This is from an exercise in algebraic topology. I am given a topological space $X$ and I am trying to show that the reduced homology modules $\tilde H(X)$ are uniquely divisible if $$\tilde H(X, \mathbb{Z}_p)=0 , p \text{ prime}.$$ The first step is to show that \begin{align*} 0\to S_*(X)\to S_*(X) \to S_*(X;\mathbb{Z}/m\mathbb{Z})\to 0 \end{align*} is short exact, which is pretty easy considering the maps $m, \pi$, multiplication and projection respectively. Here $S_*(\cdot)$ denotes the singular chain complex. From this, I guess, I can extract a long exact sequence of homology modules, using an exact triangle construction, but I am not sure how to prove that \begin{align*} 0\to H_n(X)/mH_n(X) \to H_n(X;\mathbb{Z}/m\mathbb{Z})\to T_m(H_{n-1}(X))\to 0 \end{align*} is exact, nor how to finish the proof of unique divisibility.

Remark: Unique divisibility means for every $m\in\mathbb{N}, g\in G$ $!\exists$ $g'$ such that $mg'=g$.

I would much appreciate any help or other hints as to how to finish this problem.


Look at the long exact homology sequence induced by your short exact sequence of chain complexes. You get exactness of

$\rightarrow H_n(X)\xrightarrow{\times p}H_n(X)\xrightarrow{\pi_*}H_n(X;\mathbb{Z}_p)\xrightarrow{\Delta}H_{n-1}(X)\xrightarrow{\times p}H_{n-1}(X)\rightarrow \dots$

after identifying the map induced by multiplication by $p$ with another multiplication by $p$ map. From the above you get a short exact sequence

$0\rightarrow \ker \Delta_*\rightarrow H_n(X;\mathbb{Z}_p)\rightarrow im\,\Delta\rightarrow 0$

and we have $\ker\Delta_*=im\,\pi_*\cong H_n(X)/(\ker\pi_*)\cong H_n(X)/p\cdot H_n(X)$ immediately. Now calculate $Tor(H_{n-1}(X),\mathbb{Z}_p)$ by taking the free resolution $0\rightarrow\mathbb{Z}\xrightarrow{\times p}\mathbb{Z}\rightarrow \mathbb{Z}_p\rightarrow 0$, tensoring with $H_{n-1}(X)$ and taking the kernel as usual. You get $Tor(H_{n-1}(X),\mathbb{Z}_p)=\ker(H_{n-1}(X)\xrightarrow{\times p}H_{n-1}(X))\cong im\,\Delta$ and with this you get the short exact sequence you need.

Now if $H_n(X;\mathbb{Z}_p)=0$ then from the exact sequence we must have that $H_n(X)/p\cdot H_n(X)=0$. Hence for each $x\in H_n(X)$ there exists some $x'\in H_n(X)$ such that $x=p\cdot x'$, so $H_n(X)$ is p-divisible. Now assume that $y'\in H_n(X)$ is another element with $p\cdot y'=x$. Then $p\cdot (x'-y')=x-x=0$ so the class $(x'-y')$ gives an element in $=\ker(H_{n}(X)\xrightarrow{\times p}H_n(X))=Tor(H_n(X),\mathbb{Z}_p)$. But since $H_{n+1}(X;\mathbb{Z}_p)=0$, our short exact sequence gives us that this tor group is trivial. We conclude that $x'=y'$ and therefore that $x$ is uniquely p-divisible. Since $x$ was arbitrary we get the proposition.

Sorry I wasn't less explicit with that.

  • $\begingroup$ Shouldn't $Im(\times p)=p\cdot H_i(X)$? $\endgroup$
    – user413923
    Dec 8 '19 at 3:59
  • $\begingroup$ @Daniel, rather the mistake was a typo a line earlier. Thanks for spotting the mistake. Check that you agree with everything now. $\endgroup$
    – Tyrone
    Dec 8 '19 at 16:32

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