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1. • CommentRowNumber1.
   • CommentAuthorMike Shulman
   • CommentTimeJun 13th 2013
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   Author: Mike Shulman
   Format: MarkdownItexAs discussed at [[fiber sequence]], from a map $f:A\to B$ of spaces (or objects in any pointed $(\infty,1)$-category) we obtain a long sequence $$\cdots \Omega^2 A \overset{\Omega^2 f}{\to} \Omega^2 B \overset{-\Omega\partial}{\to} \Omega fib(f) \overset{-\Omega i}{\to} \Omega A \overset{-\Omega f}{\to} \Omega B \overset{\partial}{\to} fib(f) \overset{i}{\to} A \overset{f}{\to} B $$ where minus signs denote reversal of loops. Applying $\pi_0$ to all these spaces, we obtain a long exact sequence of homotopy groups $$\cdots \pi_2(A) \overset{\pi_2 (f)}{\to} \pi_2(B) \overset{-\pi_1(\partial)}{\to} \pi_1(fib(f)) \overset{-\pi_1(i)}{\to} \pi_1(A) \overset{-\pi_1(f)}{\to} \pi_1(B) \overset{\pi_0(\partial)}{\to} \pi_0(fib(f)) \overset{\pi_0(i)}{\to} \pi_0(A) \overset{\pi_0(f)}{\to} \pi_0(B) $$ which should be the [[long exact sequence of homotopy groups]] associated to the fibration $f:A\to B$. But there are these unexpected minus signs --- I thought that the latter LES was supposed to involve $\pi_n(f)$, not $(-1)^n \pi_n(f)$? In the nonabelian case, $-\pi_1(f)$ is not even necessarily a group homomorphism! What do we do to get from this LES to the canonical one? We could insert inversion maps at appropriate places in the fiber sequence, but the fact that 3 is odd makes the pattern of where they would go somewhat obscure. Or we could observe that the image and kernel of a group homomorphism are the same as for its negative, so that replacing the latter by the former doesn't change exactness (of a sequence of pointed sets). But these seem kind of *ad hoc*. Is there a more principled solution? I haven't been able to find any reference which addresses this issue. Most algebraic topology textbooks seem to construct the LES of a fibration by way of the LES using relative homotopy groups, and those that do it this way seem to ignore the minus signs.

   As discussed at fiber sequence, from a map $f:A\to B$ of spaces (or objects in any pointed $(\infty,1)$-category) we obtain a long sequence

   $$\cdots \Omega^2 A \overset{\Omega^2 f}{\to} \Omega^2 B \overset{-\Omega\partial}{\to} \Omega fib(f) \overset{-\Omega i}{\to} \Omega A \overset{-\Omega f}{\to} \Omega B \overset{\partial}{\to} fib(f) \overset{i}{\to} A \overset{f}{\to} B$$

   where minus signs denote reversal of loops. Applying $\pi_0$ to all these spaces, we obtain a long exact sequence of homotopy groups

   $$\cdots \pi_2(A) \overset{\pi_2 (f)}{\to} \pi_2(B) \overset{-\pi_1(\partial)}{\to} \pi_1(fib(f)) \overset{-\pi_1(i)}{\to} \pi_1(A) \overset{-\pi_1(f)}{\to} \pi_1(B) \overset{\pi_0(\partial)}{\to} \pi_0(fib(f)) \overset{\pi_0(i)}{\to} \pi_0(A) \overset{\pi_0(f)}{\to} \pi_0(B)$$

   which should be the long exact sequence of homotopy groups associated to the fibration $f:A\to B$. But there are these unexpected minus signs — I thought that the latter LES was supposed to involve $\pi_n(f)$, not $(-1)^n \pi_n(f)$? In the nonabelian case, $-\pi_1(f)$ is not even necessarily a group homomorphism!

