(5-lemma) Consider a commutative diagram in a fixed abelian category of the form
where the top and bottom are exact sequences. For simplicity we denote all the differentials in both exact sequences by .
Then:
1) If and are epis and is mono, then is epi. If
2) and are monic and is epi, then is mono.
1+2) If and are isos, is epi, and is mono, then is iso.
Proof. We can embed into a category of left modules over a fixed ring (though this requires the category to be small, one can always take a smaller abelian subcategory containing the morphism in the diagram which is small). Then we can do the diagram chasing using elements in that setup. We prove only 1) as 2) is dual.
The proof of 1) is by contradiction. Suppose is not epi, hence there is which is not in the image of . Since is epi, one can choose an element such that . Now $. Since is monic that means that as well. By the exactness of the upper row, that means there is such that , hence also . We would like that be equal to but this is not so, we just see that and hence by exactness of the lower row there is such that . Since is also epi, there is such that . Now is such that
with contradiction.
Created on February 17, 2010 at 20:41:16. See the history of this page for a list of all contributions to it.