# Is there a (satisfying) proof that cellular cohomology is isomorphic to simplicial cohomology that doesn't use relative cohomlogy?

That singular and de Rham cohomologies of a smooth manifold are isomorphic has two proofs that I know of. The classical one uses Stokes' theorem to give the isomorphism explicitly. The second proof that I know uses the machinery of derived categories. Namely, instead of thinking of singular and de Rham cohomologies as the homologies of complexes of abelian groups, one thinks of them as being the hypercohomologies of complexes of sheaves (whose global sections are the abelian groups that appear in the classic definition of singular and de Rham cohomologies). The isomorphism then follows from the following two steps:

1. Showing that both complexes of sheaves are exact, implying that they are both quasi-isomorphic to $..\rightarrow 0\rightarrow \mathbb{R}\rightarrow 0\rightarrow ...$.

2. Showing that the sheaves that appear in both complexes are $\Gamma$-acyclic, implying that the hypercohomology is simply the homology of the complex after taking global sections. (In fact, showing that the sheaves are "flabby".)

I find this second proof to be very comforting, and a good way in general to show that various definitions of cohomology are isomorphic. Indeed, the same argument (mutatis mutandis) would follow for simplicial cohomology. One can also make a similar argument for algebraic de Rham in the affine case. (In the non-affine case the sheaves are not $\Gamma$-acyclic in general; in the affine case this is the algebraic version of Cartan's theorems A and B.)

Sadly, this second proof does not apply to cellular cohomology for the simple reason that the complex of abelian groups for which cellular cohomology is the homology, is not the global sections of a complex of sheaves in any natural way.

In Hatcher the proof of the isomorphism between cellular cohomology and singular cohomology goes through a rather unintuitive argument using relative cohomology. Is there a more satisfying proof? Perhaps, though not necessarily, one that uses derived categories? Alternatively, is there a proof that uses the Grothendieck group over the category of CW-complexes, or some other variation on motivic arguments?

• There's a very short proof using the spectral sequence associated to the the filtration of the space by $n$-skeletons. The $\infty$-page is the singular homology, the $E_2$ page in the cellular homology, and its clear for degree reasons that the rest of the differentials are 0. Unfortunately there isn't time before going to bed (or room in this margin) to write all the details. I recall reading it in Fuks: amazon.com/Beginners-Course-Topology-Geometric-Universitext/dp/… – Ben Webster May 13 '15 at 22:10
• Avoiding relative homology is a very bad idea by the way. The whole point is that the cellular chain groups are the homology of the $n$-skeleton rel the $n-1$-skeleton. – Ben Webster May 13 '15 at 22:13
• The Grothendieck group of finite CW complexes, suitably defined, is $\mathbb{Z}$. The isomorphism is given by taking the compactly supported Euler characteristic. It's not a very interesting group. – Qiaochu Yuan May 13 '15 at 23:02
• By the way, the proof in Hatcher is exactly the same as the spectral sequence proof Ben alludes to, just without using the word "spectral sequence". – Eric Wofsey May 14 '15 at 1:00
• @few_reps the infinite cyclic group is not nearly as interesting as the ring $\bf Z$. – Noam D. Elkies May 14 '15 at 1:30

Here is how I like best to think about this, although I'll point to a few other proofs. Consider a general space $X$ (say compactly generated). There is a natural weak homotopy equivalence $\epsilon\colon |SX|\to X$ from the geometric realization of the singular simplicial set of $X$ to $X$. Moreover, $|SX|$ is a CW complex whose cellular chains (and cochains) are isomorphic to the (normalized) singular chains and cochains of $X$. If $X$ itself is a CW complex, then $\epsilon$ is a homotopy equivalence and is homotopic to a cellular homotopy equivalence. The induced map of cellular chains is a chain homotopy equivalence between the singular chains of $X$ and the cellular chains of $X$.

Another proof just checks the Eilenberg-Steenrod axioms for both. Relative homology is irrelevant since there is a version of the axioms for the reduced homology of spaces that is equivalent to the usual axioms for pairs of spaces. There is also a version of the axioms just on CW complexes, where the excision axiom reduces to the tautology that if a CW complex $X$ is the union of subcomplexes $A$ and $B$, then $A/A\cap B$ and $X/B$ are isomorphic CW complexes. The general excision axiom reduces to this version by a purely topological argument (no use of homology in any form needed). Here is a concise reference for these statements: http://www.math.uchicago.edu/~may/CONCISE/ConciseRevised.pdf.

