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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?

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3 Answers 3

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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 $M_{2n}\vee M_{2n+1}$.

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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 $0$.

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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.

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Thanks for the reference. –  Jeff Strom Nov 19 '10 at 21:58

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