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 ith 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 ith 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.
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