Intuition behind cohomology operations

I try to unterstand cohomology operations, but I cannot get the intuition behind it. Could someone explain the intuition behind it?

My background: I have a basic understanding of homology and cohomology. I am a beginner in the subject of cohomology operations and read Mosher's "Cohomology Operations".

• Poincaré duality is a helpful tool for understand cup product on compact orientable manifold.
– user171326
Apr 5, 2017 at 14:09

You could start with Steenrod's classic article "Cohomology Operations, and Obstructions to Extending Continuous Functions* which was republished after his death. And then do a web search on the topic.

• Thank you for the source. Sternrod's article is really recommendable! After some search I can also recommend the blog amathew.wordpress.com/2011/11/16/… Apr 10, 2017 at 5:41

An intuition behind the cup product in cohomology is that is the map induced by the diagonal maps $\Delta_X \colon X \to X \times X$. These maps are natural with respect to maps $X \to Y$ and the corresponding $X \times X \to Y \times Y$, hence this product is functorial.

Now consider $X^p$, the cartesian product of $X$ with itself $p$ times, where $p$ is a prime. There is a map

$$T_X \colon X^p \to X^p$$

$$(x_1,\ldots,x_p) \mapsto (x_2,\ldots,x_p,x_1)$$

which is also natural with respect to maps $X \to Y$. The corresponding maps $X^p \to Y^p$ would give you a commutative square. The existence of the Steenrod powers, which are cohomology operations for cohomology with coefficients in $\mathbb{Z}/p$, comes from these maps.

This is just an intuition, the real details are more intricate and technical and use the smash product instead of the Cartesian product. But I believe it makes more sense when you think about it in terms of Eilenberg-MacLane spaces.

There is a natural bijection

$$H^n(X;\mathbb{Z}/p) \cong [X,K(\mathbb{Z}/p,n)]$$

where the right side denotes homotopy classes of maps $X \to K(\mathbb{Z}/p,n)$, then you can think of cohomology operations (for cohomology with coefficients in $\mathbb{Z}/p$) in terms of maps $$K(\mathbb{Z}/p,n) \to K(\mathbb{Z}/p,m)$$

Such a map would give you a way of getting an element of the $m$th cohomology of $X$ from an element of the $n$th cohomology of $X$ by composition. But we have

$$[ K(\mathbb{Z}/p,n) , K(\mathbb{Z}/p,m) ] \cong H^m(K(\mathbb{Z}/p,n);\mathbb{Z}/p)$$

and so we have a cohomology operation for each element in these groups. In fact the cohomology rings of $K(\mathbb{Z}/p,n)$ were computed by Cartan and Serre and a multiplicative basis is given by the Steenrod powers and the Bockstein.

• Great answer! For the cup product, I recommend the section "Interpretation" on en.m.wikipedia.org/wiki/Cup_product If there is no further answer in the next few days, I will accept your very intuitive answer. Thank you! :-) Apr 10, 2017 at 5:50