Limit in $\mathbb{Z} / P^{2} \to \mathbb{Z} / P $

Question

Consider the commutative diagram

in Ab, where the morphisms are the canonical ones. Let us denote the top horizontal sequence by A and the bottom one by B. Then the vertical maps induce a morphism of abelian groups $\alpha : Lim A \to Lim B$. I am going to state which of the following statement are correct:

1. $\alpha$ is bijective
2. $\alpha$ is injective but not surjective
3. $\alpha$ is surjective but not injective
4. $\alpha$ is neither injective nor surjective.

However, I am having some trouble seeing what the limits are of both sequences as I am new to homological algebra. I believe the limit of A is $\mathbb{Z}$ but I am not sure. Is this correct? I am also having some trouble seeing what the limit of B is, but I believe it could simply be 0. Is this correct?

Answer 1

The universal property of limits is the following:

Given a set $\{A_i\}_{i\in\Bbb N}$ of objects, and a diagram $$\cdots \to A_3\to A_2\to A_1$$ (where we say $f_i:A_{i+1}\to A_i$), the limit of this diagram is an object $A$ and a set of morphisms $\{g_i:A\to A_i\}_{i\in \Bbb N}$ such that

• $g_i=f_i\circ g_{i+1}$ for all $i\in \Bbb N$
• For any other object $A'$ and morphisms $\{g_i':A'\to A_i\}_{i\in \Bbb N}$, there is a unique morphism $h:A'\to A$ such that $g_i'=g_i\circ h$

Now, I could just tell you what the limits here are. They are well-known and often studied objects. But let's work through it.

You say you suspect that the limit of the top horizontal diagram is $\Bbb Z$. Does it fulfill the above? Well, with $g_i=\operatorname{id}$ it immediately fulfills everything but the last point. What about the last point? Well, we see that for any other $A'$ and $\{g_i'\}$, the commuting requirement gives $g_i=g_{i+1}$, and we see that $h=g_1$ and we indeed have that $\Bbb Z$ is the limit.

As for the bottom diagram, that's not quite as straight-forward. We can solve the injectivity without further studies of the limit itself, though. Let $P$ be the limit, and let $g_i':\Bbb Z\to \Bbb Z/p^i$ be the canonical maps (the vertical maps in your figure). By the last point of the universal property, there is a map $h:\Bbb Z\to P$, and this is the map we're interested in.

Let $x,y\in \Bbb Z$ be such that $h(x)=h(y)$. Thus we get $g'_i(x)=g'_i(y)$ for all $i$. But as $i$ grows, at some point we have $|x-y|<p^i$, and $g_i'(x)=g_i'(y)$ therefore gives $x=y$. So $h$ is injective.

For surjectivity, we first need to work a little more. Take an $x\in\Bbb Z$, and consider the sequence $x_i=g'_i(x)$. We have $f_i(x_{i+1})=[x_{i+1}]=x_i$ as elements of $\Bbb Z/p^i$. We can recover $x$ from any such sequence, because it is, in some sense, eventually constant as $i$ grows: The subsets ${g_i'}^{-1}(x_i)$ of the integers have a non-empty intersection, and that is $x$.

However, let's look at the set $A$ of all sequences $\{a_i\in \Bbb Z/p^i\}_{i\in\Bbb N}$ with the property that $f_i(a_{i+1})=a_i$. Some of them arise from integers, but far from all (consider, for instance, $a_i=\frac{p^i-1}{p-1}$ for any $p\neq 2$). And this $A$ injects into the limit, basically by construction. So clearly there are elements in the limit that don't come from $\Bbb Z$, so the map we're after is not surjective.

As a matter of fact, this $A$ is the limit we're after. It's called the $p$-adic integers. If we write numbers in base-$p$, these look like the integers, except we allow infinitely many digits to the left. The integers themselves have finitely many digits, but, say $a_i=\frac{p^i-1}{p-1}$ corresponds to the $p$-adic number $\ldots1111$.