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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.

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  • $\begingroup$ The fact that the top left corner will be cartesian is straightforward to prove. I'm pretty sure of it. $\endgroup$
    – Harry Gindi
    Commented Feb 2, 2010 at 14:13
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    $\begingroup$ You can remove $C_3$ and the two arrows through it from your diagram. $\endgroup$
    – darij grinberg
    Commented Feb 2, 2010 at 21:14

2 Answers 2

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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.

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  • $\begingroup$ I don't remember ever getting an answer to that question. $\endgroup$
    – Harry Gindi
    Commented Feb 2, 2010 at 14:41
  • $\begingroup$ Did you see that Jonathan Wise added a section on the salamander lemma to the notes he wrote up for your question? $\endgroup$
    – Tom Church
    Commented Feb 2, 2010 at 17:10
  • $\begingroup$ Nope. I'll check it out. $\endgroup$
    – Harry Gindi
    Commented Feb 2, 2010 at 17:23
  • $\begingroup$ Ah, thanks! I never was notified of the update by the system. $\endgroup$
    – Harry Gindi
    Commented Feb 2, 2010 at 17:42
  • $\begingroup$ The proof using the Salamander Lemma makes it clear you can remove C_3 from the diagram. $\endgroup$
    – Chris Schommer-Pries
    Commented Feb 3, 2010 at 14:14
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Consider the total complex of the picture (thought of as a bicomplex). Since the vertical sequences are exact, the total complex is exact. Since the bottom rows are exact, the cohomology of the total complex is the cohomology of the top row, by the spectral sequence for a bicomplex. So the top row is exact.

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  • $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$
    – Ben Webster ♦
    Commented Feb 2, 2010 at 15:04
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    $\begingroup$ So you're suggesting to try to construct the spectral sequence of a bicomplex without using the 3 x 3 lemma? $\endgroup$
    – Chris Schommer-Pries
    Commented Feb 2, 2010 at 19:51
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    $\begingroup$ I don't think contructing the spectral sequence should be a problem. There's a greater chance that the 3 x 3 lemma is used in the proof that the spectral sequence converges to the right thing, but I don't really see why. $\endgroup$
    – Ben Webster ♦
    Commented Feb 2, 2010 at 22:58
  • $\begingroup$ I'm not saying it is impossible, but my recollection is that in both the exact couple method and the usual method, when you pass to the next page of the spec. seq. and define the new differential, you usually do a diagram chase to show it is well defined. While not exactly the same as the 3 x 3 lemma, my recollection is that this diagram chase is roughly of about the same complexity. $\endgroup$
    – Chris Schommer-Pries
    Commented Feb 3, 2010 at 12:54
  • $\begingroup$ The proof above doesn't quite work, because the bottom rows are not exact if you take into account the implicit zeroes at all the other lattice points of the plane. The same remark applies to the columns, but that can be fixed by adding a fourth row of cokernels of the vertical maps entering the third row. $\endgroup$
    – Dan Grayson
    Commented Oct 30, 2013 at 23:58

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