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I'm trying to prove the following lemma by diagram chasing, but I've had no success, so I decided to ask for help here.

Let $A$, $B$, $C$, $D$, $E$, $F$, and $G$ be Abelian groups, and let $a_{1}$, $a_{2}$, $a_{3}$, $g_{1}$, $g_{2}$, $g_{3}$, $c_{1}$, $c_{2}$, $c_{3}$, $e_{1}$, $e_{2}$, $e_{3}$ be homomorphisms such that the following diagram commutes (also, $c_{1}$ and $e_{1}$ are isomorphisms, as indicated in the diagram):

enter image description here

Also, let $g_{3}a_{2} = 0$, $g_{2}c_{2} = 0$ and $g_{1}e_{2} = 0$. Prove that, if one of the "horizontal" diagonals of this hexagon is exact, i.e. if $\ker{g_{2}} = \mathrm{im}\hspace{0.1cm}c_{2}$ OR $\ker{g_{1}} = \mathrm{im}\hspace{0.1cm}e_{2}$, then $$ c_{3}c_{1}^{-1}a_{1} + e_{3}e_{1}^{-1}a_{3} = 0. $$

I tried diagram chasing, but I wasn't able to see a solution. Here's as far as I got: let $a \in A$, $b = a_{1}a$, $g = a_{2}a$, $f = a_{3}a$. Then we also have $b = g_{1}b$ and $f = g_{2}g$. Let $c = c_{1}^{-1}b$, $g' = c_{2}c$, $e = e_{1}^{-1}f$, $g'' = e_{2}e$. Then we also have $g_{1}g' = b$ and $g_{2}g'' = f$. Then, let $d' = c_{3}c$ and $d'' = e_{3}e$, and we have $g_{3}g' = d'$ and $g_{3}g'' = d''$.

Ultimately, we have $d' = c_{3}c_{1}^{-1}a_{1}a$ and $d'' = e_{3}e_{1}^{-1}a_{3}a$, so the point is to prove $d' = -d''$.

For example, suppse that the first condition holds, i.e. that $\ker{g_{2}} = \mathrm{im}\hspace{0.1cm}c_{2}$. Then, since $g_{2}g'' = g_{2}g = f$, we have $g'' - g \in \ker{g_{2}} = \mathrm{im}\hspace{0.1cm}c_{2}$, so there exists a $c' \in C$ such that $c_{2}c' = g'' - g$.

We also have $c_{3}c' = g_{3}c_{2}c' = g_{3}(g''-g) = g_{3}g'' - g_{3}g = d''$, because $g_{3}g = g_{3}a_{2}a = 0$. If there was a way to prove $c_{3}(c+c') = 0$, I'd be done, but I have no idea how to continue.

Am I on the right track? Should I approach this problem differently?

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Here's a solution from a friend (I was only missing a couple of details):

Note that $g_{1}(g'+g''-g) = g_{1}(g')+g_{1}(g'')-g_{1}(g) = b + g_{1}e_{2}e - b = b + 0 - b = 0$, and also $g_{2}(g'+g''-g) = g_{2}(g')+g_{2}(g'') - g_{2}(g) = g_{2}c_{2}c + f - f = 0$, so $g'+g''-g \in \ker{g_{1}} \cap \ker_{g_{2}}$.

If $\ker{g_{1}} = \mathrm{im}\hspace{1mm}c_{2}$, then there exists a $c'' \in C$ such that $c_{2}c'' = g'+g''-g$ (actually, $c'' = c+c'$), so $c_{1}c'' = g_{1}c_{2}c'' = g_{1}(g'+g''-g) = 0$, so since $c_{1}$ is a bijection, $c''=0$, so $c_{3}(c+c') = 0$, which is what I wanted to prove.

The other case is analogous to the first one.

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