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".
 A: 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. 
A: 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.
