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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.

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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.

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  • $\begingroup$ 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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