Consider three related CW complexes; a given CW Complex $X$, the realization of its singular complex $|X^\Delta|$, and the mapping cylinder $C_\eta$ of the comparison (evaluation) map $\eta : |X^\Delta| \to X$. The cellular (co)homology of $|X^\Delta|$ is simply the simplicial (co)homology of $X$; the inclusion $X \to C_\eta$ is a homotopy equivalence (the inclusion of a deformation retract); by a strange circumstance, the inclusion $|X^\Delta| \to C_\eta$ is a weak homotopy equivalence, because so also is the map $\eta$. Both inclusions are also cellular.

Now, so long as you are already happy that cellular (co)homology is a weak-homotopy-invariant, we now have an "explicit", if rather clunky, zigzag of equivalences relating the singular and cellular cohomology of $X$.

You're starting from knowing that simplicial homology and singular homology are the same. Let's assume that cellular homology is a homotopy type invariant (as did some previous posters). Up to homotopy equivalence, any CW complex subdivides to a simplicial complex. (The last sentence follows from the proof of Theorem 2C.6 in Hatcher, at least for finite CW complexes.) It's relatively straightforward linear algebra and combinatorics to show that subdivision doesn't change homology.

Making all details of the above precise might be unwieldy. In particular, isn't the main point of singular homology to show that simplicial/cellular homology is a homotopy type invariant? You could avoid the homotopy type problem by restricting to regular complexes (where all attaching maps are homeomorphisms), but this is a real restriction.

The usual construction of cellular homology starts with the singular homology of a topological space, then uses relative homology groups $H_n(X^n,X^{n-1})$ for a CW-complex $X$ as the $n$-th chain group, and has to prove these cellular homology groups agree with the singular homology groups, as well as that the cellular chains are free abelian groups. If you want homology with operators from $\pi_1(X,x)$, then you have to choose the base point $x$ to consider the universal cover $\widetilde{X}_x$ of $X$ based at $x$, and the cellular homology of $\widetilde{X}_x$.

Here is an indication of "some other construction", given a full account in the book Nonabelian Algebraic Topology, EMS Tracts in Mathematics Vol 15 (2011). This construction is homotopically defined, so that invariance under homeomorphism is automatic; calculation relies on a Higher Homotopy Seifert-van Kampen Theorem, HHSvKT, whose proof uses methods of strict higher homotopy groupoids, and cubical methods. This theorem has nonabelian consequences in dimensions $1$ and $2$, not otherwise available, and in higher dimensions has as but one consequence the Relative Hurewicz Theorem.

The book has an extensive Introduction giving motivation and historical background, gives many calculations, relates to many classical results, and concludes with a number of problems and problem areas.

The idea is to start with a filtered space, $X_*$, i.e. a space $X$ and an increasing sequence of subspaces $X_0 \subseteq X_1 \subseteq \cdots \subseteq X_n\subseteq X_{n+1} \subseteq \cdots \subseteq X$. We then define $C=\Pi X_*$ to consist of the fundamental groupoid $C_1=\pi_1(X_1,X_0)$ in dimension $1$, and for $n \geqslant 2$ the family of relative homotopy groups $C_n(x)= \pi_n(X_n,X_{n-1},x)$ for all $x \in X_0$. These, with boundary maps and operations of $C_1$ on $C_n$ for $n \geqslant 2$ have the structure of what is called a crossed complex.

In a sense, this is a rewrite of the classical Poincaré approach to homology which uses "formal sums", i.e. free abelian groups, but using instead clearly intuitive notions of compositions of cubes to get the algebra. The idea of using $\Pi X_*$ as an algebraic structure goes back to A.L. Blakers, (1948), and to J.H.C. Whitehead, (1949), in his paper "Combinatorial Homotopy II", which develops ideas on homotopy classification, not so well known, even today.

May 20, 2015. For an Introduction to these ideas, starting with a section on "Anomalies in Algebraic Topology", see this presentation Galway, Dec. 2014.

• This does not answer the question. – Andy Putman May 20 '15 at 19:21
• @Andy Putnam The question asked in the last paragraph of the main question seems to be about the whole structure of the usual homology theory, starting with singular theory, and continuing with computation via cellular chains. One also has to look at the whole history of algebraic topology, as glimpsed in the book and presentation cited by me. These show a direct and homotopically defined way to the cellular complex, but in a more powerful form with operators, and using crucially JHC Whitehead's notion of crossed module, a term not in Hatcher's book. – Ronnie Brown May 30 '15 at 10:09
• That's not what it asked for. It asks specifically for alternate proofs of the equivalence between singular and cellular cohomology. That is a very focused question, and you make no attempt to answer it. – Andy Putman May 30 '15 at 18:07
• My first papers resulted from asking new questions, and I can't get out of the habit in my old age! I realise this can be a little like the answer to a traveller: "If I were you I would not start from here!" In maths, this can sometimes be relevant. See also this comment of Einstein: groupoids.org.uk/einst.html Such questioning can be out of favour in a subject. – Ronnie Brown Aug 6 '16 at 20:27