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• CommentRowNumber1.
• CommentAuthorrdkw10
• CommentTimeApr 23rd 2010

I know that in singular chains $C_n(X)$ are free abelian groups generated by n-simplicies. Also, the singular cohomology has cochains $C^n(X;\mathbb{Z}) = Hom(C_n(X),\mathbb{Z})$. However, does this imply that the singular cochains $C^n(X;\mathbb{Z})$ are also free?

1. no: the direct product of an infinite number of copies of $\mathbb{Z}$ is not a free $\mathbb{Z}$-module.

• CommentRowNumber3.
• CommentAuthorrdkw10
• CommentTimeApr 23rd 2010

What is the identity element in $Hom(C_n(X),\mathbb{Z})$? Is it the hom. which sends all elements in $C_n(X)$ to $0\in\mathbb{z}$?

• CommentRowNumber4.
• CommentAuthorUrs
• CommentTimeApr 23rd 2010

At points like this it is worth to remember that fundamentally cohomology is not so much dual to homology as to homotopy: it’s all about maps out of or into objects.

• CommentRowNumber5.
• CommentAuthorUrs
• CommentTimeApr 23rd 2010
• (edited Apr 23rd 2010)

What is the identity element in $Hom(C_n(X),\mathbb{Z})$? Is it the hom. which sends all elements in $C_n(X)$ to $0\in\mathbb{z}$?

Yes, that’s the trivial cocycle.

Generally, cohomology is homotopy classes of maps from one object $X$ to another object $A$. And if that object $A$ has a singled out point $* \to A$, then the map that factors through this point $X \to * \to A$ is the trivial cocycle. If $A$ is a group object and the point “the” neutral element, then this trivial cocycle is also the neutral element in the cohomology group.

In the case in question the objects are simplicial sets, $X = Sing Y$ is the singular simplicial complex of some topological space $Y$, and the coefficient object is the Eilenberg-MacLane space $\mathcal{B}^n \mathbb{Z} = K(\mathbb{Z},n)$ regarded as a simplicial set.

• CommentRowNumber6.
• CommentAuthorUrs
• CommentTimeApr 23rd 2010

• CommentRowNumber7.
• CommentAuthorrdkw10
• CommentTimeApr 23rd 2010

Yes, I might as well explain my problem:

I want to use the exact sequence $0\rightarrow \mathbb{Z}\rightarrow \mathbb{R}\rightarrow \mathbb{R}/\mathbb{Z} \rightarrow 0$ to construct an exact sequence $0\rightarrow C^n(X,\mathbb{Z})\otimes\mathbb{Z}\rightarrow C^n(X,\mathbb{Z})\otimes\mathbb{R}\rightarrow C^n(X,\mathbb{Z})\otimes\mathbb{R}/\mathbb{Z} \rightarrow 0$, and I am stuck. I have seen how to do this with the Tor functor, but that requires $C^n(X;\mathbb{Z})$ to be free. At least as far as I could tell.

• CommentRowNumber8.
• CommentAuthorUrs
• CommentTimeApr 23rd 2010
• (edited Apr 23rd 2010)

This has a very general abstract kind of answer:

That “short” exact sequence is really to be thought of as part of a long fibration sequence

$\cdots \to \mathcal{B}^{n-1}\mathbb{R}/\mathbb{Z} \to \mathcal{B}^n \mathbb{Z} \to \mathcal{B}^n \mathbb{R} \to \mathcal{B}^n \mathbb{R}/\mathbb{Z} \to \mathcal{B}^{n+1} \mathbb{Z} \to \cdots \,.$

That this is a fibration sequence means that every pair of consecutive morphisms here fits into a homotopy pullback square

$\array{ \mathcal{B}^n \mathbb{Z} &\to& * \\ \downarrow && \downarrow \\ \mathcal{B}^n \mathbb{R} &\to& \mathcal{B}^n \mathbb{R}/\mathbb{Z} } \,.$

