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I'm stuck on the proof of Lemma 23.4 from May's Simplicial Objects in Algebraic Topology (p.99) on a seemingly harmless (and so left to reader's proof) step. Some context:

given a Kan complex $K$ with only one vertex $\phi\in K_0$ we have just defined a Kan fibration $p:P(K) \to K$, where $P(K)$ is a complex defined by: $$ P_n(K) := K_{n+1} $$ $$ \bar\partial_i := \partial_{i+1} $$ $$ \bar s_i := s_{i+1} $$ where, for clarity, the "barred" operators refer to those of $P(K)$, and the projection $$ p := \partial_0 : P_n(K) = K_{n+1} \to K_n $$ is just the $0^{th}$ face of $K$ left out in the construction. We also have a cross section $$ \sigma := s_0 : K_n \to K_{n+1}=P_n(K) $$ and the fiber complex $L(K):=p^{-1}(\phi)$. Here $\phi$ refers ambiguously also to the complex generated by $\phi$ with just one degenerate simplex, still called $\phi$, in each degree. The identification $K_{n+1}=P_n(K)$ is given in terms of a bijection $\lambda$ of sets that I omitted for readability. If you want $\bar\partial_i = \lambda \partial_{i+1} \lambda^{-1}$. We further know $P(K)$ is contractible, and assume $L(K)$ to be a group complex. The statement of the Lemma is omitted as it is not the point where I'm stuck, right after which the proof follows smoothly. The very first line of the proof reads:

$p\circ\sigma$ is the identity on $K$, so we may identify $P_n(K)$ with $L_n(K)\times \sigma(K_n)$ for each $n$.

The first obvious try has been that of using some analogue of a splitting lemma for exact sequences of pointed sets, but it failed because even if $p(x)=p(\sigma p(x))$ I have no structure on $P_n(K)$ in order to say something along the line of $``p(x\ast\sigma p(x)^{-1})=\phi"$ and get a simplex $l\in L_n(K)$ as the splitting requires.

My second try involved induction on $n$, inspired by the fact that, for dimensional reasons, $L_0(K)=P_0(K)=K_1$ and $\sigma(K_0)=\sigma(\phi)$, so there's only one possible way to define the bijection in dimension $0$, but things turned out too difficult even so.

The bottleneck still remains that, even if there's only one vertex, for $n\ge 2$ I have no structure on $P_n(K)$ or $\sigma(K_n)$ in order to "conjugate" by a path on $K$ towards $\phi$ and then use the group structure on the fiber, as one would do in classical homotopy theory. I have an intuitive image of the decomposition but can't quite get it right, I'm sure I'm missing something very trivial, it shouldn't be so problematic.

Edit 1: after some research I made sense of the notation $L$ and $P$ for the complexes in question, corresponding to the path-loop fibration. That makes clear that, even without group structure on $P(K)$, when two paths are on the same fiber one could still consider their "inverse" concatenation. I suspect this was supposed to be taken care of by having $K_0=\{\phi\}$, but it still doesn't work in higher dimension.

Edit 2: looking at the book's bibliography (a very old transcript of a seminar by Moore) this part was originally demonstrated for $K=\Gamma$ a simplicial group, which make total sense. Problem is, that the very next Theorem (23.6) makes heavy use of the group structure being just on $L(K)$ and not $K$...so this is still a problem.

Alberto Avitabile
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1 Answers1

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You can correct the situation by requiring in the axioms of Eilenberg Maclane complexes that they be group complexes.

Then there is just a modification where "if L(K) is a group complex" should be "if K is a group complex". This would naturally fit with the remark right after lemma 23.4, where Peter says "if K is a group complex, then so is L(K)", and you can prove lemma 23.4 using K as a group complex meanwhile it doesn't seem true if it isn't (Although I would be very pleased if someone could make it work). And then theorem 23.6 would be nicely proven in this new framework because K and L would be group complexes.

You might think that this approach is flawed because we then don't know if the fibers of postnikov fibrations are Eilenberg-Maclane spaces, but this isn't the case because it is not hard to define a group structure on those fibers using only the fact that their only non vanishing homotopy group is the n'th and that they are minimal complexes (observe that it must be trivial on the simplices of degree less than n, then define it on the n'th degree using the homotopy group, then define it inductively on higher degrees using the simplices of degree 1 less).

Ethan Jahan
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