In this exercise, we develop the concept of *direct limit*. Let be a nonempty paritally ordered set, and let be a collection of abelian groups. We suppose also that is *directed*; that is, for all , there exists with .

Suppose that for every pair of indices with , there is a map such that the following hold: (1) whenever and (2) for all .

Let be the disjoint union of the . Define a relation on as follows: if and only if there exists such that and .

- Show that is an equivalence relation on . We define .
- Let denote the class of in and define by . Show that if each is injective, then is also injective for all .
- Assume that the are all group homomorphisms. For , show that the operation , where is any upper bound of and , is well defined and makes an abelian group. Deduce that the are group homomorphisms.
- Prove that if all the are commutative rings with and all the are unital ring homomorphisms (that is, they send 1 to 1), then may likewise be given the structure of a commutative ring with such that the are all ring homomorphisms.
- Under the hypotheses of part (c), prove that has the following universal property: if is any abelian group such that for each there is a homomorphism with whenever , then there is a unique homomorphism such that for all .

- To show that is an equivalence, we need to verify that it is reflexive, symmetric, and transitive.
- ( is reflexive) Let . Note that , and that . Thus , and hence is reflexive.
- ( is symmetric) Suppose . Then there exists such that . Certainly , so that . Thus is symmetric.
- ( is transitive) Suppose and . Then there exist such that and such that . Since is a directed poset, there exists such that . Now . Thus , and hence is transitive.

Thus is an equivalence relation.

- Suppose that the are all injective. Choose , and let such that . Then we have . That is, for some , we have . Since is injective, . Thus is injective.
- Suppose that the are all group homomorphisms. First we show that is well defined.
Let and . Then there exists such that and such that . Now (using the directedness of ) choose arbitrary and . Finally, again choose . Now note the following.

= = = = = = = Thus , and we have . Thus is well-defined.

Next we show that is an abelian group.

- ( is associative) Let , , and be in , and let , , and . Then we have the following.

= = = = = = = = So is associative.

- Note that for all , there exists , and since the are group homomorphisms. Thus for all . Let . Now let . Then . Similarly, . Thus is an additive identity element.
- Let . Note that . Thus every element of has an additive inverse.
- Let , and let . Then . Thus is commutative.

Thus is an abelian group. Finally, we show that each is a group homomorphism. Let . Then . Thus is a group homomorphism for all .

- ( is associative) Let , , and be in , and let , , and . Then we have the following.
- Define an operator on as follows: , where is any upper bound of and in . Note the following.
- ( is well defined) Let and . Then there exist and such that and . Using the directedness of , choose , , and . Note the following.

= = = = = = = Thus , and in particular . So is well-defined.

- ( is associative) Let . Let , , and . Then we have the following.

= = = = = = = = Thus is associative.

- ( distributes over ) We will show that distributes over on the left; distributivity on the right is similar. Let , let , and let . Then we have the following.

= = = = = = = = Thus distributes over .

Thus is a ring. Moreover, we have the following.

- (If all the are commutative, then is commutative) Suppose the are all commutative. Let , and let . Then we have the following.

= = = Thus is a commutative ring.

- (If all the have and the are unital, then has a multiplicative identity) Note first that – because the are unital ring homomorphisms – whenever . Thus for all . Define . Now let . Then . Similarly, . Thus is a multiplicative identity in .
- (If for all , then in ) Suppose . Then there exists such that , so that in – a contradiction. Thus in .

Thus if the are commutative rings with , then is a commutative ring with .

- ( is well defined) Let and . Then there exist and such that and . Using the directedness of , choose , , and . Note the following.
- Suppose now that is an abelian group and that we have an indexed family of group homomorphisms such that for all .
- (Existence) Define by . We need to show that is a well defined group homomorphism.
- (Well-defined) Suppose . Then there exists such that . Since latex \rho_{i,j}$ is well defined, . Then , and we have . Hence , and is well defined.
- (Group Homomorphism) Let , and let . Then we have the following.

= = = = = =

So is a group homomorphism. Finally, note that for all and all , . Thus for all .

- (Uniqueness) Suppose that we have a group homomorphism which also satisfies for all . Then for all and all , we have . Thus we have .

- (Existence) Define by . We need to show that is a well defined group homomorphism.