proof of snake lemma using spectral
sequences∗
rm50†
2013-03-22 3:08:39
This article proves the Snake Lemma purely by appealing to the machinery
of spectral sequences. The point of this article is not really to provide a proof
of the Snake Lemma, but rather to show some of the power and utility of
spectral sequences. We work inside some abelian category, such as modules over
a ring. This article draws heavily from here, which is an excellent introduction
to spectral sequences (beware, however; it contains a number of significant and
less significant errors).
Lemma 1. (Snake Lemma) Consider the commuting diagram
0[r]A[r]B[r]C[r]00[r]A0 [r][u]α B 0 [r][u]β C 0 [r][u]γ 0
in which the rows are exact. Then there is an exact sequence
0 → ker α → ker β → ker γ → coker α → coker β → coker γ → 0
Proof. Consider the first-quadrant double complex E p,q as follows:
0
0
0
0[r]E 1,0 = A[r]a E 1,1 = B[r]b E 1,2 = C[r]c 00[r]E 0,0 = A0 [r]a [u]α E 0,1 = B 0 [r]b [u]β E 0,2 = C 0 [r]c [u]γ 0
where all other E p,q = 0. Let this be page 0 of a spectral sequence.
Computing page 1 by using the differentials in the horizontal direction, we
get
E11,0 = 0[r]E11,1 = 0[r]E11,2 = 0[r]0E10,0 = 0[r][u]E10,1 = 0[r][u]E10,2 = 0[r][u]0
since the rows are exact.
Now compute page 1 by using the differentials in the vertical direction. The
result is
E11,0 = coker α[r]E11,1 = coker β[r]E11,2 = coker γ[r]0E10,0 = ker α[r]E10,1 = ker β[r]E10,2 = ker γ[r]0
∗ hProofOfSnakeLemmaUsingSpectralSequencesi
created: h2013-03-2i by: hrm50i version:
h41944i Privacy setting: h1i hProofi h18G35i
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Page 2 of this diagram, obtained by using the maps given as the differentials,
looks like this (the arrows are the resulting differential maps on page 2):
0[rrd]0[rrd]0[rrd]0[rrd]0[rrd] ∗ ∗[rrd] ∗ [rrd] ∗ [rrd]0 ∗ [rrd] ∗ [rrd] ∗ ∗[rrd]00000
Note now that the entries marked as ∗ have stabilized, and thus must be zero
(since the cohomology of the double complex is zero, as shown at the beginning
of the proof). Also, the map between the entries marked as ∗∗ must be an
isomorphism in order that they should vanish at the next stage, for the same
reason.
The fact that the entries ∗ are all zero means that the sequences
coker α → coker β → coker γ → 0
and
0 → ker α → ker β → ker γ
are exact. The isomorphism of the entries labeled ∗∗ gives an isomorphism
between coker(ker β → ker γ) and ker(coker α → coker β). By the splicing
lemma, we get a map from ker γ → coker α, and
0 → ker α → ker β → ker γ → coker α → coker β → coker γ → 0
is exact.
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