Why steenrod commute with transgression I'm reading Hatcher's notes on spectral sequences and he mentions that steenrod squares commute with the coboundary operator for pairs (X,A) which would then explain why these operations commute with the transgression. It says it's because 
that coboundary operator can be defined in terms of suspension and we know steenrod operations commute with suspension. Does anyone know the details of this reasoning?
So...
Assuming the standard axioms of steenrod operations, how do we prove that it commutes with the coboundary operator for pairs?
Thank you,
 A: I realized that your question wasn't exactly about the Steenrod axioms themselves, but about the definition of the coboundary operator involving suspension.  In reduced homology, the boundary operator $\partial$ for the pair $(X,A)$ (where the inclusion $i:A\rightarrow X$ is a cofibration) can be defined to come from the "topological boundary map" $\partial^!$ followed by the inverse of the suspension isomorphism.  The former is itself a composition
$$ \partial^! = \pi \circ \psi^{-1}: X/A \rightarrow Ci \rightarrow \Sigma A, $$
where $Ci$ is the mapping cone of $i$, $\psi^{-1}$ is a homotopy inverse of the quotient $\psi: Ci \rightarrow Ci/CA=X/A$, and $\pi: Ci \rightarrow Ci/X=\Sigma A$.  So
$$ \partial = (\Sigma_*)^{-1} \circ \partial^!_* : \tilde{H}_q(X/A) \rightarrow \tilde{H}_q(\Sigma A) \rightarrow \tilde{H}_{q-1}(A) .$$
In fact, this is true for any reduced homology theory.  See May's "Concise Course" for details, pp. 106-7.  I'm pretty sure that the situation for cohomology is very similar.
Bottom line: In this formulation, the coboundary operator is the composition of a map induced from an actual map on spaces and the (inverse of the (?)) suspension isomorphism.  Steenrod squares commute with both of these, so they commute with the coboundary operator.
