There is a function on $\mathbb{Z}/2\mathbb{Z}$-cohomology called Steenrod squaring: $Sq^i:H^k(X,\mathbb{Z}/2\mathbb{Z}) \to H^{k+i}(X,\mathbb{Z}/2\mathbb{Z})$.  (Coefficient group suppressed from here on out.)  Its notable axiom (besides things like naturality), and the reason for its name, is that if $a\in H^k(X)$, then $Sq^k(a)=a \cup a \in H^{2k}(X)$ (this is the cup product).  A particularly interesting application which I've come across is that, for a vector bundle $E$, the $i^{th}$ Stiefel-Whitney class is given by $w_i(E)=\phi^{-1} \circ Sq^i \circ \phi(1)$, where $\phi$ is the Thom isomorphism.

I haven't found much more than an axiomatic characterization for these squaring maps, and I'm having trouble getting a real grip on what they're doing.  I've been told that $Sq^1$ corresponds to the "Bockstein homomorphism" of the exact sequence $0 \to \mathbb{Z}/2\mathbb{Z} \to \mathbb{Z}/4\mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} \to 0$. Explicitly, if we denote by $C$ the chain group of the space $X$, we apply the exact covariant functor $Hom(C,-)$ to this short exact sequence, take cohomology, then the connecting homomorphisms $H^i(X)\to H^i(X)$ are exactly $Sq^1$.  This is nice, but still somewhat mysterious to me.  Does anyone have any good ideas or references for how to think about these maps?

Here's one way to understand them.  The external cup square $a \otimes a \in H^{2n}(X \times X)$ of $a \in H^n(X)$ induces a map $f:X \times X \to K(Z_2, 2n)$.  It can be show that this map factors through a map $g:(X \times X) \times_{Z_2} EZ_2 \to K(2n)$, where $Z_2$ acts on the product by permuting the factors and $EZ_2$ can be taken to just be $S^\infty$.  If you unravel what this means, it says that our original map $f$ was homotopic to the map obtained by first switching the coordinates and then applying $f$.  It also says that this homotopy, when applied twice to get a homotopy from $f$ to itself, is homotopic to the identity homotopy, and we similarly have a whole series higher "coherence" homotopies.  Now $X \times BZ_2$ maps to $(X \times X) \times_{Z_2} EZ_2$ as the diagonal, so we get a map $X \times BZ2 \to K(2n)$.  But $BZ_2$'s cohomology is just $Z_2[t]$, so this gives a cohomology class $Sq(a) \in H^*(X)[t]$ of degree $2n$.  If we write $Sq(a)=\sum s(i) t^i$, it can be shown that $s(i)=Sq^{n-i}a$.

What does this mean?  Well, if our map $f$ actually was invariant under switching the factors (which you might think it ought to be, given that it appears to be defined symmetrically in the two factors), we could take $g$ to be just the projection onto $X \times X$ followed by $f$.  This would mean that $Sq(a)$ comes from just projecting away the $BZ_2$ and then using $a^2$, i.e. $Sq^n(a)=a^2$ and $Sq^i(a)=0$ for all other $i$.  Thus the nonvanishing of the lower Steenrod squares somehow measures how the cup product, while homotopy-commutative (in terms of the induced maps to Eilenberg-MacLane spaces), cannot be straightened to be actually commutative.  Indeed, in the universal example $X=K(Z_2,n)$, the map $f$ is exactly the universal map representing the cup product of two cohomology classes of degree $n$.

Some somewhat terse notes on this can be found here; see particularly part III. (Sorry, the link is now dead.)
