Is every long exact sequence $$\cdots\to\pi_{d+1}(B)\to\pi_d(F)\to\pi_d(E)\to\pi_d(B)\to\pi_{d-1}(F)\to\cdots$$ with topological spaces $F,E$ and $B$, where $F$ is a subspace of $E$ with inclusion map $i$, induced by a Serre fibration $p:E\to B$ ? By "induced" I mean that the maps in the sequence are given by $$p_*:\pi_d(E)\to\pi_d(B)$$ $$i_*:\pi_d(F)\to\pi_d(E)$$ and the boundary map $$\partial:\pi_{d+1}(B)\to\pi_d(F).$$

If not, what are the conditions under which this is the case?

Thank you!

  • $\begingroup$ What do you mean by "is ... induced" ? $\endgroup$ – Ralph Aug 6 '12 at 8:55
  • $\begingroup$ I edited the question and hope that it is clear now $\endgroup$ – Pierre Aug 6 '12 at 9:15
  • $\begingroup$ I understood the question differently: Given $F,E,B$, a map $i:F\to E$, and a long exact sequence of homotopy groups with the map $\pi_q(F)\to \pi_q(E)$ induced by $i$, does there exist a map $p:E\to B$ which realizes this long exact sequence? Any map can be turned into a fibration by replacing the domain by a homotopy equivalent space. So in this interpretation the question boils down to whether the homotopy fiber of $i:F\to E$ is the loop space of $B$. I don't think this is always possible. $\endgroup$ – Paul Aug 7 '12 at 4:19
  • $\begingroup$ This is precisely what I meant, thanks Paul! (You can probably re-post this as the answer) $\endgroup$ – Pierre Aug 7 '12 at 17:19
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    $\begingroup$ Related: Jan Šťovíček has a beautiful paper (arxiv.org/abs/0906.1286) where he characterizes the long exact sequences of length six which come from the snake lemma. $\endgroup$ – Mariano Suárez-Álvarez Aug 8 '12 at 2:03

Ok, so one counterexample to the other interpretation of the question would be to take a fibration $F\to E\to B'$ And let $B$ be a space with the same homotopy groups as $B'$ but not homotopy equivalent to $B$. Note that $B$ must have at least two non-trivial homotopy groups.

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You might want to look up quasifibrations, which are surjective maps $p\colon E\to B$ such that $p\colon (E,p^{-1}(b))\to (B,b)$ is a weak equivalence for all $b\in B$. Any quasifibration gives rise to a long exact sequence as in your question. There are certainly examples of quasifibrations which aren't fibrations (see Mike Shulman's comment to this question, for example).

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    $\begingroup$ While for a topological group $G$, the map $EG \to BG$ is a fibration, for a group-like topological monoid $G$ the map is only a quasi-fibration (see May, Classifying Spaces and Fibrations, 7.6). Another place where they occur is the proof of the theorem of Dold and Thom that $\pi_i(SP(X)) \cong H_i(X)$ for $SP$ the infinite symmetric product. $\endgroup$ – Lennart Meier Aug 6 '12 at 13:06

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