Point of Post: In this post we discuss when certain mapping triangles of modules may be completed.
The concept I’d like to discuss in this post is the more theoretical way of viewing mappings, in regards to “factoring through” (i.e. completing a mapping triangle). Said differently, given mappings in and out of a given module, say and when can we find a morphism which makes the resulting triangular diagram commute? We shall see that this will be of great importance to us when we start getting into some of the more advanced module theoretic topics. It mostly makes proofs easier by being able to, given a diagram, just decide whether or not we can complete it.
So, let’s get the basic set-up down. Suppose that we have three left -modules and we have the following diagram
where and (where we use, as is customary, the double-headed arrow for epimorphism). We’d like to know when we can complete this diagram in the sense that we can find a map such that the resulting diagram
Note that evidently if this were true, then given we have that and so . In other words, it’s clear that a necessary condition for the existence of such an is that . Conversely, if then the ‘obvious’ way to define a map works. Namely, we know how to get from to , and every element of is associated to an element of , namely for some . So, in a perfect world we’d just define . Of course, the problem with this is well-definedness. But, the cool thing is that if then and so so that . So, the map is well-defined and clearly satisfies . Thus, it remains to check that is an – morphism. To see this we merely note if then so that, by independence of choice, we have that . Similarly, if and then and so by well-definedness . Summing this all up gives:
Theorem: If is a ring and are left -modules. Then, the diagram may be completed to a diagram if and only if .
Of course, we know categorically that we should be able to replace the above theorem by a dual theorem by replacing all the arrows, switching inclusions, and switching to . Indeed, suppose we tried to complete the following diagram
(where we use for monomorphism) to a diagram of the form
As last time there is clearly a necessary condition, for if then and so . Thus, we see that it’s necessary that. Conversely, if this is true it’s clear that the ‘obvious’ way of defining a map should work. Namely, given any we know that and since this implies that for some , and since is injective we know that this is unique. We can then define . It’s clear that this is well-defined. More formally what we have done is used the fact that since is injective it posseses a let inverse and so what we are doing is defining by where we made vital use of . Now, since evidently it suffices to check that is a morphism. To see this it clearly suffices to prove that is a -morphism since then is the composition of -morphisms. To do this we merely note that if then so that . Similarly, if and then so that .
Theorem: Let be a ring and left -modules. Then a diagram may be extended to a diagram i and only if .
 Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.
 Rotman, Joseph J. Advanced Modern Algebra. Providence, RI: American Mathematical Society, 2010. Print.
 Blyth, T. S. Module Theory. Clarendon, 1990. Print.