I was reading this paper a while ago, and I couldn't figure out how to prove a lemma that was left as an exercise by only using universal properties and the definition of an abelian category.
I'll reproduce the diagram:
$$\ \ \matrix{
&&0&&0&&0
\cr&&\downarrow&&\downarrow&&\downarrow
\cr &&A_1 & & B_1& &C_1
\cr &&\downarrow & &\downarrow&&\downarrow
\cr &&A_2 & \to & B_2 & \to & C_2 &
\cr &&\downarrow &&\downarrow&&\downarrow
\cr 0&\to&A_3 & \to & B_3 & \to & C_3
}\ \ $$
With all rows and columns exact. (This diagram lives in an abelian category).
Show that there exists an exact sequence $A_1\to B_1\to C_1$ making the diagram commute.
Sure, it's not too hard with elements, I mean, it's just part of the snake lemma. However, proving it with universal properties is another story. By the universal property of the kernel, there are natural maps $A_1 \to B_1$, and $B_1 \to C_1$. Proving that this is exact is another story entirely. I believe I was able to show (I tried this a few months ago) that the top left corner (not counting zeros) is cartesian (a pullback square), but I still couldn't prove exactness.
I repeat, this is for a proof without elements. It should rely only on the definition of an abelian category and universal properties. If you happened not to click the link to the paper, the whole point is a proof without elements. I'd really like to see at least one proof in homological algebra actually done from the definition, just because it would be extremely instructive.
Best Answer
This should follow from the Salamander lemma, which, as you found out in your previous MO question, can be proved without the use of elements. It is a nice exercise, so I won't spell it out for you.