Loop-space functor on cohomology

Question

For a pointed space $X$ and an Abelian group $G$, the loop-space functor induces a homomorphism $\omega:H^n(X,G)\to H^{n-1}(\Omega X,G)$. More concretely, $\omega$ is given by the Puppe sequence

$$\cdots \to \Omega X \overset{\omega f}{\to} K(G,n\!-\!1) \to Mf \to X \overset{f}{\to} K(G,n)\,.$$

Equivalently, $\omega$ can also be defined as the pullback of the evaluation map $\Sigma\Omega X {\to} X$, i.e.,

$$[X, K(G,n)]\overset{\omega}{\to}[\Sigma\Omega X, K(G,n)]\simeq[\Omega X, K(G,n\!-\!1)]\,.$$

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Q1. Is $\omega$ the inverse of the transgression map in the Serre spectral sequence associated to $\Omega X\to *\to X$?

Q2. Where can I find a discussion on $\omega$'s properties, e.g., kernel and image (especially if the answer to Q1 is "no")?

Answer 1

If you are willing to read a little French, look at page 434 of Serre's "Homologie singuliere des espaces fibres" (1951). For an exercise in English, with a hint, try Exercise 2 on page 155 in Mosher and Tangora's "Cohomology Operations and Applications in Homotopy Theory" (1968). For a more detailed reference, look at Proposition 6.10 in McCleary's "User's Guide to Spectral Sequences" (1985/2001). See also Proof of the ''trangression theorem'' and On the transgression in the Serre spectral sequence here on MathOverflow.

Answer 2

For Q2, when searching it might be useful to know two things:

1. The map $\omega$ is often called the cohomology suspension (not to be confused with the suspension isomorphism in cohomology!), and
2. Elements of the kernel of $\omega$ are precisely the elements of category weight 2.

To expand on 2., as explained in Cornea-Lupton-Oprea-Tanré's book on Lusternik-Schnirelmann category, the evaluation map $\Sigma\Omega X\to X$ is equivalent to the Ganea fibration $G_1 X\to X$. The homotopy fibre of both maps is $\Omega X * \Omega X\simeq \Sigma \Omega X \wedge \Omega X$, which is $2k$-connected if $X$ is $k$-connected. This gives a range in which $\omega$ is an isomorphism, from the Serre exact sequence.

Note also that any cup product of reduced cohomology classes is the kernel of $\omega$ (obviously, since cup products vanish in a suspension). But the kernel can also contain indecomposable classes, e.g. triple Massey products.

Regarding the image of $\omega$, I'd be willing to bet that it's exactly the transgressive elements.