Torsion and homology with coefficients

Question

This is an Exercise 2.2.41 in Hatcher: From the long exact sequence (l.e.s.) of homology groups associated to the short exact sequence (s.e.s.) of chain complexes $$0 \to C_i(X) \xrightarrow{\times n} C_i(X) \to C_i(X; \mathbb Z_n) \to 0$$ deduce immediately that there are s.e.s. $$0 \to H_i(X) / nH_i(X) \to H_i(X; \mathbb Z_n) \to n-Torsion(H_{i-1}(X)) \to 0$$ where $n-Torsion(G)$ is the kernel of $G \xrightarrow{n} G$, $g \to ng$.

Use this to show that $\tilde H (X; \mathbb Z_p)=0$ for all $i$ and all primes $p$ if and only if $\tilde H_i(X)$ is a vector space over $\mathbb Q$ for all $i$.

My attempt: Consider the l.e.s. $$\ldots \to H_i(X) \xrightarrow{\times n} H_i(X) \xrightarrow{f} H_i(X; \mathbb Z_n) \xrightarrow{\partial} H_{i-1}(X) \xrightarrow{\times n} H_{i-1}(X) \to \ldots$$

Since $Im(\times n) = Kerf$, we have $Kerf= nH_i(X)$. Thus $f$ induces an injection from $H_i(X) / nH_i(X)$ to $H_i(X; \mathbb Z_n)$.

Now, since $Im \partial = Kerg =n-Torsion(H_{i-1}(X))$, we can form s.e.s. $$0 \to H_i(X) /n H_i(X) \xrightarrow{f} H_i(X; \mathbb Z_n) \xrightarrow{\partial} n-Torsion(H_i(X)) \to 0$$ Here, the first exactness is due to injectivity of $f$ and the second one holds because of $\partial (\times n)=0$.

Is there any gap in the proof?

For the only if part of the second statement is immediate because $\mathbb Q$ has no torsion and $H_i(X; \mathbb Q) / nH_i(X; \mathbb Q)=0$

For the if part, I'm stuck. Any help will be appreciated.

Answer 1

For the final part, the vanishing of $\tilde{H}_*(X,\Bbb Z_p)$ means that for all $p$, both that $\tilde H_*(X)=p\tilde H_*(X)$, and that $\tilde H_*(X)$ has no $p$-torsion, that is, multiplication by $p$ on $\tilde H_*(X)$ is both surjective and injective, that is bijective. So there's a unique multiplication by $1/p$ on $\tilde H_*(X)$ for all primes $p$, etc.

Answer 2

I do not understand the part about $f$ being surjective : it is not true, and you do not need it. I would rather write :

Since $Im(\times n) = Kerf$, so $Kerf= nH_i(X)$, thus $f$ induces and injection of $H_i(X) / nH_i(X)$ into $H_i(X; \mathbb Z_n)$, with the same image as $f$.

Now since $Im \partial = Kerg =n-Torsion(H_{i-1}(X))$, $\partial$ induces a surjection from $H_i(X; \mathbb Z_n)$ to $n-Torsion(H_i(X))$ with the same kernel as $\partial$. Thus, we can form a s.e.s. $$0 \to H_i(X) / nH_i(X) \xrightarrow{f} H_i(X; \mathbb Z_n) \xrightarrow{\partial} n-Torsion(H_i(X)) \rightarrow 0.$$ The exactness in the middle is given by $Im(f) = Ker(\partial)$.

For the converse of the second statement, use the s.e.s. and you just write for $n = p$. What does this give you?