(Anton Geraschenko) The Salamander lemma

[I’m happy to introduce Anton, our very first guest blogger.]

A couple of years ago, George Bergman gave me a copy of a fun preprint that he never got around to preparing for publication. A scan of it is posted here. It starts

The “magic” of diagram-chasing consists in establishing relationships between distant points of a diagram—exactness implications, connecting morphisms, etc.. These “long” connections are in general composits of “short” (unmagical) connections, but the latter, and even the objects they join, are frequently not visible in the diagram-chasing proof. We attempt to remedy this situation here.

If you don’t like diagram chases, it’s likely that you still won’t like them once you know the Salamander lemma. The salamanders chase the diagrams for you, but you still have to chase the salamanders. I think the salamander proofs are easier to explain (once you know the Salamander lemma), and it’s easier to see where you use the hypotheses. For example, it is totally clear that the argument for the 3\times 3 lemma can prove the “20\times 20 lemma” as well.

A bicomplex in an abelian category is a complex of complexes, i.e. an infinite grid of objects and arrows in which all the rows and columns are complexes (two consecutive arrows compose to zero) and all the squares commute. If we have a finite bicomplex, then we can always extend it by either adding zeros or various kernels and cokernels (if we want some exactness conditions). Since everything I say will take place on a bicomplex background, I’ll adopt the convention that all the arrows in the bicomplex go to the right and down, and I’ll draw the arrows in the bicomplex as dotted lines:
Given an object A in a bicomplex,
we define the vertical homology A^\parallel:=\ker d/\mathrm{im}\, c, the horizontal homology {}_=A:=\ker b/\mathrm{im}\,a, the receptor {}^\square A:=\displaystyle\frac{\ker b\cap \ker d}{\mathrm{im}\, e}, and the donor A_\square := \displaystyle\frac{\ker f}{\mathrm{im}\,a+\mathrm{im}\,c}. Note that if A=0, then all of these subquotients are zero.

Recall that to define a morphism X/Y\to Z/W, it is enough to define a morphism X\to Z such that the image of Y is contained in W. This gives us four obvious maps (called intramural maps) induced by inclusions: {}^\square A\to A^\parallel, A^\parallel\to A_\square, {}^\square A\to {}_=A, and {}_=A\to A_\square. We also see that for any morphism A\to B in a bicomplex, we have an obvious extramural map A_\square\to {}^\square B.

Salamander lemma. If A\to B is a horizontal arrow in a bicomplex, then there is a six term exact sequence C_\square\to {}_=A\to A_\square\to {}^\square B\to {}_=B\to {}^\square D as shown on the left.
The morphism C_\square \to {}_=A is the composition of the extramural map C_\square \to {}^\square A and the intramural map {}^\square A\to {}_=A. The morphism {}_=B\to {}^\square D is similar. If A\to B is a vertical arrow, then we get the six term exact sequence on the right. In either case, I’ll call this the salamander centered at A\to B. The proof of the Salamander lemma is totally obvious in any abelian category where the objects have elements; you just unwind the definitions. You can reduce to this case in the usual way (see chapter VIII, section 4 of Categories for the Working Mathematician).

Special case: if the row (resp. column) containing the morphism A\to B is exact at A and B, then the horizontal homologies {}_=A and {}_=B (resp. vertical homologies A^\parallel and B^\parallel) are zero, so the extramural map A_\square\to {}^\square B is an isomorphism.

\mathbf{3\times 3} lemma (or Nine lemma). If we have three rows and three columns as shown below, with all three columns exact, and the bottom two rows exact (at U, V, X, and Y), then the first row is exact (at A and B). Moreover, if we have the stuff in parentheses (imposing horizontal exactness at W and vertical exactness at X), then the top row is exact at C.
To prove this, we need to show that the horizontal homologies {}_=A, {}_=B, and {}_=C are zero. Using the salamander centered at B\to C, we see that {}_=B is sandwiched between 0_\square=0 and B_\square in an exact sequence. Repeatedly applying the special case (using the exactness hypotheses we have), we see that B_\square\cong 0_\square=0, as illustrated, so {}_=B is zero. Similarly, we see that {}_=A and (given the appropriate hypotheses) {}_=C are zero as well. Strictly speaking, I should put \mathrm{coker}(V\to Y) underneath Y so that I use the special case to get the isomorphism V_\square \cong {}^\square Y. Since U surjects onto X, the image of X in Y lies in the image of V, so we still have a bicomplex when we do this.

Snake lemma. If we have the bicomplex below, with the three columns exact, and the middle two rows exact, then there is a six term exact sequence A\to B\to C\to D\to E\to F.
We prove that {}_=B and {}_=E are zero as before. Now we have to construct a “connecting morphism” C\to D making the desired six term sequence exact. This is equivalent to producing an isomorphism \mathrm{coker}(B\to C)\cong \ker (D\to E). Unravelling the definition, we have that \mathrm{coker}(B\to C)=C_\square and \ker(D\to E)={}^\square D, and we have an isomorphism between these using several applications of the special case.

Four lemma. If we have the bicomplex below, with the columns exact and the middle two rows exact, then \xi(\ker \alpha)=\ker \beta and \mathrm{im}\, \alpha = \eta^{-1}(\mathrm{im}\, \beta). In particular, if \alpha is injective, then so is \beta, and if \beta is surjective, then so is \alpha.
The desired result is equivalent to proving that the top and bottom rows are exact (i.e. that the two horizontal homologies in the picture are zero), which I’ll leave as an exercise (it should be very easy; the picture has everthing you need).

Long Exact Sequence in homology. This one is a little different from the others. So far, we’ve used the special case, together with “half salamanders”. This time we’ll use a “whole salamander”. If we have the bicomplex below, with the rows exact, then there is a six term exact sequence A^\parallel\to B^\parallel\to C^\parallel\to D^\parallel\to E^\parallel\to F^\parallel.
Looking at the salamander centered at A\to D, we have 0=0_\square\to A^\parallel\to A_\square\to {}^\square D\cong 0, so A^\parallel\cong A_\square. Applying the same argument “one level down,” we have that D^\parallel\cong D_\square. Looking at the salamander centered at C\to F, we have 0\cong C_\square\to {}^\square F\to F^\parallel\to {}^\square 0=0, so {}^\square F\cong F^\parallel. Applying the same argument “one level up,” we get that {}^\square C\cong C^\parallel. Now we look at the salamander centered at B\to E, which is A_\square\to B^\parallel\to B_\square\to {}^\square E\to E^\parallel\to {}^\square F. Using the isomorphisms we just talked about (along with the isomorphisms D_\square\cong {}^\square E and B_\square \cong {}^\square C), we get the desired six term exact sequence A^\parallel\to B^\parallel\to C^\parallel\to D^\parallel\to E^\parallel\to F^\parallel.

7 thoughts on “(Anton Geraschenko) The Salamander lemma

  1. George handed me a copy of this same preprint when I was at Berkeley. Interesting stuff. (By the way, did you ask George’s permission to post this to the web? He probably won’t mind, but you should let him know about this site.)

    My favorite part of the preprint is the section on weakly-bounded double complexes. I wonder if these ideas would simplify the proof that the composition of two connecting homomorphisms along the faces of a “post” (i.e. a triple complex, which is infinite in one direction, and looks like 3×3 squares in the other two) in the two possible directions, give negatives of each other.
    (I think this is one of the last results proven in Osborne’s book.)

  2. @Pace: Yes, I asked George if I could scan and post the preprint.

    @ulfarsson: You’re right, but I don’t have the ability to edit the post. Could somebody please change it?

Comments are closed.