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Assertion: The fundamental group of a space which has an $H$- and a co-$H$ structure is trivial or infinite cyclic.

Why this is true?

(added conditions: one should probably assume that the space is connected and has the homotopy type of a CW complex)

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    $\begingroup$ What is Z(integer)? is it $\mathbb{Z}$? Or $\mathbb{Z}(n)$ for some integer $n$? Anyway, what are some examples of such spaces? This question smells like homework to me (it is certainly at the level of a graduate text). $\endgroup$
    – David Roberts ♦
    Feb 26, 2011 at 3:59
  • $\begingroup$ Are you sure this is a homework problem? I have never seen this statement, though, admittedly, I never cared about it. $\endgroup$
    – Johannes Ebert
    Feb 26, 2011 at 11:22
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    $\begingroup$ Please provide some background and motivation for your questions: gloss your notation and definitions, and explain a little about what you already know and why you want an answer to this question. Also, please clean up spelling and capitalization and so on. $\endgroup$
    – Theo Johnson-Freyd
    Feb 26, 2011 at 16:55
  • $\begingroup$ If these answers are sufficient for your purposes, why not accept one of them? $\endgroup$
    – Yemon Choi
    Oct 2, 2011 at 2:00

2 Answers 2

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Another argument which does it:

A path connected based space $X$ is a co-$H$ space if and only if the evaluation map $\Sigma \Omega X \to X$ admits a section up to homotopy. This will imply that $\pi_1(X)$ is a retract of $\pi_1(\Sigma \Omega X)$, and it's easy to see that the latter has a free fundamental group (generated by elements of $\pi_1(X)$). Now use the fact that a subgroup of a free group is free.

The rest of the argument goes as Jeff says: the fundamental group of an $H$-space is abelian.

Addendum

If in addition we assume all $X$ appearing above have the homotopy type of a finite complex, then we can actually give a complete classification.

First of all, the existence of an $H$-space structure implies that $X$ satisfies Poincare duality (this is well-known and can be deduced from the structure theorem for connected Hopf algebras; I also gave an independent proof in one of my papers--I can't remember which one).

Secondly, since $X$ is also co-$H$, there are no non-trivial cup products of classes in positive degrees.

If we combine these two statements (duality + no cup products) we infer that $X$ has cohomology concentrated in degrees $0$ and some integer $n$. If $X$ isn't a contractible, it follows from duality that $n > 0$ and $H_n(X) = \Bbb Z$. It follows that $X$ is a cohomology $n$-sphere.

Since $X$ is an $H$-space it is a simple space, so $X$ is then a homotopy $n$-sphere (proof: if $n = 1$ represent the generator of $\pi_1$ by a map $S^1 \to X$, this will be a homology isomorphism and then apply the Whitehead theorem. If $n > 1$ use the fact that $X$ must be simply connected and then use the Hurewicz theorem to represent the generator of $H_n(X)$ by a map $S^n \to X$ and argue as in the $n=1$ case).

Now, it is well known (using say, Hopf invariant one) that the $n$-sphere is an $H$-space iff $n = 0, 1,3,7$, so we've shown:

Theorem: If $X$ is both an $H$-space and a co-$H$ space and has the homotopy type of a connected finite CW complex, then $X$ has the homotopy type of a point or an $n$-sphere, where $n$ = 1,3,7.

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  • $\begingroup$ The result in the question may still hold in the full generality that is claimed here but it seems to me both arguments that $\pi_{1}(X)$ is free for path connected, co-H $X$ require an extra restriction on $X$ (locally simply connected at the basepoint does it). Otherwise you can take $X$ to be the reduced suspension of the 1-pt compactification of a countable discrete space. This is the Hawaiian earring and its fundamental group is not free. $\endgroup$
    – Jeremy Brazas
    Feb 27, 2011 at 18:09
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    $\begingroup$ I'm an algebraic topologist. I try to avoid such pathological cases. Everything for me has the homotopy type of a CW complex. $\endgroup$
    – John Klein
    Feb 27, 2011 at 20:44
  • $\begingroup$ Fair enough. I did not read this into "space." $\endgroup$
    – Jeremy Brazas
    Feb 27, 2011 at 22:20
  • $\begingroup$ @John. I don't know that how prove that $π_1(\Sigma \Omega X)$ is free. Please explain for me. $\endgroup$
    – Jino
    Feb 28, 2011 at 12:06
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    $\begingroup$ @Jino: The fundamental group of any suspension is free: first of all, if $T$ is a discrete based set, then $\Sigma T$ is a wedge of circles, so its fundamental group is free. Secondly, if $X\to Y$ is an $r$-connected map, then $\Sigma X \to \Sigma Y$ is $(r+1)$-connected. Thirdly, the map $Z\to \pi_0(Z)$ is 1-connected for any non-empty space Z, so the map $\Sigma Z\to \Sigma \pi_0(Z)$ is 2−connected. In particular, it's an isomorphism of fundamental groups. Finally,take $Z = \Omega X$ to get the result. $\endgroup$
    – John Klein
    Feb 28, 2011 at 18:07
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If $X$ is co-H, then $\pi_1(X)$ must be a free group. If $X$ is H, then $\pi_1(X)$ must be abelian. The only free group that is abelian is $\mathbb{Z}$.

Argument for the first assertion: The co-H structure (and Van Kampen) gives a factorization $$ \pi_1(X)\xrightarrow{i_*} \pi_1(X)*\pi_1(X)\xrightarrow{j_*} \pi_1(X)\times\pi_1(X) $$ of the map $\Delta_*$ induced by the diagonal $\Delta:X\to X\times X$. This shows that $\pi_1(X)$ is isomorphic to a subgroup of $G = (j_*)^{-1}( \mathrm{im}(\Delta_{*}))$, which is free on the elements $\{ x \bar x\}$ ($x$ and $\bar x$ represent the same element $x\in\pi_1(X)$ in the two summands of $\pi_1(X)*\pi_1(X)$). Now we are done because a subgroup of a free group is free.

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    $\begingroup$ That makes for a fun computation of the fundamentak group of $S^1$, provided one can show it is not zero. The least illuminating computation, I guess! :) $\endgroup$
    – Mariano Suárez-Álvarez
    Feb 26, 2011 at 21:55
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    $\begingroup$ @Jeff: Do you know of any examples of connected spaces which have both H and co-H structures, other than $S^1$? $\endgroup$
    – John Klein
    Feb 27, 2011 at 0:48
  • $\begingroup$ @John: Not that I can think of. I guess its homology with field coefficients would be a Hopf algebra with trivial diagonal, so that narrows it down considerably. $\endgroup$
    – Jeff Strom
    Feb 27, 2011 at 2:04
  • $\begingroup$ @Jeff. I don't know that your first answer : when $X$ is co-H-group, then $\pi_1(X)$ is free. How can proof this? $\endgroup$
    – Jino
    Feb 27, 2011 at 11:04
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    $\begingroup$ I guess $S^3$ and $S^7$ also qualify? But I can't think of any other examples. $\endgroup$
    – Mark Grant
    Feb 27, 2011 at 11:19

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