   What do we do to get from this LES to the canonical one? We could insert inversion maps at appropriate places in the fiber sequence, but the fact that 3 is odd makes the pattern of where they would go somewhat obscure. Or we could observe that the image and kernel of a group homomorphism are the same as for its negative, so that replacing the latter by the former doesn’t change exactness (of a sequence of pointed sets). But these seem kind of ad hoc. Is there a more principled solution?

   I haven’t been able to find any reference which addresses this issue. Most algebraic topology textbooks seem to construct the LES of a fibration by way of the LES using relative homotopy groups, and those that do it this way seem to ignore the minus signs.

2. • CommentRowNumber2.
   • CommentAuthorUrs
   • CommentTimeMar 1st 2016
   • (edited Mar 1st 2016)
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   Author: Urs
   Format: MarkdownItexI just happen to come back to this issue. Correct me if I am wrong, but I think the $n$Lab entry was wrong: the long exact fiber sequence given by iteratively adjoining homotopy pullback squares does _not_ have any funny loop reversal. $$ \array{ \Omega A &\longrightarrow& \ast \\ \downarrow^{\mathrlap{\Omega f}} && \downarrow \\ \Omega B & \longrightarrow & ker(f) &\longrightarrow& \ast \\ \downarrow && \downarrow^{\mathrlap{g}} && \downarrow \\ \ast &\longrightarrow & A &\stackrel{f}{\longrightarrow}& B } \;\;\;\;\; \simeq \;\;\;\;\; \left( \array{ \ast &\longrightarrow& \ast \\ \downarrow && \downarrow \\ ker(f) &\longrightarrow& \ast \\ \downarrow^{\mathrlap{g}} && \downarrow \\ A &\stackrel{f}{\longrightarrow}& B \\ \uparrow && \uparrow \\ \ast &\longrightarrow& \ast } \right) \stackrel{\underset{\longleftarrow}{\lim}}{\mapsto} \left( \Omega A \stackrel{\Omega f}{\longrightarrow} \Omega B \right) $$ But the thing is that this is _not_ the same as iterating the homotopy fiber construction, because $$ \array{ ker(g) &\longrightarrow& \ast \\ \downarrow &\swArrow& \downarrow \\ ker(f) &\stackrel{g}{\longrightarrow}& A } \;\;\; \simeq \;\;\; \array{ \Omega B &\longrightarrow& \ast \\ \downarrow &\swArrow& \downarrow \\ ker(f) &\stackrel{g}{\longrightarrow}& A } \;\;\;\; \simeq \;\;\;\; \left( \array{ \Omega B &\longrightarrow& ker(f) \\ \downarrow &\swArrow& \downarrow^{\mathrlap{g}} \\ \ast &\longrightarrow& A } \right)^{-1} $$ Now what the classical textbooks do is not to take the pasting composites of homotopy pullbacks to be the definition of the long fiber sequence, but they take the iterated homotopy fiber construction as the definition. If one does that, then one finds one has to throw in loop reversal. E.g. Switzer's book, item 2.57. I have fixed that in the entry [here](https://ncatlab.org/nlab/show/fiber+sequence#ContinuingToLongFiberSequences). (Sorry for replying to Mike's ancient message here this way. Certainly Mike has long answered this question for himself already. But in looking for a thread suitable to announce this edit, this one here seems the appropriate one.)

   I just happen to come back to this issue.

   Correct me if I am wrong, but I think the $n$Lab entry was wrong:

   the long exact fiber sequence given by iteratively adjoining homotopy pullback squares does not have any funny loop reversal.

