Question: Why do we use continuous maps from the standard $n$-simplices $\Delta_n$, and not another space like $B^n$ or $I^n$, when defining singular homology?

I am taking courses in convex optimization and homology theory at the same time. I was expecting what I learned in convex analysis about simplices to be relevant and applied frequently to the study of singular homology, since the latter is defined in terms of simplices, but this does not seem to be the case. Thus I am confused about why simplices are used in defining singular homology in the first place, when singular homology does not seem to take advantage of the properties of simplices.

For example, higher-dimensional cubes $I^n$ or closed balls $B^n$ (I'm not sure which) are used when defining the higher homotopy groups, and for $n=0,1$ the standard simplices, cube, and ball all coincide (point and unit interval, respectively). This allows one to interpret $0$- and $1$-chains in a space $X$ as points and/or paths in $X$, respectively. However, $\Delta_n$ and $I_n$ (or $B_n$) no longer coincide for $n \ge 2$, so it is unclear whether the analogous identifications still hold, see e.g. here.

Likewise, one uses closed spheres $B^n$ for defining CW complexes, so if there was a theory using continuous maps from the $B^n$ (instead of from the $\Delta^n$), perhaps it would be easier to apply to the important class of spaces that CW complexes?

In other words, it seems like there is a big tradeoff for using the simplices $\Delta_n$ instead of the more natural choices of spheres $B^n$ or cubes $I^n$, so there should be a good reason justifying this tradeoff -- however none of my books address this issue, leaving me very confused.

Also, simplices play a very big role in convex analysis (to the best of my knowledge), and can be shown within the framework of convex analysis to have many special properties. However, when working with singular homology, it never seems like anyone ever uses these special properties as identified in convex analysis -- indeed, I have never seen an algebraic topology book with an interlude or appendix explaining facts from convex analysis relevant to understanding simplices (and thus seemingly also singular homology) -- why is that?

Likewise, in convex analysis one also studies cubes and cross-polytopes as well as their images under affine transformations. Why don't we also consider continuous maps from those spaces?

One can also define in addition to simplicial complexes complexes from cubes (cubical complexes) as well as "cell complexes built from arbitrary compact convex polytopes [which] are likewise easy to work with" (Ghrist, Elementary Applied Topology, p. 27). Why don't we ever work with continuous maps from arbitrary compact convex polytopes (including cubes and cross-polytopes) or continuous maps from arbitrary convex bodies (compact and convex spaces, including spheres)?

From Massey, A Basic Course in Algebraic Topology, p. 148:

If $X$ is a compact, connected, orientable n-dimensional manifold, then $H_n(X)$ is infinite cyclic, and $H_q(X)=\{0\}$ for all $q>n$. In some vague sense, such a manifold is a prototype or model for nonzero $n-$dimensional homology groups.

Would considering continuous maps from $I^n$ or $B^n$ or any other graded sequence of compact convex spaces not produce a homology theory, the way continuous maps from simplices $\Delta_n$ do? Even if they did, would these homology theories not be homotopy invariant?

Related questions: (1)(2)(3)(4)(5)

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    $\begingroup$ IIRC Massey uses $I^n$. But, he needs to mod out non-degenerate cases, where the mapping does not depend at all on one or more coordinates. IIRC the reason for that extra complication is that otherwise you get "wrong" homology for a one-point-space, if you don't do that! For example the constant map from $I^2$ to a single point would then have zero boundary, but not be a boundary itself. $\endgroup$ Nov 10, 2016 at 11:58
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    $\begingroup$ Aside: for CW-complexes, there is cellular homology. $\endgroup$
    – user14972
    Nov 11, 2016 at 20:55
  • $\begingroup$ @Hurkyl I was just thinking about this -- is cellular homology for CW-complexes like "non-singular homology"? Since the $n-$cells are actually homeomorphic to an $n-$simplex/ball/cube, rather than the case of singular homology, when the singular $n-$simplices are just the images of continuous maps, but not necessarily of embeddings, so they need not be actual $n-$dimensional manifolds with or without boundary. Is cellular homology for CW-complexes where one can identify the chain groups with formal sums of actual/proper $n-$dimensional manifolds with or without boundary? $\endgroup$ Nov 12, 2016 at 10:32

3 Answers 3


nLab has a page on cubical sets with connections with the following excerpt:

The ordinary cube category is a test category. This means that bare cubical sets carry the structure of a category with weak equivalences whose homotopy category is that of ∞-groupoids.

But the category of cubes with connection is even a strict test category (Maltsiniotis, 2008). This means that under geometric realization (see the discussion at homotopy hypothesis) the cartesian product of cubical sets with connection is sent to the correct product homotopy type.

The lack of this property for cubical sets without connection was one of the original reasons reasons for abandoning Kan’s initial cubical approach to combinatorial homotopy theory in favour of the simplicial approach; the implications of this new result have yet to be thought through. Another reason was that cubical groups were in general not Kan complexes; however cubical groups with connection are Kan complexes. See the paper by Tonks listed below.

What I take away from this is that, while cubical sets do correctly capture the homotopy type of a space, their category is defective in a way that the category of simplicial sets is not, and the foundations of the subject were developed before it was discovered how to repair the notion of cubical set.