Now, forming cocycles in cohomology on any object $X$ is just forming the oo-categorical hom-object $\mathbf{H}(X,-)$. But as in ordinary category theory, the hom-functor preserves limits in the second argument, hence in particular preserves pullbacks. this means that applying $\mathbf{H}(X,-)$ to the above diagram yields the homotopy pullback diagram

$\array{ \mathbf{H}(X,\mathcal{B}^n \mathbb{Z}) &\to& * \\ \downarrow && \downarrow \\ \mathbf{H}(X,\mathcal{B}^n \mathbb{R}) &\to& \mathbf{H}(X,\mathcal{B}^n \mathbb{R}/\mathbb{Z}) }$

hence the fibration sequence

$\mathbf{H}(X,\mathcal{B}^n \mathbb{Z}) \to \mathbf{H}(X,\mathcal{B}^n \mathbb{R}) \to \mathbf{H}(X,\mathcal{B}^n \mathbb{R}/\mathbb{Z})$

Taking connected components of these hom-spaces yields the exact (in the middle) sequence of pointed sets

$\pi_0 \mathbf{H}(X,\mathcal{B}^n \mathbb{Z}) \to \pi_0 \mathbf{H}(X,\mathcal{B}^n \mathbb{R}) \to \pi_0 \mathbf{H}(X,\mathcal{B}^n \mathbb{R}/\mathbb{Z})$

which is the bit of the long exact sequence in cohomology that you are after

$\pi_0 \mathbf{H}(X,\mathbb{B}^n \mathbb{Z}) = H^n(X,\mathbb{Z}) \to \pi_0 \mathbf{H}(X,\mathbb{B}^n \mathbb{R}) = H^n(X,\mathbb{R}) \to \pi_0 \mathbf{H}(X,\mathbb{B}^n \mathbb{R}/\mathbb{Z}) = H^n(X,\mathbb{R}/\mathbb{Z})$
• CommentRowNumber9.
• CommentAuthorrdkw10
• CommentTimeApr 23rd 2010

Wow, thanks for the (generalized) answer. Although I must admit that I need several hours to understand it ;) . So, is what you are doing something like taking the short es $0\rightarrow \mathbb{Z}\rightarrow\mathbb{R}\rightarrow \mathbb{R}/\mathbb{Z}\rightarrow 0$ and forming the following s.e.s:

$0\rightarrow C_n(X)\otimes \mathbb{Z}\rightarrow C_n(X)\otimes\mathbb{R}\rightarrow C_n(X)\otimes\mathbb{R}/\mathbb{Z}\rightarrow 0$

which is possible since $C_n(X)$ is free. Then, taking the Hom of this to give

$0\rightarrow Hom(C_n(X),\mathbb{Z})\otimes \mathbb{Z}\rightarrow Hom(C_n(X),\mathbb{Z})\otimes\mathbb{R}\rightarrow Hom(C_n(X),\mathbb{Z})\otimes\mathbb{R}/\mathbb{Z}\rightarrow 0$,

which gives the desired result?

• CommentRowNumber10.
• CommentAuthorUrs
• CommentTimeApr 23rd 2010
• (edited Apr 23rd 2010)

Okay, the important point for what you want is that we have an adjunction

$(free \dashv forget) : sAb \stackrel{\overset{F}{\leftarrow}}{\underset{U}{\to}} sSet$

between abelian simplicial groups and simplicial sets, where $U$ is the forgetful functor that sends a simplicial group to its underlying Kan complex simplicial set, and where $F$ is the free simplicial group functor, that sends a simplical set $S$ to $\mathbb{Z}[S]$, the simplicial abelian group which in degree $n$ is the free abelian group on the set $S_n$.

The cohomology that you are looking at is that of the simplicial set $Sing X$ (the fundamental infinity-groupoid of the topological space $X$).

So in particular we have that

$\mathbb{Z}[Sing X]_n = C_n(X)$

are the $n$-chains.

