Zoran Skoda
five lemma

(5-lemma) Consider a commutative diagram in a fixed abelian category of the form

A 1 A 2 A 3 A 4 A 5 f 1 f 2 f 3 f 4 f 5 B 1 B 2 B 3 B 4 B 5\array{ A_1 & \to & A_2 & \to & A_3 & \to & A_4 &\to & A_5\\ \downarrow f_1 &&\downarrow f_2 &&\downarrow f_3 &&\downarrow f_4 &&\downarrow f_5 \\ B_1 & \to & B_2 & \to & B_3 & \to & B_4 &\to & B_5 }

where the top and bottom are exact sequences. For simplicity we denote all the differentials in both exact sequences by dd.

Then:

1) If f 2f_2 and f 4f_4 are epis and f 5f_5 is mono, then f 3f_3 is epi. If

2) f 2f_2 and f 4f_4 are monic and f 1f_1 is epi, then f 3f_3 is mono.

1+2) If f 2f_2 and f 4f_4 are isos, f 1f_1 is epi, and f 5f_5 is mono, then f 3f_3 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 f 3f_3 is not epi, hence there is bB 3b\in B_3 which is not in the image of f 3f_3. Since f 4f_4 is epi, one can choose an element a 4A 4a_4\in A_4 such that f 4(a 4)=d(b)f_4(a_4) = d(b). Now $0=d 2b=df 4(a 4)=f 5d(a 4)0 = d^2 b = d f_4 (a_4) = f_5 d (a_4). Since f 5f_5 is monic that means that da 4=0d a_4 = 0 as well. By the exactness of the upper row, that means there is a 3A 3a_3\in A_3 such that da 3=a 4d a_3 = a_4, hence also df 3(a 3)=f 4d(a 3)=f 4(a 4)=dbd f_3 (a_3) = f_4 d (a_3) = f_4(a_4) = d b. We would like that f 3(a 3)f_3(a_3) be equal to bb but this is not so, we just see that d(bf 3(a 3))=0d (b-f_3(a_3)) = 0 and hence by exactness of the lower row there is bB 2b'\in B_2 such that db=bf 3(a 3)d b' = b-f_3(a_3). Since f 2f_2 is also epi, there is a 2A 2a_2\in A_2 such that f 2(a 2)=bf_2(a_2) = b'. Now da 2+a 3A 3d a_2+a_3\in A_3 is such that

f 3(da 2+a 3)=df 2(a 2)+f 3(a 3)=db+f 3(a 3)=bf 3(a 3)+f 3(a 3)=bf_3 (d a_2 + a_3) = df_2(a_2)+f_3(a_3) = db'+f_3(a_3) = b - f_3(a_3)+f_3(a_3) = b

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.