MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Write $M_n = S^n \cup_2 D^{n+1}$. I know, as a matter of folklore, that the identity map $\mathrm{id}_M$, considered as an element of the group $[M_n, M_n]$, has order $4$ (for $n > 3$, let's say).

What is a reference for this? And is there a simple and pretty argument?

share|cite|improve this question
up vote 3 down vote accepted

Alternatively: if the order were $2$ then $M_n\wedge M_n$ would be $M_{2n}\vee M_{2n+1}$. The mod $2$ cohomology of $M_n$ has a generator $a$ in degree $n$ and a generator $b=Sq^1(a)$ in degree $n+1$ and nothing else. It follows that the cohomology of $M_n\wedge M_n$ has generators $a\otimes a$, $a\otimes b$, $b\otimes a$ and $b\otimes b$ and we find using the Cartan formula that $Sq^2(a\otimes a)=b\otimes b$. However, $Sq^2$ is zero on the cohomology of $M_{2n}\vee M_{2n+1}$.

share|cite|improve this answer
A textbook reference for this argument is Example 4L.7 on page 495 of my Algebraic Topology book. – Allen Hatcher Nov 19 '10 at 11:24

If the identity map from $X$ to itself, considered as a stable map, is killed by $m$, then for any generalized cohomology theory $E$ every reduced cohomology group $E^nX$ is killed by $m$. But when $X=\mathbb RP^2$ and $E$ is real $K$-theory there is a class not killed by $2$: if $L$ is the nontrivial rank $1$ vector bundle then $L\oplus L$ is not stably trivial because its second Stiefel-Whitney class is not $0$.

share|cite|improve this answer

I don't know who first proved this fact, but a standard reference would be this paper of Toda's. It is stated as Theorem 4.1. Toda's proof is essentially the same as Tom's answer.

share|cite|improve this answer
Thanks for the reference. – Jeff Strom Nov 19 '10 at 21:58

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.