Why are we interested in cohomology?

Solution 1:

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.

Solution 2:

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).