The Eilenberg-MacLane object $\mathcal{B}^n \mathbb{Z}$ (for example) is a simplicial set with the structure of a simplicial abelian group, so for emphasis I write $U(\mathcal{B}^n \mathbb{Z})$.

\begin{aligned} H^n(X,\mathbb{Z}) & \simeq \pi_0 sSet(Sing X, U \mathcal{B}^n \mathbb{Z}) \\ & \pi_0 sAb( \mathbb{Z}[Sing X], \mathcal{B}^n \mathbb{Z} ) \end{aligned}

and then using the Dold-Kan correspondence

$\cdots \subset Hom_{Ab}(C_n(X), \mathbb{Z}) \,.$

I should explain this in more detail. But I am getting tired. In fact, what I really should do is type all this into some nLab entry. But large parts of this story you can find here and there. For instance at abelian sheaf cohomology.

• CommentRowNumber11.
• CommentAuthorrdkw10
• CommentTimeApr 24th 2010

I am still having a hard time following what you are saying. My experience in this area is only dealing with the standard introductory topics of singular chains, homology and singular cohomology. Is there anyway to address my question sticking to this? But still, I really appreciate you taking the time to explain this in a more general setting. Hopefully in some time I will be able to fully appreciate what exactly you are saying.

• CommentRowNumber12.
• CommentAuthorHarry Gindi
• CommentTimeApr 24th 2010

No, it doesn't. Take some space such that C_1 is the infinite direct sum of copies of Z. Dualizing, we flip this out to a direct product, but the direct product is not free, as domenico noted...

• CommentRowNumber13.
• CommentAuthorrdkw10
• CommentTimeApr 24th 2010

Thanks marry, I can follow what you are saying. Do you have any hints on the other problem though?

• CommentRowNumber14.
• CommentAuthorHarry Gindi
• CommentTimeApr 24th 2010

The nForum really isn't for homework help...

• CommentRowNumber15.
• CommentAuthorUrs
• CommentTimeApr 24th 2010

I have expanded the Definition-section of singular cohomology now.

I did this nPOV-style, of course. Maybe somebody wants to open a subsection “Pedestrian description” with more explicit descriptions. Me, personally, I have to go to bed now. :-)

• CommentRowNumber16.
• CommentAuthorrdkw10
• CommentTimeApr 24th 2010

Harry, this is NOT a homework problem. I have been reading through a couple of papers by Freed on TQFTs (as my other posts state) and getting help here (mainly from Urs) when problems arise. In the paper TQFTs from finite gauge group Freed says that “using the fact that $H^n(BG,\mathbb{R}) = 0$ for finite $G$ and the short exact sequence $0\rightarrow \mathbb{Z}\rightarrow \mathbb{R}\rightarrow\mathbb{R}/\mathbb{Z}\rightarrow 0$, we get that $H^n(BG,\mathbb{R}/\mathbb{Z})\cong H^{n+1}(BG,\mathbb{Z})$. Now, even though this is inconsequential in what follows in his paper, I still would like to derive it. I am Sorry if my question sounded like a hw problem.

• CommentRowNumber17.
• CommentAuthorzskoda
• CommentTimeApr 24th 2010
• (edited Apr 24th 2010)

Thank you for your activity. Are you a physicist or a mathematician ? I was just doing some bibliographies in nlab, so let me mention some books having practical topology in the context of QFT for physicists:

• R.S. Ward, R.O. Wells, Twistor geometry and field theory, Cambridge Univ. Press 1990.
• Charles Nash, Differential topology and quantum field theory 1991. (the book also builds up on a similar book on math background by Nash and Sen)
• Albert Schwarz, Quantum field theory and topology, Grundlehren der Math. Wissen. 307, Springer 1993.

and a good wider bibliography is compiled by Kapustin here.

• CommentRowNumber18.
• CommentAuthorrdkw10
• CommentTimeApr 24th 2010

Dear Zskoda,

I am studying physics mainly, but also several courses in mathematics.

