We can construct the Bockstein spectral sequence in the following way. First we start with the exact sequence $0 \rightarrow \mathbb{Z} \rightarrow \mathbb{Z} \rightarrow \mathbb{Z}_p \rightarrow 0$ and tensor this with the (free abelian) chain complex of a topological space $C_*(X)$. This yields the exact sequence $0 \rightarrow C_*(X) \rightarrow C_*(X) \rightarrow C_*(X) \otimes \mathbb{Z}_r$ which induces a long exact sequence via the snake lemma. Taking direct sums we can collapse this into an exact couple and produce derived couples $A_r$, $B_r$ where the $B_r$ are the pages of the spectral sequence. Now I've been reading the wikipedia page for Bockstein spectral sequences and I'm a little bit confused at the last part of their argument.
They take the initial short exact sequence in this post: $0 \rightarrow \mathbb{Z} \rightarrow \mathbb{Z} \rightarrow \mathbb{Z}_p \rightarrow 0$ and then they tensor it with $A_r$. This yields the short exact sequence: $$A_r \rightarrow A_r \rightarrow A_r \otimes \mathbb{Z}_p \rightarrow 0$$ and then they note that the kernel of the first map is just $Tor(A_r, \mathbb{Z}_p)$ which yields the exact sequence: $$0 \rightarrow Tor(A_r, \mathbb{Z}_p) \rightarrow A_r \rightarrow A_r \rightarrow A_r \otimes \mathbb{Z}_p \rightarrow 0.$$ This short exact sequence tells us that, if we call the map between the $A_r$s, $p$, then $ker(p) = Tor(A_r, \mathbb{Z}_p)$ and $coker(p) = A_r \otimes \mathbb{Z}_p$.
But then they assert that after expanding the exact couple back into a long exact sequence we can see:
$$0 \rightarrow (p^{r-1}H_n(X)) \otimes \mathbb{Z}_p \rightarrow B_r \rightarrow Tor(p^{r-1}H_{n-1}(X), \mathbb{Z}_p) \rightarrow 0.$$
Can anyone explain to me, in some decent amount of detail, where this last SES comes from?
If you unfold the exact couple into a long exact sequence, you get $$ ... \to A_r^n \xrightarrow{\times p} A_r^n \xrightarrow{j} B_r^n \xrightarrow{k} A_r^{n-1} \xrightarrow{\times p} A_r^{n-1} \to ... $$ The image of $k$ is the kernel of $\times p$, which is $\mathrm{Tor}(A_r, \mathbb{Z}/p)$. The kernel of $j$ is the image of $\times p$, which is $A_r \otimes \mathbb{Z}/p$. So there is a short exact sequence $$ 0 \to A_r \otimes \mathbb{Z}/p \to B_r \to \mathrm{Tor}(A_r, \mathbb{Z}/p) \to 0. $$ Now what is $A_r$? It's the image of $\times p: A_{r-1} \to A_{r-1}$, and therefore it's the image of the map $\times p^{r-1}$ on $A_1$, the integral homology.