Let be a partially ordered set (not necessarily directed) and suppose that for each , we have a group .
Suppose further that for every with , there is a map such that (1) whenever and (2) for all .
Define . is called the inverse or projective limit of the .
- Suppose all of the are group homomorphisms. Prove that is a subgroup of .
- Assume the hypotheses in (a) and let with the usual ordering. For each let be the projection to the th component. Show that if each is surjective, then so is for each .
- Show that if the are all commutative rings with and all the are unital ring homomorphisms then can be given the structure of a commutative ring with such that the are all ring homomorphisms.
- Under the hypotheses in (a), prove that has the following universal property: if is any group such that for each there is a homomorphism with whenever , then there is a unique group homomorphism such that for all .
- Consider . (Our groups are currently nonabelian, so that the identity is denoted by 1.) For all , since is a group homomorphism we have . Thus . In particular, is not empty.
Now let , and consider . Note that if , we have . Thus . By the Subgroup Criterion, .
- Fix and . Now construct an element of as follows. If , let . Let . Now using induction, suppose is defined for all . Let be any element of such that ; such an element exists because is surjective.
I claim that . To see this, let . We proceed by induction on .
For the base case, let . Then . Suppose now that for some , whenever , . Now consider . If , then . If , then . If , then . Thus . Certainly then , so that is surjective. (Note that this construction depends essentially on induction- that is, we needed the index set to be linearly ordered. Question: what if is a larger ordinal?)
- Consider the multiplication on inherited from . Let . Since for all we have , is closed under this operator. Thus in fact is a subring of . Certainly then the are ring homomorphisms.
- Suppose we have such a group and a family of group homomorphisms such that whenever . Define by (that is, the th component of is ). Certainly then , so that for all . Suppose now that we have a homomorphism such that for all . Then for all , we have , so that . Thus , and hence is unique with this property.