• CommentRowNumber19.
• CommentAuthorHarry Gindi
• CommentTimeApr 24th 2010
• (edited Apr 24th 2010)

rdkw, any chance you could move this topic to the General Discussion -> Mathematics, physics, and philosophy section? The nForum category is mainly reserved for meta-stuff about the nForum itself.

Harry, this is NOT a homework problem.

Sorry, I just figured that it might be since it looks like a homework problem from a first semester algebraic topology course, and with finals coming up...

• CommentRowNumber20.
• CommentAuthordomenico_fiorenza
• CommentTimeApr 24th 2010
• (edited Apr 24th 2010)

if $G$ is finite, then the nerve of $G$ is finite in each dimension, and so $C_n(BG,\mathbb{Z})$ is a finite rank free$\mathbb{Z}$ module. Hence, also $C^n(BG,\mathbb{Z})$ is free, and your original argument applies.

(I must admit that for once I agree with the biting comments by Harry: me too was sure rdkw10 question was an homological algebra/algebraic topology exercise: there’s no indication of the question being relevant to a going-on study of Freed’s papers, and having stated the precise problem in the original question would have allowed for an immediate relevant answer. So I found quite impolite crying out loud this was NOT an homework problem. I usually don’t like Harry direct style, which seldom makes me uncomfortable (I’m the kind of person who avoids discussions as much as possible), but I agree with him this time.)

2. as far as concerns the topic classification, this surely should not be n-Forum. however also General Discussion -> Mathematics, physics, and philosophy does not convince me, I’d like that place for more general discussions. what about creating a new subcategory General Discussion -> Questions & answers, with tag “This is not a place for having homeworks done!”?

clearly a discussion could start in Questions & answers, and then become of more general interest (or vice versa), but this is not a problem at all since we can move discussions from a category to another.

• CommentRowNumber22.
• CommentAuthorrdkw10
• CommentTimeApr 24th 2010

Dear Domenico,

Thanks for clearing things up! However, I must admit that I am a little confused with you response afterwords. This isom. is relevant to the article of Freed, it is the whole reason why we can consider exponentiated actions. Also, I did not mean to sound “quite impolite crying out loud this was NOT an homework problem”, I mearly wanted to give Harry a place where he could see exactly what I was talking about (so he, and anyone else concerned, could see that this came from a paper). I can understand were you are coming from about not wanting to solve hw problems, I agree.

3. Dear rdkw10,

Sorry for having been impolite. I by no means wanted to say that your question is not relevant to understanding Freed’s statement, but to point out that your post #1 above actually had no reference to this, and so looked like an exercise of the kind you can find in a book on algebraic topology after the definition of singular cochains. I sincerely apologize again for having been rude and having written things in a way could be misunderstood.

I’m glad you found the math content of my comment useful :)

• CommentRowNumber24.
• CommentAuthorrdkw10
• CommentTimeApr 24th 2010

Dear Domenico,

Yes, I agree totally. I should have given a little background material first. I apologize for any confusion that might have caused, and in the future I will definitely explain why I am asking a question if it could be interpreted as a hw problem. Thank you again for your help.

• CommentRowNumber25.
• CommentAuthorUrs
• CommentTimeApr 24th 2010
• (edited Apr 24th 2010)

I have added now to fiber sequence a new Examples-section Integral versus real cohomology which is about the general role that the fiber sequence $\mathbf{B}^n \mathbb{Z} \to \mathbf{B}^n \mathbb{R} \to \mathbf{B}^n \mathbb{R}/\mathbb{Z}$ plays in cohomology.

• CommentRowNumber26.
• CommentAuthorrdkw10
• CommentTimeApr 24th 2010

Great, thanks! I will go and look.

• CommentRowNumber27.
• CommentAuthorDavidRoberts
• CommentTimeApr 25th 2010

NB: the last line of the section you (Urs) linked to has $Z/R$ as coeffs instead of $R/Z$.