  • $\begingroup$ This is an interesting perspective. I will have to do some research though before I understand what cubical and simplicial sets/categories are -- so far I literally only know the definitions of the most basic cases, i.e. a cube and simplex. $\endgroup$ Nov 12, 2016 at 10:35
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    $\begingroup$ @William: A simplicial set is an algebraic structure that, on one extreme, can be used to hold all of the data in the maps $\Delta^n \to X$, and fro the opposite extreme can viewed as a recipe for constructing a space out of simplicies. Also, as an algebraic structure, they are one of the popular ways for outside subjects to make a connection to algebraic topology. $\endgroup$
    – user14972
    Nov 14, 2016 at 3:08

The use of simplices is technically easier but sometimes one has no choice but to use cubes, as for Jean-Pierre Serre's spectral sequence set-up. For this purpose, Serre had to develop an approach with cubes in the 1950s. Since then spectral sequences have become standard tools in many areas of mathematics, so one certainly can't claim cubes have not been used instead of simplices.

  • $\begingroup$ Sorry for claiming that -- I didn't know it was false. Why is the use of simplices technically easier? Why and how do the two methods lead to different results? (i.e. how does one distinguish between the two methods when simplices and cubes are homeomorphic -- assuming that they are homeomorphic -- are they?) $\endgroup$ Nov 10, 2016 at 12:40
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    $\begingroup$ Serre showed that the theories are equivalent, i.e. produce the same homology and cohomology theories. This is more complicated than observing that the simplex and the cube are homeomorphic, because the boundary maps for singular chains are defined in terms of the combinatorics of the boundary of the basic object, be it simplex or cube. $\endgroup$ Nov 10, 2016 at 12:51
  • $\begingroup$ why should one need to use cubes for spectral sequences? i think one can construct them without cubes $\endgroup$
    – Juan Fran
    Nov 10, 2016 at 16:11
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    $\begingroup$ @JuanFran, I am confident in your research abilities. I was merely reporting on how Serre did it. It stands to reason that a spectral sequence for a fibration would involve cubes rather than simplices, because the simplest fibration is after all a product, like a square. $\endgroup$ Nov 10, 2016 at 16:17
  • $\begingroup$ I see, Well I find it an interesting point $\endgroup$
    – Juan Fran
    Nov 10, 2016 at 17:18

In Massey's book, where cubes are used instead of simplices, in order to define the chain group, one has to first define the group of degenerate $n-$"cubes" $D_n(X)$ and then mod out the space of all singular $n-$"cubes" $Q_n(X)$ by the degenerate ones $D_n(X)$ in order to finally arrive at the $n-$dimensional chain group $C_n(X)=Q_n(X)/D_n(X)$. In contrast, Vick's book, using the simplicial approach, defines $C_n(X)$ to directly be the space of continuous maps from $n-$simplices, without having to mod out a "degenerate" subspace first -- this suggests that such a construction/step is only necessary when using the cubical approach but not the simplicial one, and thus the reason why many authors prefer the simplicial approach may be because it is technically simpler as a result.

The following quotes from this essay by Hurewicz seem like they may be relevant somehow:

Homotopy classes of mappings of an $n-$dimensional polytope $P$ into the sphere $S^n$ are in one-to-one correspondence with the elements of the $n-$dimensional cohomology group of $P$ with integers as coefficients.

To determine the homotopy groups of a given space is, generally speaking, an extremely difficult problem (even for finite polytopes) which so far has been solved only in a few special cases. In this respect there is a significant difference between homotopy and homology. When a polytope $P$ is broken up in two subpolytopes $Q$ and $R$, there is a relatively simple relation (Meyer Vietoris theorem restated recently in terms of the so-called exact sequences) between the homology invariants of the polytopes $P, Q, R$ and the intersection $Q \cap R$. No analogous relation exists for homotopy groups. This is tied up to the fact that a continuous image of the n-sphere in $P$ cannot be decomposed into "small" spherical images, the way a simplicial chain can be decomposed into "small parts." Therefore the basic process of homology theory consisting in decomposing a space into smaller pieces with simpler homology structure has no counterpart in homotopy theory.

Although I am not sure if this entirely relevant, since Hurewicz is discussing spheres $S^n$, which have no boundary, and not the balls $B^n$, which have the spheres $S^{n-1}$ as boundary, and the spheres $S^{n-1}$ are homeomorphic (I think?) to the connected sum of two $n-1$-balls $B^{n-1}$. In other words it seems like the problem Hurewicz is describing exists only for "boundary-less" spheres, but not for balls.

This distinction between balls and spheres seems especially relevant in the following passage:

Let $Y$ be a path-connected space and let $H_n(Y)$ be the $n-$dimensional homology group of $Y$ based on singular chains, with integers as coefficients. A continuous image of $S^n$ in $Y$ can be regarded as a singular $n-$cycle. Since two homotopic spherical images determine homologous singular cycles, one obtains a "natural" homomorphism of $\pi_n(Y)$ into $H_n(Y)$.

Since spheres are essentially "boundary-less" balls (since they are the connected sum of two balls along their boundary, I think) this might suggest that, since the image of "boundary-less" balls lies in the kernel of the boundary map (which is what being a singular $n-$cycle means I think) then the image of arbitrary balls would just be an arbitrary chain, not necessarily one with zero boundary (one in the kernel of the boundary map). In other words, that singular homology using $n-$balls is the same thing as singular homology using $n-$simplices or $n-$cubes, and that simplices are generally used out of respect for historical tradition and because the choice doesn't impact the resulting theory. But I'm not sure of this at all.

Arcwise connected spaces whose homotopy groups in dimensions less than or equal to $n$ vanish are called $n$-connected. This property is equivalent to the condition that every continuous image of an arbitrary $n$-dimensional polytope in $F$ be homotopic to a single point.

I just realized that the following quote from Hatcher's Algebraic Topology (p. 101) might be relevant, although it seems to more directly address cellular, simplicial, and cubical complexes rather than singular homology itself, per se:

Arbitrary polyhedra can always be subdivided into special polyhedra called simplices... so there is no loss of generality, though initially there is some loss of efficiency, in restricting attention entirely to simplices.


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