The beginning of homological algebra involves lots of diagram chasing, for proving most of the theorems. This gets repetitive after a while. To make things more interesting and satisfactory, one would like to remove the mechanical dredge work as far as possible.

One possible approach, as advocated in Lang's Algebra book, is to first prove the snake lemma by diagram chasing, and then reduce everything else as much possible to the snake lemma. This is mostly satisfactory. But some proofs, such as that of five lemma(as far as I am able to reconstruct now), require a small amount of diagram chasing in addition to using snake lemma.

There seems to be an alternate approach using the salamander lemma, as given in the references in this question. Salamander lemma can be used to prove the other important lemmas. But according to the description in Anton Geraschenko's post, it is like chasing salamanders around, rather than chasing elements.

Are there other possible approaches to founding homological algebra avoiding extensive diagram chasing? That is, is it possible to base homological algebra on some other lemma/proposition, than the two possibilities mentioned above?

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    $\begingroup$ I should point out that with the Salamander Lemma, you rarely chase a whole salamander, so the chasing is considerably easier. You usually use exactness hypotheses to determine that bits of salamanders are isomorphisms, and then string together these isomorphisms. They usually make a straight line between the two things you're trying to show are isomorphic (e.g. 0 and some cokernel). There are lots of example in this blog post: sbseminar.wordpress.com/2007/11/13/… $\endgroup$ May 16, 2010 at 17:49
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    $\begingroup$ Based on one of your other posts and your remark there that Mayer--Vietoris is not intuitive because it relies on the snake lemma (whereas it can in fact be thought of quite geometrically, and is then quite intuitive), I would suggest that rather than trying to avoid diagram chasing, you instead try to understand some aspects of it more geometrically, and thus build up your intuition. E.g. the homology long exact sequence for a pair of spaces can be derived via some diagram chase from the associated short exact sequence of singular chains, but it can also be thought very geometrically ... $\endgroup$
    – Emerton
    May 17, 2010 at 4:22
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    $\begingroup$ Namely, if you have an n-chain in $X$ whose boundary lies in $A$, then passing to that boundary gives an $n-1$-chain in $A$. (This is the boundary map in the long exact sequence.) If you in fact get zero as the answer, then the chain you had was a cycle on $X$ (this is exactness at $H^n(X,A)$), ... . I think it is very worthwhile to develop a geometric understanding of this kind for the long exact sequence of a pair, as well as for Mayer--Vietoris. $\endgroup$
    – Emerton
    May 17, 2010 at 4:25

3 Answers 3


One other way you can prove all diagram chasing results (that I know of) is to first prove that there is a spectral sequence associated with a double complex (actually, there are two!). This involves some diagram chasing, but you only do it once. I guess you could also argue that you still have to chase the spectral sequences.

As an example, here's how you'd prove that a short exact sequence of chain complexes induces a long exact sequence in homology. We start with the double complex with exact rows

    ↑      ↑      ↑
0 → Ai+1 → Bi+1 → Ci+1 → 0
    ↑      ↑      ↑
0 → Ai  →  Bi  →  Ci  → 0
    ↑      ↑      ↑

(I'm assuming the columns are bounded below, but it doesn't matter since this result is "local": you can truncate above and below the point you're interested in and then prove the result)

We can regard this as the $E_0$ page of two different spectral sequences, depending on whether we decide to start with the vertical or horizontal arrows as our differential. Both spectral sequences have to abut to the homology of the total complex associated to this double complex.

First, use the horizontal arrows. Since the rows are exact, when we take homology, we get zero everywhere, so the $E_1$ page is identically zero. Nothing interesting is going to happen now: we've gotten to $E_\infty$. So the homology of the total complex is zero.