@rdkw10 (pls ask questions if the following is too brief - it will help fill out the nlab)

once you have the long exact sequence in cohomology (and here we are I think using sheaf cohomology - but the technical details are not important for a first reading - it’s just that the coeff. groups are topological, so ordinary cohomology doesn’t work), then the coeffs. $\mathbb{R}$ are what is called a soft sheaf: they are like the dual of a space with no interesting cohomology with any coeffs. (by ’interesting’ I mean away from dimension zero), in that all cohomology of paracompact spaces (e.g. fin. dim. manifolds, and I’m pretty sure the space $BG$ at hand) with $\mathbb{R}$ as coefficients is trivial in nonzero dimensions. Then every third term in the long exact sequence is zero, and this implies the other terms pair up to be isomorphic. To get the long exact sequence is a major result, but there are lots of places to read about this: Hatcher’s book Algebraic Topology (free!) is one and I presume you have that. If this is the point on which you need more help then I can supply pedestrian details.

• CommentRowNumber28.
• CommentAuthorHarry Gindi
• CommentTimeApr 26th 2010
• (edited Apr 26th 2010)

There's a trick to getting the long exact sequence on cohomology from an SES on (co)chain complexes which allows you to reduce to two much easier lemmas.

Apply the sharp 3x3 lemma after reducing the complex to four degrees to show that you have left exact sequences (0->A->B->C is exact) on kernels (of the differentials) and right exact sequences on cokernels (of differentials). Then look at the induced maps of cokernels into kernels. The kernels and cokernels of these maps will be the (forgive my perhaps inverted indexing) nth and n+1st cohomology groups. The result follows by application of the snake lemma. This is true for any exact sequence of chain complexes taking values in an abelian category.

The direct proof in Hatcher is less enlightening than this proof in my opinion. If you'd like to see the conceptual proof for the snake lemma (the 3x3 lemma is handled as a part of the proof of the snake lemma), here is a proof of it for an abelian category working directly from the axioms (it is also much more conceptual):

http://www.math.ubc.ca/~jonathan/snake.pdf

Also note that where I said to use the 3x3 lemma, you could use the snake lemma and forget about the snake map.

• CommentRowNumber29.
• CommentAuthorUrs
• CommentTimeApr 26th 2010

and here we are I think using sheaf cohomology

Oh, are we? I thought the question was about singular cohomology. But It would certainly be good to discuss this for sheaf cohomology in the entry fiber sequence, too.

• CommentRowNumber30.
• CommentAuthorDavidRoberts
• CommentTimeApr 26th 2010
• (edited Apr 26th 2010)

Is singular cohomology with topological group coefficients ok? I know it’s not ok for some class of spaces (those with not enough maps into them by simplices), but if we are working with manifolds, then the damage isn’t too bad, we can use the isomorphism to sheaf cohomology (I don’t know a direct proof of the triviality of $H^k(M,\mathbb{R})$ for $k\geq 1$ for singular cohomology, but this is a poor excuse)

• CommentRowNumber31.
• CommentAuthorUrs
• CommentTimeApr 26th 2010
• (edited Apr 26th 2010)

Is singular cohomology with topological group coefficients ok?

What does “ok” mean? I just thought the question was about singular cohomology. Hence the title of this thread! :-)

But I am happy to discuss sheaf cohomology…

• CommentRowNumber32.
• CommentAuthorDavidRoberts
• CommentTimeApr 26th 2010
• (edited Apr 27th 2010)

I mean, behaves like it should: $H^1(M,G)$ classifies bundles etc. I was just addressing the part of the question I know. If we discuss sheaf cohomology let’s do it in another thread.

• CommentRowNumber33.
• CommentAuthorUrs
• CommentTimeApr 26th 2010

I think you have a point. Possibly the question wasn’t actually meant to be about singular cohomology, but was meant to be in a context where $\mathbb{R}$-cohomology is always trivial.