Prove that there is a long exact sequence $\ldots → H_k(X) \overset{2}\to H_k(X) → H_k(X;\Bbb Z/2) → H_ {k−1} (X) → · · ·$

Question

Consider the exact sequence $0 \to \Bbb Z \overset{2}{\to} \Bbb Z → \Bbb Z/2 → 0$. Prove that there is a long exact sequence

$\ldots → H_k(X) \overset{2}{\to} H_k(X) → H_k(X; \Bbb Z/2) → H_ {k−1} (X) → · · ·$

Here $H_k(X)$ are the kth Homology groups (mentioning this following the comment by AndreasBlass) .

(Pardon my editing skills in Exact sequences , it's simply the map $2 \times$ homomorphism from the left most object to the next object )

Let, $0 → \Bbb Z \overset{2}{\to} \Bbb Z → \Bbb Z/2 → 0 \ldots (*)$

Then tensoring $(*)$ with $H_k(X)$ gives, $$H_k(X) \otimes\Bbb Z \overset{2}{\to} H_k(X) \otimes\Bbb Z →H_k(X) \otimes \Bbb Z/2 → 0 $$ $$\implies H_k(X) \overset{2}{\to} H_k(X) →H_k(X) \otimes \Bbb Z/2 → 0 \ldots(**)$$

But we also have from Universal coefficient theorem, $$0\to H_k(X) \otimes \Bbb Z/2 \to H_k(X;\Bbb Z/2) \to Tor(H_{k-1}(X);\Bbb Z/2) \to 0 \dots(***)$$

Then I tried combining $(**)$ and $(***)$,$$H_k(X) \overset{2}{\to} H_k(X) \to H_k(X;\Bbb Z/2) \to Tor(H_{k-1}(X);\Bbb Z/2) \to 0 $$

First of all, I don't know whether at all this makes any sense, anyway I am stuck here!

Thanks in advance for help!

Answer 1

What you are left to show is that $Tor(H_{k−1}(X);ℤ/2)= H_{k-1}(X)$

Lemma: if G is abelian, then $Tor(G;\mathbb{Z}/n) = \ker(G\overset{n}{\to} G)$.

By lemma $Tor(H_{k-1};\mathbb{Z}/2)= \ker(H_{k-1}\overset{2}{\to} H_{k-1})$

We get the exact sequence $ H_k(X) \overset{2}{\to} H_k(X) \to H_k(X; \mathbb{Z}/2) \to Tor(H_{k-1}(X);\mathbb{Z}/2)\to H_{k-1}(X) \overset{2}{\to} H_{k-1}(X)\to ...$

You get a map $\phi: H_k(X; \mathbb{Z}/2) \to H_{k-1}(X)$. The image of $\phi$ is "Tor" which is the kernel of $\overset{2}{\to}$