Now what if we use the vertical arrows? We get the $E_1$ page

0 → Hi+1(A) → Hi+1(B) → Hi+1(C) → 0

0 → Hi(A)  →  Hi(B)  →  Hi(C)  → 0

We'd like to prove that the rows are exact in the middle and that kernel of $H^{i+1}(A)\to H^{i+1}(B)$ is isomorphic to the cokernel of $H^i(B)\to H^i(C)$. Let's flip to the $E_2$ page and see what happens. Let $K^i=\ker(H^{i+1}(A)\to H^{i+1}(B))$, $M^i=\mathrm{cok}(H^i(B)\to H^i(C))$, and let $L^i$ be the homology of the above row at $H^i(B)$. We get the $E_2$ page

0   Ki+1   Li+1   Mi+1   0
   ↘     ↘     ↘
       ↘      ↘      ↘
0   Ki     Li     Mi   0

(pardon my ascii art: those are meant to be arrows going two spots to the right and one spot down)
Note that on the $E_{\ge 3}$ pages, the differentials will be too long to connect anything in these three columns to anything other than zero. Since we must abut to zero, this $E_2$ page is the "last chance" for the homology to vanish. It follows that the sequences $0\to L^i\to 0$ and $0\to K^{i+1}\to M^i\to 0$ must be exact, so $L^i=0$ and $K^{i+1}\cong M^i$, which is exactly what we wanted to show.

Note that I was able to completely ignore the question of what the differentials on higher pages were; I just had to know that they exist.

Diagram-chasing results almost always assume that the rows are exact and then make assertions about the homology groups you get from the columns, so you can always run one spectral sequence and immediately get that the homology of the total complex is zero, then run the other spectral sequence and note when the spectral sequence has it's "last chance" to cancel all homology groups.

  • $\begingroup$ hmm ... proving fundamental results of homological algebra with spectral sequences, whose theory is based on fundamental results of homological algebra ... $\endgroup$ May 16, 2010 at 22:47
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    $\begingroup$ @Martin: Which results do spectral sequences depend on? The development of spectral sequences in (say) Lang (books.google.com/books?id=Fge-BwqhqIYC&pg=PA814) is only a couple of pages long and doesn't rely on any external results. In fact, I was wrong in the first paragraph: proving that there is a spectral sequence associated to a double complex doesn't involve any diagram chasing. As far as I can tell, the only reason we don't learn spectral sequences earlier is because we think they're scary, not because they require a lot of background. $\endgroup$ May 16, 2010 at 23:44
  • $\begingroup$ I agree with Anton's observation. Even the more general spectral sequence associated to an exact couple ( en.wikipedia.org/wiki/Spectral_sequence#Exact_couples ) relies on very very little machinery. It involves little or no diagram chasing. $\endgroup$ May 17, 2010 at 2:36
  • $\begingroup$ I thought that you need to use the long exact sequence in homology in order to produce an exact couple out of a filtered differential object. I don't know if there's a really direct way to get from the spectral sequence of an exact couple to the spectral sequence(s) of a double complex. $\endgroup$ May 17, 2010 at 3:57

But isn't doing homological algebra without diagram chasing akin to doing geometry without circles, or doing algebra without multiplication? It's just the nature of the subject, a big part of it (and a fairly pleasant, I might add).

A standard solution is to use Freyd's Embedding theorem for abelian categories, which turns on diagram chase for all abelian categories. Taking it on faith and proving separately later is not ideal, but to me it seems better than the alternatives.

  • $\begingroup$ Yes, it is the nature. Besides, diagram chasing in abelian categories can be made precise with the notion of elements (-> Mac Lane). But I think the original question refers just to $R$-Mod. At least it makes sense. $\endgroup$ May 16, 2010 at 22:50
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    $\begingroup$ I agree with this (+1). But I wanted to say that the question is still worth taking seriously: questions of the form "Can we do X without Y?" have often turned out to be very fruitful in mathematics (especially if the answer is "no"!), even if Y is a very natural and effective way to do X. $\endgroup$ May 17, 2010 at 1:56

Checkout chapter "Abstract Homological Algebra" on the book "Basic Homological Algebra", by M. Scott Osborne (http://books.google.com/books?id=0_3swQG0hpsC). Everything is done very close to the axioms of Abelian Category, whithout chasing. Morphisms are constructed only through factorization properties.

  • $\begingroup$ That chapter seems to depend on having enough projective objects in the category, in a certain sense defined there. But Mac Lane did it without projective objects in his section "Diagram Lemmas" in his chapter "Abelian Categories" in his book "Categories for the Working Mathematician". $\endgroup$ Oct 31, 2013 at 23:28

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