   $$\array{ \Omega A &\longrightarrow& \ast \\ \downarrow^{\mathrlap{\Omega f}} && \downarrow \\ \Omega B & \longrightarrow & ker(f) &\longrightarrow& \ast \\ \downarrow && \downarrow^{\mathrlap{g}} && \downarrow \\ \ast &\longrightarrow & A &\stackrel{f}{\longrightarrow}& B } \;\;\;\;\; \simeq \;\;\;\;\; \left( \array{ \ast &\longrightarrow& \ast \\ \downarrow && \downarrow \\ ker(f) &\longrightarrow& \ast \\ \downarrow^{\mathrlap{g}} && \downarrow \\ A &\stackrel{f}{\longrightarrow}& B \\ \uparrow && \uparrow \\ \ast &\longrightarrow& \ast } \right) \stackrel{\underset{\longleftarrow}{\lim}}{\mapsto} \left( \Omega A \stackrel{\Omega f}{\longrightarrow} \Omega B \right)$$

   But the thing is that this is not the same as iterating the homotopy fiber construction, because

   $$\array{ ker(g) &\longrightarrow& \ast \\ \downarrow &\swArrow& \downarrow \\ ker(f) &\stackrel{g}{\longrightarrow}& A } \;\;\; \simeq \;\;\; \array{ \Omega B &\longrightarrow& \ast \\ \downarrow &\swArrow& \downarrow \\ ker(f) &\stackrel{g}{\longrightarrow}& A } \;\;\;\; \simeq \;\;\;\; \left( \array{ \Omega B &\longrightarrow& ker(f) \\ \downarrow &\swArrow& \downarrow^{\mathrlap{g}} \\ \ast &\longrightarrow& A } \right)^{-1}$$

   Now what the classical textbooks do is not to take the pasting composites of homotopy pullbacks to be the definition of the long fiber sequence, but they take the iterated homotopy fiber construction as the definition. If one does that, then one finds one has to throw in loop reversal. E.g. Switzer’s book, item 2.57.

   I have fixed that in the entry here.

   (Sorry for replying to Mike’s ancient message here this way. Certainly Mike has long answered this question for himself already. But in looking for a thread suitable to announce this edit, this one here seems the appropriate one.)

3. • CommentRowNumber3.
   • CommentAuthorCharles Rezk
   • CommentTimeMar 5th 2016
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   Author: Charles Rezk
   Format: MarkdownItexThe "classical" version of the fiber sequence has one big advantage: it is invariant under shifts. Take the classical fiber sequence $\dots \to \Omega B\to \mathrm{fib}(f) \xrightarrow{g} A \xrightarrow{f} B$, and remove the "$f$". Then you are looking at another classical fiber sequence, starting at $g$ instead of $f$. In the "pasting" version, this isn't true. If you try to write down an equivalence between "pasting fiber sequence starting at $f$ with $f$ removed", and "pasting fiber sequence starting at $g$", you have to use sign reversing maps at some of the objects (assuming you have chosen a standard way to identify them with $\Omega^?(\text{one of the first three objects})$). I think it works out that the 12-fold shift of the pasting fiber sequence is equivalent to the original one without using sign reversing maps, but not before. If you are trying to get signs to come out right, that is a lot more confusing, I think.

   The “classical” version of the fiber sequence has one big advantage: it is invariant under shifts. Take the classical fiber sequence $\dots \to \Omega B\to \mathrm{fib}(f) \xrightarrow{g} A \xrightarrow{f} B$, and remove the “$f$”. Then you are looking at another classical fiber sequence, starting at $g$ instead of $f$.

   In the “pasting” version, this isn’t true. If you try to write down an equivalence between “pasting fiber sequence starting at $f$ with $f$ removed”, and “pasting fiber sequence starting at $g$”, you have to use sign reversing maps at some of the objects (assuming you have chosen a standard way to identify them with $\Omega^?(\text{one of the first three objects})$). I think it works out that the 12-fold shift of the pasting fiber sequence is equivalent to the original one without using sign reversing maps, but not before.

   If you are trying to get signs to come out right, that is a lot more confusing, I think.

4. • CommentRowNumber4.
   • CommentAuthorUrs
   • CommentTimeMar 8th 2016
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   Author: Urs
   Format: MarkdownItexThanks, good point. I'll eventually edit the entry a bit more.

   Thanks, good point. I’ll eventually edit the entry a bit more.