Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

So when defining a Cohomology theory of a spectrum you define the addition structure via the pinch map. I.E. to define addition on $[X,E_n]$ look at $f,g \in [\sum X, E_{n+1}]$ let $\iota: \sum X \to \sum X \vee \sum X$ then $f+g=\iota^*(f\vee g)$. Identity is given by the constant map and inverses are apparently given by reversing the direction of $S^1$ in $\sum X=S^1 \wedge X$: $-f(s,x)=f(-s,x)$.

What is the homotopy between the $f-f$ and the identity a constant map? I would imagine this is the same for the higher homotopies.

Should I use adjointness? I would rather have a direct homotopy.

share|improve this question

1 Answer 1

up vote 1 down vote accepted

Yes, inverses are indeed given by reversing the $S^1$ coordinate. This can be described in exactly the same way as one proves that reversing a path representing an element of the fundamental group gives the inverse. Namely, given $f:\Sigma X \rightarrow E_{n+1}$, consider each $s\in S^1 = [0,1]/\partial[0,1]$ as determining a map $X \rightarrow E_{n+1}$ (so that at $s=basepoint$ the map is constant). You can define the homotopy from $(f\vee (-f)) \circ \iota$ to the constant map by drawing the usual square $[0,1]^2\in \mathbb{R}^2$ with the constant map throughout the triangle $y \geq |2x-1|$. (Think of the horizontal direction as the suspension coordinate $s$, of $X$ as not really being pictured (or being "depth", if you'd like), and of the vertical direction being the time coordinate of the homotopy.) From this description you can work out the precise formula if you like, although I doubt it'll be illuminating.

What do you mean by "the higher homotopies"?

share|improve this answer
$\pi_n$ and so on, –  Sven Sep 27 '11 at 15:10
Yes; in fact more generally $[\Sigma^n X, Y]$ is a group for $n \geq 1$ and is an abelian group for $n\geq 2$. Inverses (and the proof that they are inverses) go exactly as they do for higher homotopy groups (which are the special case $X=S^0$). A cute way to say this is that $\Sigma X$ is a cogroup object in the homotopy category of topological spaces (and $\Sigma^2 X$ is a cocommutative cogroup object) -- this just says that it satisfies the usual diagrams for a group object (en.wikipedia.org/wiki/Group_object) but with all the arrows reversed. The addition on... –  Aaron Mazel-Gee Sep 27 '11 at 16:22
$[\Sigma^n X, Y]$ comes from the comultiplication map $\Sigma^n X \rightarrow \Sigma^n X \vee \Sigma^n X$, and will be commutative if the comultiplication is cocommutative. You should check out Switzer's "Algebraic Topology: Homotopy and Homology" if you enjoy this sort of perspective. –  Aaron Mazel-Gee Sep 27 '11 at 16:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.