# 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,

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Check e.g. Mosher & Tangora, they spell this out in more detail. It seems like it's generally best to "shut the hood" (of the ol' Steenrod algebra car) as soon as possible and just be happy with the axiomatic characterization, though. – Aaron Mazel-Gee Nov 16 '10 at 8:45

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.

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Thank you that was the answer I was looking for. – Analyst2 Nov 18 '10 at 18:54
This is a version of the geometric boundary theorem. – Sean Tilson Dec 2 '10 at 6:12
Cool, what's that? – Aaron Mazel-Gee Dec 27 '10 at 9:04
Aaron, sorry for the long delay. The idea is that taking the boundary of a class in a LES actually comes from some "geomtric" map/construction. It is exactly your last paragraph. They have come up in some other places as well. – Sean Tilson Aug 8 '11 at 4:15
@AaronMazel-Gee In the discussion by May you describe, he says that one can easily prove (by a diagram chase) that this composite is the axiomatically given connecting homomorphism. However, I don't see what exactly that means or how it would be proved. Presumably one would show that it fits into the standard long exact sequence, but I'm not sure how to do that in the case of a genaralized reduced homology theory (i.e., from the axioms alone). Could you comment on that? – rj7k8 Aug 15 '15 at 10:12