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Here is what I would call the transgression theorem. Let $X$ be a pointed space and $\Omega X$ its loop space. There are two maps $H_{p}(\Omega X) \to H_{p+1}(X)$ which should be the same. I am looking for an easy proof.

The first map is by the following procedure:

$$ H_{p}(\Omega X) \cong H_{p+1}(\Sigma \Omega X) \to H_{p+1}(X); $$

the first is the suspension isomorphism (o.k., I should assume $p>0$), the second is given by the evaluation $\Sigma \Omega X \to X$. The other map is given in terms of the homological Leray-Serre spectral sequence for the path-loop fibration $\Omega X \to PX \to X$ ($PX$=path space, $PX \simeq pt$). Note that the differential $d^{p+1}:E^{p+1}_{0,p}\to E^{p+1}_{p+1,0}$ is an isomorphism because $E^{p+2}_{0,p}= E^{p+2}_{p+1,0}=0$.

Now consider the composition

$$H_{p}(\Omega X) \to H_{0}(X, H_p (\Omega X))=E^{2}_{0,p} \to \ldots \to E^{p+1}_{0,p} \stackrel{(d^{p+1})^{-1}}{\to} E^{p+1}_{p+1,0} \subset E^{2}_{p+1,0}$$ $$= H_{p+1} (X, H_0(\Omega X)) \to H_{p+1}(X).$$

(everything makes perfect sense for nonsimply connected $X$, using local coefficient systems).

Here is my question:

  1. are these maps indeed equal (up to sign)?

  2. is there a proof of this, which does not involve chasing differentials in the spectral sequence and the Hurewicz theorem?

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Some of the formulas don't show up nicely for me. Tex Bug? – Thierry Zell Jan 20 '11 at 16:19
I inserted some "`"'s and a linebreak to fix the tex-problems. – Andreas Thom Jan 20 '11 at 16:49
@johannes: Have you tried looking at Hatcher's spectral sequences book? I think he proves this in a way that satisfies your requirements. – Dylan Wilson Jan 20 '11 at 16:53
Here's a link:, and the proof should be in chapter 1 – Dylan Wilson Jan 20 '11 at 16:53
@Dylan: the proof in Hatcher's book uses a concrete description of the $d^p$-differential. I was looking for a proof that avoids this and uses the naturality of the spectral sequence instead. – Johannes Ebert Jan 21 '11 at 10:09

They are equal up to sign.

If $F\to E\to B$ is a Hurewicz fibration, where $B$ is well-pointed, then we have a factorization $E\to E/F \to B$ and we have the Barratt-Puppe extension $E/F \to \Sigma F$. This gives a diagram $$ B \quad \overset{a}\leftarrow \quad E/F \quad \overset{b} \to \quad \Sigma F $$ and when the transgression is defined it is given by the homomorphism these maps induce on homology.

More precisely, if $x \in H_k(\Sigma F)$ lifts to an element $y$ of $H_k(E/F)$ via $b_\ast$, then one defines the transgression of $x$ as $a_*y$. To make this well-defined, you need to take into account the indeterminacy of the lifts $y$ (this is the image of the map $H_k(E) \to H_k(E/F)$ and you must quotient out by this indeterminacy--but this won't matter in your case--cf. below).

(You can see this e.g., in McCleary's book, p. 185)

Now consider the path-loop fibration $\Omega X \to PX \to X$. In this case $E/F$ is identified with $\Sigma \Omega X$ and the map $E/F\to \Sigma F$ is identified with $\pm$ identity. Thus the lift is unique in this case (no indeterminacy) and the transgression is the map $\Sigma \Omega X \to X$ up to sign on homology.

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should I take this to mean you are warming up to spectral sequences? – Sean Tilson Mar 4 '11 at 23:21

I don't have much time so this will be sketchy, but I think you can prove it for the cohomology spectral sequence by looking at the universal case of $K(\mathbb{Z},n)$ (and the homology case should follow by dualizing). When $X=K(\mathbb{Z},n)$, both of the groups involved in the two maps are $\mathbb{Z}$, and you know that both maps need to be isomorphisms, so they must agree up to sign. For general $X$, you can then use that any class in $H^n(X)$ is pulled back from $K(\mathbb{Z},n)$ under some map together with the naturality of the spectral sequence.

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I had this argument in mind. However, I was unable to make the step "follow by duality" precise. – Johannes Ebert Mar 4 '11 at 19:24
For purposes of dualizing, any homology class in $H_n(X)$ is pushed forward from a homology class of a space $Y$ such that $H_n(Y)$ is finitely generated and free (proof: glue together the simplices that make up the chain in $X$ to obtain an $n$-dimensional complex $Y$). The dualizing argument should then work on $Y$, and by naturality you then get the result for $X$. – Eric Wofsey Mar 4 '11 at 21:41

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