# Why Cohomology Groups?

Why do we need cohomology groups? Homology groups are easier to compute and given two topological spaces, there is an isomorphism in homology groups if and only if there is an isomorphism in cohomology groups. So why do I need them?

• Well say if you have some space $X$ that you know all you want to know about and some construction on another space $Y$ that gives a homotopy class of maps $f:X \to Y$. If you want to study this construction, its natural to look at the image of fixed homology classes in $H_*(X)$ under $f_*$. But then what do you do if you instead get a map $Y\to X$? A concrete place where this happens is the study of vector bundles using characteristic classes. – PVAL-inactive Jan 21 '16 at 18:34
• math.stackexchange.com/questions/39707/… – Cheerful Parsnip Jan 21 '16 at 18:35
• Homology groups are not always easier to compute. In fact, in certain cases one uses duality in order to compute homology groups in terms of known cohomology groups – H1ghfiv3 Jan 21 '16 at 18:39
• (1) Cohomology comes equipped with a natural cup structure; (2) Poincare duality naturally involves cohomology; (3) cohomology is naturally contravariant with rather than covariant, which makes more sense in many cases. – anomaly Jan 21 '16 at 18:44
• A different reason is that cohomology is "dual" to homotopy. That is, $H^n(X; G) \cong [X, K(G, n)]$ where $[X, Y]$ is the set of based homotopy classes of maps $X \to Y$, which has a natural group structure with $Y = K(G, n)$ since $K(G, n)$ is an homotopy-associative, homotopy-commutative $H$-space with a homotopy-inverse. – Balarka Sen Jan 21 '16 at 18:47

Fundamental reason why cohomology is more powerful is that singular cohomology has a natural ring structure over them. If $\varphi$ and $\psi$ are $R$-valued $m$ and $n$-cochains on a topological space $X$, then one can define an $(m+n)$ cochain $\varphi \smile \psi$ by

$$(\varphi \smile \psi)([v_0, \cdots, v_{m+n}]) = \varphi([v_0, \cdots, v_m]) \psi([v_m, \cdots, v_{m+n}])$$

This pairing descends down to a pairing $H^n(X;R) \times H^m(X; R) \to H^{n+m}(X;R)$ on cohomology. Thus, $H^*(X;R) := \bigoplus_i H^i(X;R)$ automatically has a graded ring structure.

The intuitive reason is homology is about equivalence classes of chains on $X$, whereas cohomology is about equivalence classes of ring valued functions over chains in $X$. Functions can be multiplied (multiply the values pointwise) whereas simplices can't.

This ring structure is a powerful invariant. For example, $\Bbb{CP}^2$ and $S^2 \vee S^4$ have the same (co)homology groups in all dimensions. However, they are not homotopy equivalent as the integral cohomology rings differ: Cup square of generator of $H^2$ in $H^*(S^2 \vee S^4) \cong H^*(S^2) \oplus H^*(S^4)$ is zero, while cup square of generator of $H^2$ in $H^*(\Bbb{CP}^2)$ is nontrivial.

(Note that this also proves that the Hopf map is not nullhomotopic, thus proving $\pi_3(S^2)$ is nontrivial. Indeed, a generalization of this idea is the Hopf invariant.)

• Homology has a coproduct, and in good situations that structure gives exactly the same information as the cup product. – Mariano Suárez-Álvarez Jan 26 '16 at 12:36
• @MarianoSuárez-Alvarez Hmm, I have indeed heard of that, but I don't know much about it. Can you give me a reference for that? (+1 on both your comment and answer) – Balarka Sen Jan 26 '16 at 12:41
• Mariano is right. However, the fact that the coproduct on cohomology comes really much more natural is important here. – H1ghfiv3 Feb 2 '16 at 14:12

Quote from Elements of Algebraic Topology by J. R. Munkres:

One answer is that cohomology appers naturally when one studies the problem of classifying, up to homotopy, maps of one space in other. Another is that cohomology is involved when one integrates differential forms on manifolds. Still another answer is [...] that the cohomology groups have an additional algebraic structure -- that of a ring -- and that this ring will distinguish between spaces when the groups themselves not.

Others have given the first reasons that arose historically, and that relate to a first course in algebraic topology. But a further point comes up from a more advanced perspective: sheaf cohomology has a natural definition using injective resolutions, and these exist in any category of sheaves. You can define sheaf homology by projective resolutions but these do not always exist. See https://mathoverflow.net/questions/5378/when-are-there-enough-projective-sheaves-on-a-space-x

So étale cohomology and other derived functor cohomologies are more natural than the corresponding homologies.

• Quote:" these [injectives] exist in any category of sheaves". The category of coherent sheaves has not enough injectives, but fortunately the category of quasi-coherent sheaves has enough injectives. +1! – User3773 Jan 22 '16 at 16:44
• Yes, @Cla points to an important example showing not every category made of sheaves has enough injectives. But for any topological space, or even for any Grothendieck topology, the category of sheaves on that has enough injectives. – Colin McLarty Jan 23 '16 at 1:26

Cohomology appears naturaly. Lots of problems are solved locally and then the local solutions glued together to construct global solutions, and cohomology is the precise description of how to do that.

• Maybe you can elaborate a bit about the local-global problem and cohomology? I am interested. – Balarka Sen Jan 26 '16 at 12:49
• push. teach us! – syzygy Jun 26 '16 at 0:34
• I make the same request as Balarka Sen. – rmdmc89 Jul 17 '18 at 19:20