Zoran Skoda five lemma

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

$$\array{ A_1 & \to & A_2 & \to & A_3 & \to & A_4 &\to & A_5\\ \downarrow f_1 &&\downarrow f_2 &&\downarrow f_3 &&\downarrow f_4 &&\downarrow f_5 \\ B_1 & \to & B_2 & \to & B_3 & \to & B_4 &\to & B_5 }$$

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

Then:

1) If $f_2$ and $f_4$ are epis and $f_5$ is mono, then $f_3$ is epi. If

2) $f_2$ and $f_4$ are monic and $f_1$ is epi, then $f_3$ is mono.

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

$$f_3 (d a_2 + a_3) = df_2(a_2)+f_3(a_3) = db'+f_3(a_3) = b - f_3(a_3)+f_3(a_3) = b$$

with contradiction.