A basic construction in homotopy is Puppe sequences. Given a map $A \stackrel{f}{\to} X$, its homotopy cofiber is the map $X\to X/A=X \cup_f CA$ from $X$ to the mapping cone of $f$. If we then take the cofiber again, something remarkable happens: $(X/A)/X$ is naturally homotopy equivalent to the suspension $\Sigma A$ of $A$. This isn't hard to see geometrically; a nice picture and discussion can be found in pages 397-8 of Hatcher. If we iterate this, we end up with a sequence $$A \to X \to X/A \to \Sigma A \to \Sigma X \to \Sigma(X/A) \to \Sigma^2 A \to \cdots$$ in which each map is the homotopy cofiber of the previous map. If we then apply a functor which sends cofiber sequences to exact sequences, we get a long exact sequence. This can be understood as the origin of long exact sequences of cofibrations in (co)homology, using the fact that $H^n(X)=H^{n+1}(\Sigma X)$.

One subtlety of this construction is that under the natural identifications of $(X/A)/X$ and $((X/A)/X)/(X/A)$ with $\Sigma A$ and $\Sigma X$, the map $\Sigma A\to\Sigma X$ is not the suspension of the original map $f$, but rather its negative (i.e., $-1\wedge f: S^1\wedge A=\Sigma A \to \Sigma X=S^1\wedge X$, where -1 is a map of degree -1). The geometric explanation for this can neatly be seen in Hatcher's picture, where the cones are successively added on opposite sides, so the suspension dimensions are going in opposite directions.

However, you usually don't need to worry about this sign issue. First, since a map of degree -1 is a self-homotopy equivalence (even homeomorphism) of $\Sigma A$, we could just change our identification of $(X/A)/X$ with $\Sigma A$ by such a map and then we would just have $\Sigma f:\Sigma A \to \Sigma X$ (note though that then we are not changing how we identify the next space in the sequence with $\Sigma X$, which breaks some of the symmetry of the picture). Alternatively, if we only care about the Puppe sequence because of the long exact sequences it gives us, we could note that an exact sequence remains exact if you change the sign of one of its maps.

My question is: is there any situation where these signs really do matter and have interesting consequences? Might they be somehow connected to the signs that show up in graded commutative objects in topology?

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    $\begingroup$ If I can add a question to this one: are they any famous examples of errors which result from mishandling the signs in the Puppe sequence? $\endgroup$ – Charles Rezk Nov 18 '09 at 3:56

There is a specific sort of situation I know about where that sign matters. Suppose you have $f:X \rightarrow Y$ and $g:Z \rightarrow W$ cofibrations (if the maps are not cofibrations, all the same things work - you just replace the quotient spaces by mapping cones). You extend both maps to their Dold-Puppe sequences, so you get the sequences

$X \rightarrow Y \rightarrow Y/X \rightarrow \Sigma X \rightarrow \Sigma Y \ldots$


$Z \rightarrow W \rightarrow W/Z \rightarrow \Sigma Z \rightarrow \Sigma W \ldots$

Now suppose you have maps $a: X \rightarrow W$ and $b: Y \rightarrow W/Z$ making the obvious square commute up to homotopy. You can then extend these to make a commutative ladder from the first Dold-Puppe sequence to the second. (Notice that the sequences are deliberately offset from each other by one spot.)

Using the usual parameters and the obvious choices of homotopies you will get a square involving $Y/X, \Sigma X, \Sigma Z, \Sigma W$. This square will commute if it includes the map $-\Sigma g: \Sigma Z \rightarrow \Sigma W$, but not generally with the map $\Sigma g$. (To check all this, I recommend doing the Dold-Puppe sequences with mapping cones rather than quotient spaces but keeping the homotopy equivalences with the quotient spaces in mind, which is the only way I know to calculate what the right maps should be.)

At this point, if you were feeling stubborn, you could replace the map in your ladder $\Sigma X \rightarrow \Sigma Z$ with $-1$ times that map, and that would allow you to have used $\Sigma g$ in the square I mention in the above paragraph, but that creates other issues; if you choose not to simply use suspensions of your original maps to go from one Dold-Puppe sequence to the other then you run into problem when you are mapping between Dold-Puppe sequences without the shift of this example.

I hope this helps unravel Greg's answer (which is correct - you need the sign to get good mapping properties).

Of course one sees exactly the same phenomenon in the category of chain complexes of abelian groups (where homotopy is chain homotopy) and other such categories. I agree with Theo and Mark that one thinks about the suspension as "odd" (in the sense of parity not the sense of peculiar).

The published paper that Mark refers to that has an error of exactly this sort (which is unfortunately fundamental to the paper) is by Lin Jinkun in Topology v. 29, no. 4, pp. 389-407. I read this paper in preprint form in 1988 and missed this error, but discovered it in 1992 when reading another paper by the same author with the same error. In the Topology paper the error is made in diagram 4.4 on the right hand square (proof of Lemma 4.3).

