15

The five lemma is an extremely useful result in algebraic topology and homological algebra (and maybe elsewhere). The proof is not hard - it is essentially a diagram chase.

Exercise 1.1 in McCleary's "Users Guide to Spectral Sequences" has the problem of proving the five-lemma using a spectral sequence.

Now I know this can be done using a spectral sequence of a double complex, but this is not yet introduced yet. In fact we don't really know too much about spectral sequences after Chapter 1. We know about filtration on a graded space, what a spectral sequence is, and how to set it up. We have showed that is collapses at certain pages under some appropriate conditions. There is a large section about bigraded algebras/spectral sequences and we know about reconstucting $H^*$ from knownledge of $E^{\ast,\ast}_\infty$

But it doesn't immediately strick me how to use just this knowledge to prove the five lemma. Maybe McClearly is just presuming his readers are smart enough to figure out to construct a total complex (i.e. summing along the diagonal)?

Or am I missing something obvious?

t.b.
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Juan S
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  • I have tagged this spectral-sequences. This tag has only been used once before, although there are numerous spectral sequence questions on this site, so feel free to remove if not required – Juan S Jul 15 '11 at 08:25
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    I removed the (algebraic-topology) tag, as this is definitely not a question about algebraic topology – t.b. Jul 15 '11 at 09:55

3 Answers3

15

Consider a commutative diagram

            r      r'
0 <--- M' <--- M <--- M'' <--- 0
       |       |      |
       | f'    | f    | f''
       V       V      V
0 <--- N' <--- N <--- N'' <--- 0
            k      k'

with exact rows and with $f'$ and $f''$ being isomorphisms.

This can be viewed, by completing with zeroes, as a double complex. Since the complex has finitely many non-zero components, the two standard spectral sequences that arise from the filtration by rows and the filtration by columns both converge to the homology of the total complex, which I shall denote $X$.

The first spectral sequence ${}^IE$, which arises from the filtration by rows, has zero ${}^IE^1$ term, exactly because we have assumed that the rows are exact. Since ${}^IE$ converges to $H_\bullet(X)$, we see that $H_\bullet(X)=0$, that is, that the total complex is exact.

Now the second spectral sequence ${}^{II}E$, which arises from the filtration by rows, has $1$st term ${}^{II}E_1$ as in the following diagram

                   R            R'
0 <---  ker f'   <---  ker f  <---  ker f'' <--- 0



0 <--- coker f'' <--- coker f <--- coker f'' <--- 0
                   K            K'

with the horizontal maps $R$, $R'$, $K$ and $K'$ induced by the maps $r$, $r'$ and $k$, $k'$ in the original diagram. Since $f'$ and $f''$ are isomorphisms, this is really

0 <--- 0 <---  ker f  <--- 0 <--- 0



0 <--- 0 <--- coker f <--- 0 <--- 0

It follows that all the differentials in the ${}^{II}E_1$ term are zero, so the ${}^{II}E_2$ in fact equals ${}^{II}E_1$ as a graded object. Moreover, the way the non-zero objects in ${}^{II}E_2$ are placed implies immediately that the differentials in all the terms ${}^{II}E_r$ with $r\geq2$ vanish, so that ${}^{II}E_\infty={}^{II}E_1$. Finally, the shape of ${}^{II}E_\infty$ and the fact that ${}^{II}E$ converges to zero imply that ${}^{II}E_1$ is itself zero. In other words, $\ker f$ and $\operatorname{coker}f$ are zero: this, of course, tells us that $f$ is an isomorphism.

(If we don't assume that $f'$ and $f''$ are isomorphisms, the ${}^{II}E_1$ has non-trivial differentials, and the spectral sequence ${}^{II}E$ only degenerates at ${}^{II}E_3$. If you write down exactly what this means, you will obtain the Snake Lemma—: the one non-zero differential in ${}^{II}E_2$ is the connecting homomorphism)

Mariano Suárez-Álvarez
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  • Dropping the the exactness requirement at $M''$ and $N'$ also gives you the part of the snake lemma that is commonly left out, namely that $\operatorname{Ker}{r'} = \operatorname{Ker}{R'}$ and dually $\operatorname{Coker}{K} = \operatorname{Coker}{k}$. – t.b. Jul 15 '11 at 18:32
1

Upon reflection, I think it can be argued without really thinking too much about double complexes (but essentially using the same argument).

So start by just imaging the complex with just the maps between the rows as the $E_0$ page of some spectral sequence. Since it is exact, it collapses to 0 at the $E_1$ page and so we know that whatever we are calculating is zero.

Then rotate things 90 degrees and do the same thing (so this time think about maps between the columns) as the $E_0$ page of some sequence. Then take the homologies which gives you a series of kernels along the bottom row, and cokernels in the first row. Since you know which functions are mono/epi you know which of these are zero. Get to the $E_2$ page and everything collapses by a degree argument. Then make the assumption (I'm not sure I see a simple justification for this) that the two spectral sequence are giving the same thing to argue that that whatever is left in the $E_2$ page of the spectral sequence must be zero - which gives what you that the required map must be an isomorphism.

Juan S
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1

There is a different way of doing this. The details are left as an exercise for the interested reader.

Suppose, as in my other answer, that we have a commutative diagram

            r      r'
0 <--- M' <--- M <--- M'' <--- 0
       |       |      |
       | f'    | f    | f''
       V       V      V
0 <--- N' <--- N <--- N'' <--- 0
            k      k'

with exact rows and $f'$ and $f''$ isomorphisms.

Consider the complex

                   f
... ---> 0 ---> M ---> N ---> 0 ---> ....

endowed with the increasing filtration such that the $0$th layer is given by the subcomplex

... ---> 0 ---> M' ---> N' ---> 0 ---> ....

and whose $1$th layer is the whole thing.

Construct the spectral sequence associated to this filtered complex. It converges. Now indentify explicitely what the $0$th page of the spetral sequence is (hint: it is ssentially the maps $f'$ and $f''$...) and show that the $E^2$ page is zero. Conclude that $f$ is an isomorphism.

Mariano Suárez-Álvarez
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