## Basic properties of inverse limits

Let $I$ be a partially ordered set (not necessarily directed) and suppose that for each $i \in I$, we have a group $A_i$.

Suppose further that for every $i,j \in I$ with $i \leq j$, there is a map $\mu_{j,i} : A_j \rightarrow A_i$ such that (1) $\mu_{j,i} \circ \mu_{k,j} = \mu_{k,i}$ whenever $i \leq j \leq k$ and (2) $\mu_{i,i} = 1$ for all $i \in I$.

Define $\varprojlim A_i = \{ (a_i) \in \prod_I A_i \ |\ \mu_{j,i}(a_j) = a_i\ \mathrm{whenever}\ i \leq j \}$. $\varprojlim A_i$ is called the inverse or projective limit of the $A_i$.

1. Suppose all of the $\mu_{j,i}$ are group homomorphisms. Prove that $\varprojlim A_i$ is a subgroup of $\prod_I A_i$.
2. Assume the hypotheses in (a) and let $I = \mathbb{N}$ with the usual ordering. For each $i \in I$ let $\mu_i : \varprojlim A_i \rightarrow A_i$ be the projection to the $i$th component. Show that if each $\mu_{j,i}$ is surjective, then so is $\mu_i$ for each $i$.
3. Show that if the $A_i$ are all commutative rings with $1 \neq 0$ and all the $\mu_{j,i}$ are unital ring homomorphisms then $\varprojlim A_i$ can be given the structure of a commutative ring with $1 \neq 0$ such that the $\mu_i$ are all ring homomorphisms.
4. Under the hypotheses in (a), prove that $\varprojlim A_i$ has the following universal property: if $D$ is any group such that for each $i \in I$ there is a homomorphism $\pi_i : D \rightarrow A_i$ with $\pi_i = \mu_{j,i} \circ \pi_j$ whenever $i \leq j$, then there is a unique group homomorphism $\pi : D \rightarrow \varprojlim A_i$ such that $\mu_i \circ \pi = \pi_i$ for all $i$.

1. Consider $(1) \in \prod_I A_i$. (Our groups are currently nonabelian, so that the identity is denoted by 1.) For all $i \leq j$, since $\mu_{j,i}$ is a group homomorphism we have $\mu_{j,i}(1) = 1$. Thus $(1) \in \varprojlim A_i$. In particular, $\varprojlim A_i$ is not empty.

Now let $(a_i),(b_i) \in \varprojlim A_i$, and consider $(a_ib_i^{-1}) \in \prod_I A_i$. Note that if $i \leq j$, we have $\mu_{j,i}(a_jb_j^{-1}) = \mu_{j,i}(a_j)\mu_{j,i}(b_j)^{-1} = a_ib_i^{-1}$. Thus $(a_i)(b_i)^{-1} \in \varprojlim A_i$. By the Subgroup Criterion, $\varprojlim A_i \leq \prod_I A_i$.

2. Fix $k \in \mathbb{N}$ and $a \in A_k$. Now construct an element $\alpha = (a_i)$ of $\prod A_i$ as follows. If $i < k$, let $a_i = \mu_{k,i}(a)$. Let $a_k = a$. Now using induction, suppose $a_i$ is defined for all $k \leq i \leq t$. Let $a_{t+1}$ be any element of $A_{t+1}$ such that $\mu_{t+1,t}(a_{t+1}) = a_t$; such an element exists because $\mu_{t+1,t}$ is surjective.

I claim that $\alpha \in \varprojlim A_i$. To see this, let $i \leq j$. We proceed by induction on $j$.

For the base case, let $j = i = 0$. Then $\mu_{j,i}(a_j) = a_i$. Suppose now that for some $t \geq 0$, whenever $i \leq j \leq t$, $\mu_{j,i}(a_j) = a_i$. Now consider $\mu_{t+1,i}(a_{t+1})$. If $t+1 < k$, then $\mu_{t+1,i}(a_{t+1}) = \mu_{t+1,i}(\mu_{k,t+1}(a))$ $= \mu_{k,i}(a) = a_i$. If $t+1 = k$, then $\mu_{t+1,i}(a_{t+1}) = \mu_{k,i}(a) = a_i$. If $t+1 > k$, then $\mu_{t+1,i}(a_{t+1}) = \mu_{t,i}(\mu_{t+1,t}(a_{t+1})$ $= \mu_{t,i}(a_t)$ $= a_i$. Thus $\alpha \in \varprojlim A_i$. Certainly then $\mu_k(\alpha) = a$, so that $\mu_k$ is surjective. (Note that this construction depends essentially on induction- that is, we needed the index set $\mathbb{N}$ to be linearly ordered. Question: what if $I$ is a larger ordinal?)

3. Consider the multiplication on $\varprojlim A_i$ inherited from $\prod A_i$. Let $(a_i),(b_i) \in \varprojlim A_i$. Since for all $i \leq j$ we have $\mu_{j,i}(a_jb_j) = \mu_{j,i}(a_j) \mu_{j,i}(b_j) = a_ib_i$, $\varprojlim A_i$ is closed under this operator. Thus in fact $\varprojlim A_i$ is a subring of $\prod A_i$. Certainly then the $\mu_i$ are ring homomorphisms.
4. Suppose we have such a group $D$ and a family of group homomorphisms $\pi_i : D \rightarrow A_i$ such that $\pi_i = \mu_{j,i} \circ \pi_j$ whenever $i \leq j$. Define $\pi : D \rightarrow \varprojlim A_i$ by $\pi(x) = (\pi_i(x))$ (that is, the $i$th component of $\pi(x)$ is $\pi_i(x)$). Certainly then $(\mu_i \circ \pi)(x) = \mu_i((\pi_i(x)))$ $= \pi_i(x)$, so that $\mu_i \circ \pi = \pi_i$ for all $i$. Suppose now that we have a homomorphism $\psi : D \rightarrow \varprojlim A_i$ such that $\mu_i \circ \psi = \pi_i$ for all $i$. Then for all $x \in D$, we have $\mu_i(\psi(x)) = \pi_i(x)$, so that $(\pi(x))_i = \pi_i(x) = (\psi(x))_i$. Thus $\psi = \pi$, and hence $\pi$ is unique with this property.