In Hatcher's Algebraic Topology, a $\Delta$-complex structure on a topological space $X$ is defined as follows. Here, $\Delta^n$ denotes the standard $n$-simplex in $\mathbb{R}^{n+1}$, and $\overset{\circ}{\Delta}{}^n$ denotes its interior.

A $\mathbf\Delta$-complex structure on a space $X$ is a collection of maps $\sigma_\alpha : \Delta^n \to X$, with $n$ depending on the index $\alpha$, such that:

1. The restriction $\sigma_\alpha | \overset{\circ}{\Delta}{}^n$ is injective, and each point of $X$ is in the image of exactly one such restriction $\sigma_\alpha | \overset{\circ}{\Delta}{}^n$.
2. Each restriction of $\sigma_\alpha$ to a face of $\Delta^n$ is one of the maps $\sigma_\beta : \Delta^{n-1} \to X$. Here we are identifying the face of $\Delta^n$ with $\Delta^{n-1}$ by the canonical linear homeomorphism between them that preserves the ordering of the vertices.
3. A set $A \subset X$ is open iff $\sigma_\alpha^{-1}(A)$ is open in $\Delta^n$ for each $\sigma_\alpha$.

I can place a $\Delta$-complex structure on the sphere $S^2$ using three $0$-simplices, three $1$-simplices, and two $2$-simplices, essentially by gluing the two $2$-simplices along their boundary edges so that each simplex is a hemisphere and the (common) boundary is the equator.

Question: Is there a $\Delta$-complex structure on the sphere $S^2$ using fewer than three $0$-simplices?

I do not believe that this is possible, since I don't see how one can possibly glue the edges of some simplices onto fewer than three vertices to get a sphere; but, I don't know how to write down a rigorous argument.

posted 8 months ago

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The Amplitwist‭

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