# Why homology with coefficients?

I am currently studying a bit of homology theory (on topological spaces). Let $H_n(X)$ denote the singular homology groups of the topological space $X$, then as you know we can define the singular homology with coefficients in the abelian group $G$ by $$H_n(X;G)=H_n(S_*(X)\otimes G)$$ Now, I understand the definitions and computations, proofs etc., but I would like to understand the following:

1. Why is this interesting to study? Does it make computations easier somewhere (for some groups $G$)? Does it allow us to prove interesting theorems we can't prove using $G=\mathbb{Z}$?
2. What are interesting examples of use of homology with coefficients?
3. Is the theory of "degree modulo $2$" (treated e.g. in Milnor's book Topology from the differentiable viewpoint) in truth just degree theory on $H_n(X;\mathbb{Z}_2)$?
• First of all you want to be careful about what is meant by homology with coefficients. It is not true that $H_n(X,\Bbb{Z}) \otimes G$ is the same as first tensoring your chain complex and then calculating the homology there. There is an obstruction given by a Tor term in universal coefficients! – user38268 Dec 27 '13 at 13:23
• @Benja Ah, I didn't know it. I removed the bracketed sentence. Thanks for the comment, would you care to give me a reference on the subject? – Daniel Robert-Nicoud Dec 27 '13 at 14:48
• See my answer math.stackexchange.com/questions/47100/… for some examples of why $\mathbb Z_2$ coefficients are useful. – Cheerful Parsnip Dec 27 '13 at 18:11
• @DanielRobert-Nicoud Look at en.wikipedia.org/wiki/Universal_coefficient_theorem. If you're not sure about the Tor thing, you can learn about it by reading some posts on this site. It's just the left derived functor of the tensor product. – user38268 Dec 28 '13 at 10:02

You can find the construction of homology with general coefficients and the universal coefficient theorem in Hatcher's Algebraic Topology, which is available free from his website.

The answer to the second part of your first question is yes, especially in the case that we take $G$ to be a field, most often finite or $\mathbb{Q}$, or $\mathbb{R}$ in differential topology. Homology over a field is simple because $\operatorname{Tor}$ always vanishes, so you get e.g. an exact duality between homology and cohomology. Homology with $\mathbb{Z}_2$ coefficients is also the appropriate theory for many questions about non-orientable manifolds-their top $\Bbb{Z}$-homology is zero, but their top $\Bbb{Z}_2$ homology is $\Bbb{Z}_2$, which leads to the degree theory in Milnor you were mentioning.

Cohomology with more general coefficients than $\mathbb{Z}$ is even more useful than homology. For instance it leads to the result that if a manifold $M$ has any Betti number $b_i(M)<b_i(N)$, where $b_i$ is the rank of the free part of $H_i$, there's no map $M\to N$ of non-zero degree. This has lots of quick corollaries-for instance, there's no surjection of $S^n$ onto any $n$-manifold with nontrivial lower homology! Edit: This is obviously false, and I no longer have any idea whether I meant anything true.

But in the end $H_*(X;G)$ is more of a stepping stone than anything else; it gets you thinking about how much variety there could be in theories satisfying the axioms of homology. It turns out there's almost none-singular homology with coefficients in $G$ is the only example-but if we rid ourselves of the "dimension axiom" $$H_*(\star)=\left\{\begin{matrix}\mathbb{Z},*=0\\0,*>0\end{matrix}\right.$$ then we get a vast collection of "generalized (co)homology theories," beginning with K-theory, cobordism, and stable homotopy, which really do contain new information. In some cases, so much new information that we can't actually compute them yet!

• I don't understand your point about $S^n$ not surjecting onto an $n$-manifold with non-trivial lower homology. For example, here is a smooth surjection from $S^2$ to $S^1\times S^1$: First, the projection map $S^2\rightarrow D^2$ to a disc, given by $(x,y,z)\mapsto(x,y)$ is definitely smooth. Now, a small modification of the usual square-with-edges-identified picture gives a smooth surjection from $D^2$ to $S^1\times S^1$. (Of course, as you said, the degree will be $0$, but isn't this still a surjective smooth map?) (And sorry to be 3.5 years late!) – Jason DeVito Apr 25 '17 at 23:22
• @JasonDeVito I also do not understand Kevin's point! – Kevin Arlin Apr 26 '17 at 5:49
• I kept thinking about this a bit more, and I think the following is true: For given smooth manifolds $M$ and $N$ of dimension $m$ and $n$ respectively, assume $N$ is connected and $m\geq n$. Then there is a smooth surjection $M\rightarrow N$. Sketch: Map $M$ to $S^{m}$ by collapsing everything outside a small ball to a point, map $S^m$ to $D^m$ as in the previous comment. If $m > n$, project $D^m$ down to $D^n$ by just forgettnig coordinates. Equip $N$ with an auxillary Riemannian metric and imagine $D^n$ is a large ball in $T_p N$. Exponentiating gives a surjection by Hopf-Rinow. – Jason DeVito Apr 26 '17 at 17:06

First of all, on the level of complexes $C_n(X;A)=C_n(X)\otimes A$, but (in general) $H_n(X;A)\ne H_n(X)\otimes A$. Nevertheless, if you know ordinary homology you can compute homology with any coefficients — so these groups don't contain any really new information (see 'universal coefficient theorem' for details).

