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Consider the CW structure on $\mathbb{RP}^n$ given by one cell in every dimension. This gives rise to the cellular complex $C_\bullet(\mathbb{RP}^n)$ which is generated by a single element $c_i$ for every degree $0\le i\le n$ and with boundary map: $$\partial c_i=\cases{0&if $i$ is even\\2c_{i-1}&if $i$ is odd}$$ which gives us the usual homology. If we dualize this complex (with respect to $\mathbb{Z}$), we have the cochain compex $C^\bullet(\mathbb{RP}^n)$ which is generated by the maps: $$\begin{array}{rcrcl}c_i^*&:&C_i(\mathbb{RP}^n)&\longrightarrow&\mathbb{Z}\\&&c_i&\longmapsto&1\end{array}$$ The coboundary is given by: $$\delta c_i^*=c_i^*\circ\partial=\cases{2c_{i+1}^*&if $i$ is even\\0&if $i$ is odd}$$ And this gives a cohomology that is different from the usual singular cohomology, as we can see for example from the fact that $H^0=0$.

Now, if I haven't done stupid mistakes in the above, this would show that cellular cohomology is not (always) isomorphic to singular cohomology, while this is always true in homology. Why does this happen?

I have been told that if we have a CW structure such that the cellular complex can be embedded into the singular complex (via a chain map), then the cohomologies of the dual complexes will be isomorphic. Is this always true? Where can I find a proof of this fact? Does the inclusion of complexes induce an isomorphism in homology (so that we have an isomorphism in cohomology because of the universal coefficients theorem and $5$-lemma)?

Sorry if that's a lot of questions for one post.

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  • $\begingroup$ Cellular cohomology is isomorphic to singular cohomology on CW complexes, see Hatcher p.202. Why do you think $H^0(\mathbb{RP}^n) = 0$? $\endgroup$
    – Emilio Ferrucci
    Aug 17, 2014 at 8:45
  • $\begingroup$ @EmilioFerrucci Because I get that $\ker(\delta:C^0\to C^1)=0$, since $\delta c_0^*(c_1)=c_0^*(\partial c_1)=c_0^*(2c_0)=2$, and thus $\delta c_0^*=2c_1^*$. DId I do something wrong somewhere? $\endgroup$
    – Daniel Robert-Nicoud
    Aug 17, 2014 at 9:31
  • $\begingroup$ Ah yes maybe I see the problem: it looks like you switched even and odd in $\partial$ (and therefore in $\delta$). You can find the computation on p.144 of Hatcher, where he uses the fact that the degree is the sum of local degrees to compute the cellular boundary map directly from the definition in the even and odd cases. Does this help, and do you still need help with the questions in the last paragraphs? $\endgroup$
    – Emilio Ferrucci
    Aug 17, 2014 at 9:46
  • $\begingroup$ (Notice your $\partial$ would not give rise to the usual homology, since $\partial \colon C_1 \to C_0$ would have an image and $H_0$ would be $\mathbb{Z}_2$) $\endgroup$
    – Emilio Ferrucci
    Aug 17, 2014 at 9:58
  • $\begingroup$ @EmilioFerrucci Ah, thanks! (What a dumb mistake...) As for the other questions, I would be glad if you could provide me with an answer, because I didn't find one on Hatcher (but I might have overlooked something). $\endgroup$
    – Daniel Robert-Nicoud
    Aug 17, 2014 at 10:30

1 Answer 1

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As it came out in the comments, your doubt about cellular cohomology of $\mathbb{RP}^n$ not being isomorphic to singular cohomology was because you switched the even and odd cases in $\partial$, as seen on Hatcher (current online edition) p.144.

In fact, cellular (co)homology is always isomorphic to singular (co)homology for CW complexes: Hatcher is again a good reference for this, see pages 139 for homology and 203 for cohomology. The proof for homology doesn't involve chain maps, quasi isomorphisms or chain homotopies, rather it hinges on an ad hoc diagram chasing argument, in which one uses the fact that the chain groups of the cellular chain complex are the relative singular homology gruops $H_i(X^i,X^{i-1})$. With $\mathbb{Z}$ coefficients it is true that there is a quasi isomorphism (indeed a chain homotopy) of the two complexes, since it os a general fact that for complexes of abelian groups, isomorphic homology is a sufficient condition for there to exist a chain homotopy (see here). I'm not sure whether this holds for some larger class of modules (like modules over PIDs), but for chain complexes of modules over a generic ring this is false in general, and I'm not sure if it turns out to be true in the case of singular and cellular complexes of CW complex (this is what the OP seems to be asking here).

Again, the standard proof of the equivalence of singular or cellular cohomology doesn't operate with maps at the level of chain complexes, and is similar to the proof for homology, to the extent that the bulk of the argument is just handled by using the universal coefficient theorem and the homological case, if I remember correctly. Hatcher not only shows that $H^\bullet(X;G) \simeq H^\bullet_{CW}(X;G)$, but also that the cellular cochain complex is the dual complex of the cellular chain complex, which isn't evident from the definition.

I hope all of this answers your questions. I could add details, but Hatcher is really a very good reference for all of this.

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