I've been studying algebraic topology for over half a year now and came across alot of different topics of it (fundamental groups, Van Kampen, singular homology, homology theory, Mayer Vietoris, universal coefficient theorem, knot theory etc.) and recently we started to study cohomolgy in our lecture.

We defined cohomology, proved the universal coefficient theorem and with that we were able to prove quite alot of analogous results which we already proved for homology.

My questions:

Why do we want to study cohomology? Are there some advantages of computing the cohomology of a given topological space compared to computing its homology group? Are there some really suprising/fascinating results which heavily depend on cohomological properties/theorems?


3 Answers 3


A major issue is the multiplicative structure that is around in cohomology. This allows you to distinguish spaces, which have the same homology. As an example, the $X:=\mathbb CP^2$ and $Y:=S^2\vee S^4$. Then both $X$ and $Y$ are CW-complexes with one cell in dimensions $0$, $2$, and $4$. Hence in both cases the homology with integral coefficients is $\mathbb Z$ in degrees $0$, $2$, and $4$ and $0$ in all other dimensions. This readily implies that also the cohomology groups with integral coefficients are $\mathbb Z$ in degrees $0$, $2$, and $4$ and $0$ in all other dimensions. However, for $X$ the square of a generator of $H^2$ is a generator for $H^4$, whereas for $Y$ this square is zero. (The result for $X$ follows easily, for example, from the fact that the square of the Kähler form on $\mathbb CP^2$ is a volume form, for $Y$ it is kind of obvious that there will be no relation between $H^2$ and $H^4$.)

This shows that $\mathbb CP^2$ is not homotopy equivalent to $S^2\vee S^4$, which in turn implies that the two attaching maps $S^3\to S^2$ used for the two spaces cannot be homotopic, i.e. that the Hopf-fibration is not null-homotopic.

Another issue is that in some situation the fact that cohomology is a contravariant functor is extremely useful. For example, for a topological group $G$ there is a classifying space $BG$ which carries a principal $G$-bundle $EG\to BG$. This has the property that for any sufficently nice space $X$ any principal bundle over $X$ can be written as a pullback $f^*EG$ and $f^*EG\cong g^*EG$ if and only if $f$ and $g$ are homotopic. So a principal bundle gives you a classifying map $f:X\to BG$. Using this map, one can now pull back chomology classes from $BG$ to cohomology classes on $X$. Theses are canonically associated to the bundle, since homotopic maps induce the same pullback in cohomology. This is the topological version of the theory of characteristic classes and it would not work out with homology.

  • 1
    $\begingroup$ I'll just point out that although homology has the corresponding dual structure to some extent (coalgebra), it has problems when working over a ring that isn't a field. But, the cup product always works! Also, due to our historical bias, algebra is easier than coalgebra. $\endgroup$ Apr 13, 2016 at 11:46
  • $\begingroup$ I just want to add that another possibility to see why a generator of $H^{2}\left( \mathbb{C}P^2\right)$ does not square to zero is by using the non-singularity of the cup product pairing which involves Poincaré Duality. $\endgroup$
    – noctusraid
    Aug 21, 2016 at 13:40

There are several elementary points that can be made :

  • On the other hand, why consider only homology ? I don't see why one would be more natural than the other (actually for me cohomology seems more natural because I'm used to subjects where cohomology natuarally appears).
  • There are cohomology theories that are clearly useful and natural and are related to singular cohomology : de Rham cohomology, group cohomology and Galois cohomology, sheaf cohomology for instance. They often appear when you are interested in deriving a left exact functor and not a right exact one, which... happens.
  • There are duality theorems (all kind of variants of Poincaré duality) that make use of cohomology, so even if ultimately you are interested in homology, studying cohomology can be useful.
  • Cohomology natually carries a sort of algebra structure given by the cup product, which is really helpful in a lot of situations.

I'll give an explicit example coming from algebra since this is what I understand best. Take $G$ a finite group, and $X = K(G,1)$ the corresponding Eilenberg-MacLane space (so $\pi_1(X)=G$ and $\pi_i(X)=0$ if $i>1$). Then you can write $H_n(G,A) := H_n(X,A)$ and $H^n(G,A):= H^n(X,A)$ for any abelian group $A$.

This is a special case of group homology/cohomology (namely the case where $G$ acts trivially on $A$). And though group homology is quite useful (for instance Hurewitz's theorem says that $H_1(G,\mathbb{Z}) = G^{ab}$), group cohomology appears way more often, so singular cohomology in this context is more useful.

Now if you don't care about groups you may not be happy with that, but even if you're only interested in topology, clearly $X = K(G,1)$ is an interesting space since it's "the space that only has $\pi_1(X) = G$ in its homotopy". So understanding maps $Y\to X$ is clearly a natural question.

And if $G$ is abelian you have the result that for any CW-complex $Y$, $[Y,K(G,n)] \simeq H^n(Y,G)$, so singular cohomology (with coefficients) classifies maps to Elenberg-MacLane spaces (up to homotopy).

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    $\begingroup$ While I fully agree with this answer, I think it might be worth mentioning that the concern of the OP is natural when coming from topology, where cohomology seems to be much less natural than homology (unlike when coming from a more algebraic perspective where the two are practically interchangeable). $\endgroup$ Apr 12, 2016 at 10:30
  • $\begingroup$ @TobiasKildetoft you're right, it seems unnatural. I have the feeling I didn't gain much by learning cohomology yet from the topological point of view. So to answer Captain Lama's question 'On the other hand, why consider only homology ?' I'd say: 'Because I don't know anything cohomology is capable of which homology is not' $\endgroup$
    – noctusraid
    Apr 12, 2016 at 14:02
  • $\begingroup$ I tried to put in a convincing example. $\endgroup$ Apr 12, 2016 at 14:29
  • $\begingroup$ Studying maps into Eilenberg MacLane spaces (aka cohomology classes), leads to the Postnikov decomposition, which is a sort of a dual to a cellular decomposition, reversing the point of view. This connects to things like obstruction theory, in which cohomology plays a key role. What could be more topological than trying to lift maps? $\endgroup$ Apr 13, 2016 at 15:33

a) Cohomology should be regarded as a graded-commutative algebra, therefore is a finer invariant. This allows for some quite interesting definitions, e.g. the Hopf invariant.

b) If $M$ is a manifold of suitable dimension, then one has $b_{n-k}=b_k$ for the Betti numbers (proved via Poincare duality, which involves cohomology). It is sometimes easier to compute cohomology and use duality Theorems to deduce information about homology.

c) Characteristic classes

d) The de Rham theorem, establishing links with differential geometric cohomology.

e) Cohomology is a representable functor.

f) (related to c) Intersection theory.

g) cohomology sounds more sophisticated than homology, therefore allowing for better bragging among the commoners

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    $\begingroup$ +1 for g)! Meanwhile I've encountered the Poincare duality and the cup product structure, these were quite convincing. $\endgroup$
    – noctusraid
    May 24, 2016 at 15:03
  • $\begingroup$ also: you can prove Serre's Theorem concerning homotopy groups of sphere using the Cup product structure of the Serre spectral sequence $\endgroup$
    – Juan Fran
    May 24, 2016 at 15:12

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