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  • $\begingroup$ I should add that I learned of the example that I mentioned from Hal, so I'm glad he gave this definitive answer. $\endgroup$ – Mark Hovey Dec 1 '09 at 23:37

There was a case where a paper proved a very strong result and the referee found an unfixable hole in the proof caused by forgetting the sign in the Puppe sequence! As others have said, it is analogous to the sign in the tensor product of chain complexes:

$d(x \otimes y) = dx \otimes y + (-1)^k x \otimes dy$

if x has degree k. It matters.

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    $\begingroup$ Is it possible to give an idea about what said result was without violating any confidentiality? $\endgroup$ – Tyler Lawson Nov 18 '09 at 13:35
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    $\begingroup$ There are some detailed computations of the behavior of beta elements and products thereof in the Adams-Novikov spectral sequence at odd primes. It was one of those. This was quite some time ago, maybe the early 1990's or even late 1980's. $\endgroup$ – Mark Hovey Nov 18 '09 at 15:57

This is not exactly an answer but shows that these signs do matter in general: Example 4.21, page 32, of B. Iversen's ‘Cohomology of sheaves’ (Universitext, Springer-Verlag, Berlin, 1986) shows that you cannot change the signs of one map in a distinguished triangle in a triangulated category without breaking its distinguishedness—the example is in the homotopy category of complexes of abelian groups. You can swap two signs, though.

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  • $\begingroup$ This is basically what I was going to post - for more detail you can look at my answer in the context of the homotopy and derived categories of abelian groups here - mathoverflow.net/questions/4653/… $\endgroup$ – Greg Stevenson Nov 18 '09 at 3:56

Basically my interpretation of Mariano's answer is as follows: one needs to be consistent with the signs in the Puppe sequence in order for the good properties one wants to hold (namely the mapping axiom when using cofiber sequences as triangles, I am not sure off the top of my head if this is a problem in the enhanced versions), but at least in my opinion in some sense which consistent choice of signs one makes is just convention.

This requirement leads one to the sign changes in the axiom concerning rotation of triangles in a triangulated category. It is this which in some sense is the source of the graded-commutative behaviour one observes (so really it comes from the signs in the Puppe sequence as one source of motivation). For instance one can formalize this behaviour by considering the "central ring" $Z^*(\mathbf{T})$ of a triangulated category $\mathbf{T}$ (it might not form a set hence the "") whose graded pieces are $$Z^i(\mathbf{T}) = \{\eta\colon Id \to \Sigma^i \; \vert \; \eta\Sigma = (-1)^i\Sigma\eta \}$$ i.e., natural transformations from the identity functor to powers of the suspension functor which commute or anticommute with suspension depending on the degree. Composition makes this into a graded-commutative "ring" which has a natural action on the graded hom-sets of $\mathbf{T}$, where we need the graded-commutativity to ensure that the sign switch on suspension is taken care of and so everything plays nicely with triangles. So if you have interesting natural transformations in higher degrees you have a natural graded-commutative action.

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If you take a commuting square of maps and use homotopy cofibers to extend outward to a big grid, you eventually run into some anticommuting squares, which are really supercommuting squares. I think this answers your last question in the affirmative. Whether this answers your other questions really depends on your standards, I suppose.

The anticommuting square is mentioned in Faisceaux Pervers and possibly also the triangulated category chapter in Weibel's Homological Algebra.

Edit: I think there is a way to write a Mayer-Vietoris sequence as chains on the totalization of a big grid like this. Then you need the signs to make it work. Here is an attempt at a diagram:

$$\require{AMScd}\begin{CD} A @>>> B @>>> B/A @>>> \Sigma A \\ @VVV @VVV @VVV @VVV \\ C @>>> D @>>> D/C @>>> \Sigma C \\ @VVV @VVV @VVV @VVV \\ C/A @>>> D/B @>>> X @>>> \Sigma(C/A) \\ @VVV @VVV @VVV @VVV \\ \Sigma A @>>> \Sigma B @>>> \Sigma(B/A) @>>> \Sigma^2 A \\ \end{CD}$$

The bottom right square anticommutes, but the rest commute. The maps on the bottom and right edges have minus signs.

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    $\begingroup$ Ah, so maybe the point is that we should think of the suspension as an "odd" operation, that somehow got (super)commuted past something? (co)homology theories are basically super, no? $\endgroup$ – Theo Johnson-Freyd Nov 18 '09 at 6:30
  • $\begingroup$ Suspension shifts the dimension of singular chains up by one. One then runs into orientation considerations when gluing (as in Eric's question), and it changes their parity when looking at products. $\endgroup$ – S. Carnahan Nov 18 '09 at 15:31
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    $\begingroup$ Yes, I think Theo is exactly right. In homotopy categories that come from a monoidal model category, commuting S^m (the m-fold suspension of the unit) past S^n always introduces a sign of (-1)^{mn}. This was a conjecture in my model categories book that was resolved by Denis-Charles Cisinski. Essentially one should think of monoidal model categories as algebras over simplicial sets, so whatever happens in simplicial sets to the unit will also happen in any monoidal model category. $\endgroup$ – Mark Hovey Nov 18 '09 at 16:03

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