## Exact Sequences and Homology (Pt. IV)

**Point of Post: **This is a continuation of this post.

Ok, now on to the general case. What the above really told us was that to prove that we get a well-defined map we really need to show that and that . Of course, there is no containment in general abelian categories, and thus we need to replace these containments by finding canonical maps and canonical maps .

We begin by constructing a map . To do this we consider the diagram

Where it should be obvious what the maps are. Now, note that and so by the definition of the kernel we get a unique arrow such that .

To construct the arrow we use the trick that since we are dealing with abelian categories we only have to define a map . To do this we consider the diagram

So, if we can show that . That said, we can put in the following maps

$latex\text{ }$

Note then that by construction we have commutativity of that square and so since by definition. Thus, we get a well-defined arrow and thus using the natural isomorphisms which comes from the definition of an abelian category we get maps . Now, finally we consider the following diagram

This diagram is commutative, whicn enables one to quickly check that the composition

is zero, which (by definition of cokernels) gives us a map which we shall denote , , or just when the is clear.

One can check, of course, that if we are dealing with that as one would hope.

It’s easy to run through the construction of to find that and . Moreover, note that to compute one has to find , but it’s easy to check that this is just

Which tells us that . Finally, noting the obvious fact that we may conclude that:

**Theorem: ** is an additive functor .

We call the * homology functor.*

*Remark: *The above was absolutely horrible. Namely, while doing the construction this way is nice in that it is a) actually explicit in the sense that we produce the arrows we are looking for and b) shows a very theoretic, categorical way in which is defined it is just a horrid diagram chase. So, why did I actually do it? Due to the metatheorem concerning abelian categories one can “theoretically” just prove this result for for all rings , note that this construction only involves “local” ideas, and say that we can therefore pull the result from . While this is an invaluable tool/idea, one which we shall be using often, it is good to see (at least once) that one can (if one is determined enough) actually “construct” the maps that the metatheorem guarantees us. Anyways, besides the somewhat messy nature of the chase it is actually quite simple if one merely keeps the case in mind and follows their nose.

**References**

[1] Weibel, Charles A. *An Introduction to Homological Algebra*. Cambridge [England: Cambridge UP, 1994. Print.

[2] Schapira, Pierre. “Categories and Homological Algebra.” Web. <http://people.math.jussieu.fr/~schapira/lectnotes/HomAl.pdf>.

[3] Rotman, Joseph. *An Introduction to Homological Algebra*. Dordrecht: Springer, 2008. Print.

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