Main reason to consider (co)homology with non-trivial coefficients, I believe, that sometimes these groups (and extra structure on these groups) are easier to compute.

For example, when you compute $H_n(X;\mathbb Z/2)$ you can forget about all problems with signs etc (a concrete example: it's quite easy to compute (co)homology of $SO(n)$ with coefficients $\mathbb Z/2$ but very hard even to describe answer for ordinary (co)homology group); when you compute homology with rational coefficients, you can forget about all torsion-related phenomena.

Somewhat more advanced example: the algebra of stable cohomological operations mod p (see 'Steenrod squares' / 'Steenrod powers' ) is much easier to describe than the algebra of integral cohomological operations. (All this reminds slightly situation in number theory, where many problems are easier to solve mod p or maybe rationally than in integers.)

Also some constructions (obstructions, characteristic classes...) naturally 'live' in (co)homology with non-trivial coefficients (see e.g. 'Stiefel–Whitney classes').

Finally, (co)homology group with coefficients in a field with char 0 have different analytical description ('de Rham cohomology' etc) and (at least in some situations) some extra structue (e.g. 'Hodge structures' in cohomology of projective complex manifolds).

You already probably know the advantages of thinking of homology as being a functor of $X$. By allowing an abelian group $G$ as coefficients, it becomes a functor of two variables, the space $X$ and the group $G$. Now you can exploit the $G$-functoriality as well.

For example, the short exact sequence of groups $$0 \to \mathbb Z \to \mathbb Z \to \mathbb Z/p\mathbb Z \to 0$$ (the second arrow is mult. by $p$) gives rise to the short exact sequence of cohomology groups $$0 \to H^i(X,\mathbb Z)/p H^i(X,\mathbb Z) \to H^i(X,\mathbb Z/p\mathbb Z) \to p\text{-torsion subgroup of } H^{i+1}(X,\mathbb Z) \to 0.$$ (Here I have switched from homology to cohomology, since I find that latter a little easier to think about, and easier to state these sorts of results about.)

So studying cohomology with coefficiens in $\mathbb Z/p\mathbb Z$ is closely related to studying torsion in cohomology with $\mathbb Z$-coefficients. If $p$ is a prime (which is the case I was imagining, although the above is valid for any integer $p$), then $\mathbb Z/p\mathbb Z$ is a field, and so we can investigate torsion in cohomology by studying cohomology with coefficients in a (finite) field. The latter investigation has technical advantages, e.g. Poincare duality if $X$ is a manifold.

Thinking of cohomology as a functor of $G$ as well as of $X$ leads to the idea of sheaf cohomology (a sheaf of abelian groups is something like an abelian group that can vary from point to point over $X$). The possibility of taking $G$ to be a non-constant sheaf greatly increases the flexibility of cohomology as a tool.

Another example of utility of coefficients:

If $E \to B$ is a fibre bundle with fibre $F$, and $B$ is connected and simply connected, then there is a spectral sequence (the Leray spectral sequence) $$E_2^{i,j} := H^i(B,H^j(F,G) ) \implies H^{i+j}(E,G),$$ relating the cohomology of the base and fibre to that of the total space.

Note that even if we take $G$ to be $\mathbb Z$, the cohomology of the base has coefficients in the cohomology groups $H^j(F,\mathbb Z)$, which are likely not just equal to $\mathbb Z$.

If $B$ is not simply connected, then there is a similar spectral sequence, but the coefficients are a locally constant, but typically non-constant, sheaf on $B$.

• Thanks for your answer. I an familiar with the basics of category theory, but (unfortunately?) the course of algebraic topology I took didn't delve too much into that point of view. Maybe in the second part of the course... Anyway, do you have a reference for a book looking at algebraic topology from a categorical setting? – Daniel Robert-Nicoud Dec 28 '13 at 0:30
• @DanielRobert-Nicoud: Dear Daniel, Any alg. top. textbook should have some functorial/categorical discussion (e.g. because one of the important tools in applying (co)homology is not just the groups themselves, but the maps between them induced by maps between spaces). I wouldn't suggest going out of your way to learn more category theory; this is (in my opinion) a subject best learnt largely by osmosis as you see various categorical notions appearing in different contexts. What I would suggest is finding examples of spaces with torsion in their (co)homology, and then computing with ... – Matt E Dec 28 '13 at 1:59
• ... various coefficients and seeing how the answers are related to one another. (E.g. see concretely how the $\mathbb F_p$-Betti number differs from the $\mathbb Q$-Betti number when there is $p$-torsion in the homology.) As you learn more topology/geometry, your understanding of the role of (co)homology will mature, as will your understanding of the role of different choices of coefficients. Regards, – Matt E Dec 28 '13 at 2:01