This question arose out of my attempts to understand another question. The most popular construction for the chain complex for defining singular homology uses the $n$-simplex.

But it is also possible to use other spaces. For example, one can use the $n$-cubes instead, as done in certain books. Also it occurs to me that one can use the discs $D^n$, with orientation specified at the boundary. I haven't checked all the details; but I am hopeful that it can be made to work and made to prove that this homology is the same as the homology constructed with cubes or with simplexes.

So, why is the simplex the most used choice? Granted, it has a certain symmetry in all directions and so it is aesthetically somehow more satisfying. But are there other reasons?

Re to Greg Kuperberg: The details go roughly like the following. An $n$-disc is the unit ball in $\mathbb R^n$. The boundary map is $D^n$ going to $S^{n-1}$ on the boundary, but so that this boundary is a union of two discs $D^{n-1}$ but equipped with opposite orientation, ie the parts above and below the equator. Orientation for an $n$-disc is a choice of direction into the center, or going away from the center. Orientation for a $1$-disc is just a choice of direction. A bit more checking is needed to fix the orientation, but I am hopeful that it can be done. Assuming this, and after extending to chains, clearly the images are contained in kernels. Thus homology can be defined. It will be more difficult to prove that this is (after sorting out minor discrepancies) essentially the same as the homology given by simplexes. There is no nice barycentric subdivision, which was a problem. But, since the $n$-disc is homeomorphic to the $n$-simplex, I was hoping that at least an ugly proof of equivalence could be constructed.

This question arose out of my attempts to understand another question. The most popular construction for the chain complex for defining singular homology uses the $n$-simplex.

But it is also possible to use other spaces. For example, one can use the $n$-cubes instead, as done in certain books. Also it occurs to me that one can use the discs $D^n$, with orientation specified at the boundary. I haven't checked all the details; but I am hopeful that it can be made to work and made to prove that this homology is the same as the homology constructed with cubes or with simplexes.

So, why is the simplex the most used choice? Granted, it has a certain symmetry in all directions and so it is aesthetically somehow more satisfying. But are there other reasons?

Re to Greg Kuperberg: The details go roughly like the following. An $n$-disc is the unit ball in $\mathbb R^n$. The boundary map is $D^n$ going to $S^{n-1}$ on the boundary, but so that this boundary is a union of two discs $D^{n-1}$ but equipped with opposite orientation, ie the parts above and below the equator. Orientation for an $n$-disc is a choice of direction into the center, or going away from the center. Orientation for a $1$-disc is just a choice of direction. A bit more checking is needed to fix the orientation, but I am hopeful that it can be done. Assuming this, and after extending to chains, clearly the images are contained in kernels. Thus homology can be defined. It will be more difficult to prove that this is (after sorting out minor discrepancies) essentially the same as the homology given by simplexes. There is no nice barycentric subdivision, which was a problem. But, since the $n$-disc is homeomorphic to the $n$-simplex, I was hoping that at least an ugly proof of equivalence could be constructed.

This question arose out of my attempts to understand another question. The most popular construction for the chain complex for defining singular homology uses the $n$-simplex.

But it is also possible to use other spaces. For example, one can use the $n$-cubes instead, as done in certain books. Also it occurs to me that one can use the discs $D^n$, with orientation specified at the boundary. I haven't checked all the details; but I am hopeful that it can be made to work and made to prove that this homology is the same as the homology constructed with cubes or with simplexes.

So, why is the simplex the most used choice? Granted, it has a certain symmetry in all directions and so it is aesthetically somehow more satisfying. But are there other reasons?

This question arose out of my attempts to understand another question. The most popular construction for the chain complex for defining singular homology uses the $n$-simplex.

But it is also possible to use other spaces. For example, one can use the $n$-cubes instead, as done in certain books. Also it occurs to me that one can use the discs $D^n$, with orientation specified at the boundary. I haven't checked all the details; but I am hopeful that it can be made to work and made to prove that this homology is the same as the homology constructed with cubes or with simplexes.

So, why is the simplex the most used choice? Granted, it has a certain symmetry in all directions and so it is aesthetically somehow more satisfying. But are there other reasons?

Re to Greg Kuperberg: The details go roughly like the following. An $n$-disc is the unit ball in $\mathbb R^n$. The boundary map is $D^n$ going to $S^{n-1}$ on the boundary, but so that this boundary is a union of two discs $D^{n-1}$ but equipped with opposite orientation, ie the parts above and below the equator. Orientation for an $n$-disc is a choice of direction into the center, or going away from the center. Orientation for a $1$-disc is just a choice of direction. A bit more checking is needed to fix the orientation, but I am hopeful that it can be done. Assuming this, and after extending to chains, clearly the images are contained in kernels. Thus homology can be defined. It will be more difficult to prove that this is (after sorting out minor discrepancies) essentially the same as the homology given by simplexes. There is no nice barycentric subdivision, which was a problem. But, since the $n$-disc is homeomorphic to the $n$-simplex, I was hoping that at least an ugly proof of equivalence could be constructed.