For a sphere $S^n$, the diagonal map $\Delta:S^n\to S^n\wedge S^n$ sending $x\mapsto x\wedge x$ is null-homotopic. This is the homotopy group $\pi_n(S^n\wedge S^n)=\pi_n(S^{2n})$ is trivial.

I'm wondering if there are other examples of this phenomenon. That is, is there a non-contractible based space $X$ which is not a sphere, such that the diagonal map $\Delta:X\to X\wedge X$ is null-homotopic?

  • $\begingroup$ Any contractible space? $\endgroup$ Feb 22, 2011 at 6:52

3 Answers 3


Questions like this are well studied by homotopy theorists, in particular those working in Lusternik-Schnirelmann category (warning: nothing to do with the objects and morphisms kind of category!). In particular, what you are asking for is examples of spaces with weak category equal to 2. The weak category of a pointed space $X$ is the least $k$ such that the reduced diagonal $\triangle_k\colon X\to X\wedge X\wedge\cdots\wedge X = X^{\wedge k}$ is null-homotopic. The weak category was introduced by Berstein and Hilton in this paper, as an approximating invariant for LS-category: note that $\mathrm{wcat}(X)\leq\mathrm{cat}(X)$ for any space. (Warning 2: Nowadays, the prevailing convention seems to be that these invariants are normalised, so that the weak category of a contractible space is zero.)

Note that the weak category of any (non-contractible) suspension $\Sigma Y$ is $2$. In the paper of Berstein and Hilton linked above they construct spaces $Z$ with weak category $2$ and LS-category $3$ (which therefore cannot be suspensions).

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    $\begingroup$ To make the conclusion explicit: if $X=\Sigma Y$, then $\Delta:X\to X\wedge X$ is nullhomotopic. $\endgroup$ Feb 22, 2011 at 11:06
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    $\begingroup$ And the converse is false. $\endgroup$
    – Mark Grant
    Feb 22, 2011 at 16:45
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    $\begingroup$ @Mark: But see John Klein's answer for a case when the converse is true. $\endgroup$
    – Tom Church
    Feb 23, 2011 at 2:18
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    $\begingroup$ About the first warning: Also nothing to do with Baire category. $\endgroup$ Dec 13, 2022 at 23:03

Take an $n$-connected CW complex $X$ of dimension $<2n$. Then $X \wedge X$ is $2n$-connected, thus the diagonal map has to be null-homotopic.


To add to the above: an $r$-connected CW complex $X$ of dimension $\le 3r$ has the homotopy type of a suspension $\Leftrightarrow$ its reduced diagonal map $X \to X \wedge X$ Is null-homotopic. This is the Berstein-Hilton-Ganea theorem.


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