## Direct Limits of Directed Sets

**Point of Post: **In this post we make note of a simple fact about what the direct limit of a directed set over a directed system looks like.

*Motivation*

Most of the examples we have discussed up until this point concerning direct limits have involved directed systems not over the bare minimum preordered set–no most of the time they are over directed sets. So what? Why does being over a directed set make anything better? Well, that is the content of the post. Namely, we shall see that if we take the directed limit of a directed system over a directed set (opposed to a general preordered set) we know that all the elements “look nice” (in a sense to soon be made clear). Of course, this begs the question as to why, considering this new information, we wouldn’t just from the get-go restrict our attention to directed systems over directed sets? Well, put bluntly, we’ll miss out on some good stuff. In other words, allowing us to considered directed systems over general preordered sets (and their subsequent direct limits) allows us to bring under the umbrella of “direct limit” some ‘degenerate’ cases which are very useful. First and foremost in my mind is the coproduct of a set of modules which is obtained by defining the trivial preorder on . Thus, we don’t want to throw out the possibility of discussing direct limits of directed sets over preordered sets, but we’d at least like (since, as we said, ‘most times’ the preordered set is really a directed set) know how to reap the benefits of the additional properties of directed sets when they arise.

*Direct Limits Over Directed Sets*

What we really want to show is that every element of a direct limit when considering directed systems over directed sets “looks nice” in the sense that it’s the image of the of the put-in maps . Put more seriously:

**Theorem: ***Let be a directed set and a directed system over . Then, if is a direct limit of this system, then for every there exists some such that for some .*

**Proof: **Since the unique isomorphism between and satisfies the appropriate identities between and it suffices to prove this for . To do this we note that by definition every element of can be written as . But, since is directed we can find some with for each . But, we know that and so it easily follows that

**References:**

[1] Dummit, David Steven., and Richard M. Foote. *Abstract Algebra*. Hoboken, NJ: Wiley, 2004. Print.

[2] Rotman, Joseph J. *Advanced Modern Algebra*. Providence, RI: American Mathematical Society, 2010. Print.

[3] Blyth, T. S. *Module Theory.* Clarendon, 1990. Print.

[4] Lang, Serge. *Algebra*. Reading, MA: Addison-Wesley Pub., 1965. Print.

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