Algebraic Topology – Difference Between Homology and Cohomology

algebraic-topology

Homology and cohomology are similar because the latter is the former acted by $\text{hom}$ functor, and we also have

Theorem Let $C$ and $D$ be free chain complexes; let $\phi:C\to D$ be a chain map. If $\phi$ induces homology isomorphisms in all dimensions, then $\phi$ induces cohomology isomorphism in all dimensions.

That is to say, if we can not use homology to distinguish two space, so do cohomology.

However my teacher told us that the essential difference between them is the ring structure on the cohomology. So my questions are

  • Can we introduce a cup product on homology? If not, why?

  • Why is the ring structure essential difference between them?

Any advice is helpful. Thx.

Best Answer

Let $X$ be a set and $k$ a field. A toy model for homology is the free $k$-vector space $k[X]$ on $X$, while a toy model for cohomology is the $k$-vector space $k^X$ of functions $X \to k$. (In fact these are the homology and the cohomology, respectively, of $X$ regarded as a discrete space.) For example:

  • $k[X]$ is covariant with respect to maps of sets, while $k^X$ is contravariant.
  • $k^X$ is the vector space dual of $k[X]$ (analogous to universal coefficients for cohomology).
  • We can identify $k^X$ and $k[X]$ when $X$ is finite but not in general (analogous to Poincare duality).
  • $k^X$ has an algebra structure while $k[X]$ instead has a coalgebra structure.

Thinking about the case where $X$ is an infinite set is particularly enlightening. There you see that the basic difference is that homology classes are "compactly supported": they live on finite subsets of $X$, while cohomology classes can live on all of $X$. This is an important intuition for understanding the statement of Poincare duality on a not-necessarily-compact manifold.

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