[Edit: The previous paragraph is based on a faulty computation of one of the other cases, which has since been corrected. As such, case (2) below could also yield $S^3$, but the writer is unsure of whether it does or not.]
(1) If you identify $[012]\sim[013]$ and $[123]\sim[023]$ you get
$$H_n(X) = \begin{cases}\mathbb{Z}&\text{ if } n=3\\0&\text{ if } n=2\\0&\text{ if } n=1\\\mathbb{Z}&\text{ if } n = 0.\end{cases}$$
(2) If you identify $[012]\sim[123]$ and $[013]\sim [023]$ you get
$$H_n(X) = \begin{cases}\mathbb{Z}&\text{ if } n=3\\ 0&\text{ if } n=2\\ 0 &\text{ if } n =1\\ \mathbb{Z} & \text{ if } n = 0.\end{cases}$$
(3) If you identify $[012]\sim [023]$ and $[013]\sim[123]$ you get
$$H_n(X) = \begin{cases}0&\text{ if } n=3\\ \mathbb{Z}/2\mathbb{Z} &\text{ if } n = 2\\ 0&\text{ if } n=1\\ \mathbb{Z} & \text{ if } n=0.\end{cases}$$
