Re: Snake lemma-Five lemma

In article <1126050288.212909.136470@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
singau <singau_online@xxxxxxxxxxx> wrote:

>in Lang's Algebra, the proof of the five lemma is an exercise. I am
>able to prove it, but Lang also gives a hint: to show surjectivity, one
>can use the snake lemma. I don't see how i can invoke the snake lemma.
>can anyone tell me?

Here's the snakeiest proof I can come up with:

Take the diagram from the Five Lemma:

M_1 ---> M_2 ---> M_3 ---> M_4 ---> M_5
| | | | |
| | | | |
V V V V V
N_1 ---> N_2 ---> N_3 ---> N_4 ---> N_5

with exact rows; call the vertical maps f_1, f_2, f_3, f_4, f_5. We
want to show that (a) if f_1 is surjective, and f_2,f_4 are
monomorphisms, then f_3 is a monomorphism; and that (b) if f_5 is a
monomorphism, and f_2,f_4 are surjective, then f_3 is surjective.

Call the maps on the first row g_1 through g_4, and the maps on the
second row h_1 through h_4. Consider the following diagram:

0 ---> Coker(g_1) ---> M_3 ---> Ker(g_4) ---> 0
| | |
| | |
V V V
0 ---> Coker(h_1) ---> N_3 ---> Ker(h_4) ---> 0

where the vertical maps are f_2', f_3, and f_4', with f_2', f_4' the
maps induced in homology by taking the corresponding chain of M's to
the chain of N's. For example, mapping 0->M1->M2->0 to 0->N1->N2->0
using (f1,f2) induces f_2'.

This map has the associated exact sequence

(*) Ker(f_2') --> Ker(f_3) --> Ker(f_4')
--> Coker(f_2') --> Coker(f_3)--> Coker(f_4)'

To prove (a), note that f_4' is injective (it is a restriction of
f_4). Consider the diagram

M_1 ---> M_2 ---> Coker(g_1) ---> 0
| | |
V V V
0->N_1/Ker(h_1)->N_2 --> Coker(h_1)

with the obvious vertial maps; the first map is p, f_1 followed by the
canonical projection. By the Snake Lemma, we have an exact
sequence

Ker(f_2) --> Ker(f_2') --> Coker(p).

We know Ker(f_2)=0; since p is a composition of surjective maps, it is
surjective. So this exact sequence degenerates into 0->Ker(f_2')->0,
showing Ker(f_2')=0. Now from (*) it follows that Ker(f_3)=0, which is
what we needed to prove.

For (b), since f_2 is surjective we have that f_2' is surjective (it
is the composition of f_2 with the projection onto the cokernel). Now
consider

Ker(g_4) ---> M_4 ---> Im(g_4) ---> 0
| | |
V V V
0-> Ker(h_4) ---> N_4 ---> N_5

where the first vertical map is f_4', and the last is the restriction
of f_5.

Applying the Snake Lemma, we get

Ker(f_5|_{Im g_4}) --> Coker(f_4') --> Coker(f_4)

is exact. We know f_5 is injective, so the first term is 0. And we are
given that f_4 is surjective, so the last term is 0. Hence the
cokernel of f_4' is trivial. From (*) we deduce that the cokernel of
f_3 is also trivial, so f_3 is surjective, as claimed.


--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================

Arturo Magidin
magidin@xxxxxxxxxxxxxxxxx

.

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