Explanation of example 3F.7 in Hatcher The section I am refering to is the following example on page 314 of Hatcher's Algebraic Topology:

I'm a bit confused by his statement about relations and can't quite see what he is trying to say. Could someone help me interpret this example? I understand his approach, it's just a bit terse when he says "the terms after $y_1(1,0,...)$ ... and in particular allow one to reduce an arbitrary sequence $(b_0,b_1,\cdots )$ to a unique sequence"
 A: Here's a different way of getting the same result, if it helps...
The maps $\mathbb{Z} \rightarrow \mathbb{Z}/p^j$ give a map of diagrams between $\cdots \rightarrow \mathbb{Z} \stackrel{1}{\rightarrow} \mathbb{Z}\stackrel{1}{\rightarrow} \mathbb{Z}$ and $\cdots \rightarrow \mathbb{Z}/p^3 \rightarrow \mathbb{Z}/p^2 \rightarrow \mathbb{Z}/p$ with kernel given by the diagram $\cdots \rightarrow p^3\mathbb{Z} \rightarrow p^2 \mathbb{Z} \rightarrow p\mathbb{Z}$. Identifying each of these with $\mathbb{Z}$ via multiplication by $p^n$ we get the inverse system in the example. So we have an exact sequence of inverse systems which we'll denote by $0 \rightarrow K \rightarrow A \rightarrow B \rightarrow 0$ which gives rise to an exact sequence (essentially by definition of derived functor...)
$$
0 \rightarrow \text{lim }K \rightarrow \text{lim }A \rightarrow \text{lim }B \rightarrow \text{lim}^1 K \rightarrow \text{lim}^1 A
$$
Now, $\text{lim }B = \mathbb{Z}_p$, by definition, $\text{lim}^1A = 0$ since all the maps are the identity, $\text{lim }K = 0$ as Hatcher says (nothing is infinitely divisible by $p$), and $\text{lim }A = \mathbb{Z}$ by observation, so we have
$$
0 \rightarrow \mathbb{Z} \rightarrow \mathbb{Z}_p \rightarrow \text{lim}^1K \rightarrow 0
$$
which is what we wanted to show.
