5
$\begingroup$

Let $M$ be a compact connected topological manifold of dimension $n>1$. Suppose the corepresented functor $[M,-]\colon Top_{\ast}\rightarrow Set$ lifts to monoids or equivalently that $M$ is a co-h-space: I must prove this implies that the manifold is simply connected and its homology is given by $H_k(M)\cong \begin{cases} \mathbb{Z} \quad \text{if} \ k \in \{0,n\} \\ 0 \quad \text{else.} \end{cases}$

I don't have much confidence with co-h-spaces but here I really don't know what approach I have to follow: my first idea was to prove that $\pi_1(M)=1$, form this deduce that $M$ is orientable and try to use Poincarè duality to work out the other homology groups. But form the assumption of $M$ to be co-h-space I can only deduce that $\pi_1(M)$ is a free group and nothing else.

Thanks for any help and advice.

$\endgroup$
  • $\begingroup$ Consider $M=S^1$. $\endgroup$ – archipelago Jun 4 '15 at 21:48
  • $\begingroup$ Ok, to avoid trivial cases I can add the assumption n>1. $\endgroup$ – N.B. Jun 4 '15 at 21:50
  • 1
    $\begingroup$ Note that your conclusions imply that $M$ is homeomorphic to $S^n$. $\endgroup$ – archipelago Jun 4 '15 at 22:00
  • 1
    $\begingroup$ If you can show that $M$ is simply connected everything would be fine: A co-H-space has trivial cup products. Using Poincaré duality, one gets the desired structure of the homology groups. $\endgroup$ – archipelago Jun 4 '15 at 22:30
  • 1
    $\begingroup$ @Dan A homology iso between simply connected spaces is a weak equivalence. Every topological manifold has the homotopy type of a CW complex. I don't think I said anything wrong, but would be happy to get corrected. $\endgroup$ – archipelago Jun 5 '15 at 6:27
4
$\begingroup$

I should have found a proof of the statement.

First I state the following claims that are well known results about co-h-spaces (for a reference see the literature cited in the comments).

Claim 1: if X is a co-h-space then $\pi_1(X)$ is free.

Claim 2: if X is a co-h-space then taken $\alpha \in H^{p_1}(X; G_1)$, $\beta \in H^{p_2}(X; g_2)$ the cup product $\alpha \cup \beta=0$ in $H^{p_1+p_2}(X; G_1 \otimes G_2)$.

By 1) we have $\pi_1(M) \cong F_r$ and $H_1(M)\cong \mathbb{Z}^r$. Now we know that avery compact manifold is always $\mathbb{Z}/_{2 \mathbb{Z}}$ orientable so we can use the Poincarè duality with coefficients mod 2. It can be proved that if the PD holds then for M without boundary connected compact $R$-orientable manifold then the pairing $H^{k}(M;R)\otimes H^{n-k}(M;R)\rightarrow R$ given by $\alpha \otimes \beta \mapsto\langle \alpha \cup \beta, [M]\rangle$ is perfect.

Therefore we have using 2) for $n>2$ that $H^k(M; \mathbb{Z}/_2)=0$ if $0<k<n$. Now consider the s.e.s. $0 \rightarrow \mathbb{Z} \xrightarrow{2}\mathbb{Z}\rightarrow \mathbb{Z}/_{2 \mathbb{Z}}\rightarrow 0$ and by the associated Bockstein sequence we have $0=H_2(M;\mathbb{Z}/_{2 \mathbb{Z}})\rightarrow H_1(M)\xrightarrow{2}H_1(M)\rightarrow H_{1}(M; \mathbb{Z}/_{2 \mathbb{Z}})=0$ (we need the assumption $n>2$ to get $H_2=H_1=0$ by the PD).

We have proved therefore $\mathbb{Z}^r \xrightarrow{2} \mathbb{Z}^r$ is an iso implying $r=0$ and the fundamental group of $M$ is trivial. This implies that $M$ is $\mathbb{Z}$-orientable (i.e. the classical orientation) and now we can use PD with integer coefficients. Reasoning just as before (use the pairing and claim 2)) we get $H_k(M)=0$ for $0 <k<n$ and trivially $H_0(M)=H_n(M)=\mathbb{Z}$ proving finally the statement of the exercise.

For the case $n=1,2$ we can easily use the classification theorems of $1,2$-dimensional manifolds to get the claim.

If $n=1$ being $M$ compact $M$ must be homeomorphic to the circle $S^1$ which, being the reduced suspension of $S^0$ is a co-h-space.

If $n=2$ recall we proved $H_1(M)=\mathbb{Z}^r$ thus $M$ cannot be a not-orietable surface (since their $H_1$ has torsion): we have $M \cong \Sigma_g$ for some $g \in \mathbb{N}$. We have therefore that $M$ is orientable and as above we argue $H_1(M)\cong H^{1}(M)=0$ leaving the sphere $S^2$ as only possible candidate but $S^2= \Sigma S^1$ a reduced suspension then it is actually a co-h-space.

$\endgroup$
  • $\begingroup$ Well done! I hope my comments have been helpful. $\endgroup$ – archipelago Jun 7 '15 at 20:46
  • $\begingroup$ Indeed: thanks for your help. $\endgroup$ – N.B. Jun 7 '15 at 20:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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