# Wild Spectral Sequences Ep. 2: Five, Isomorphism!

Last time on Wild Spectral Sequences, we conquered the snake lemma using a spectral sequence argument. This time, we meet a new beast: the five lemma. The objective is the usual: prove the five lemma using spectral sequences.

Recall that the five lemma states that given a diagram

in an abelian category, if the rows are exact and $a,b,d,e$ are isomorphisms, then so is $c$. Actually, the hypotheses are too strong. It suffices to have $b,d$ isomorphisms, $a$ an epimorphism and $e$ a monomorphism. One can deduce this via J. Leicht's "strong four lemma" (which we might try and prove via a spectral sequence too) or just by using the regular diagram-chasing proof of the five lemma.

### Bring on the Spectral Sequence!

As usual, the technique for these diagram proofs is the same: use the two spectral sequences associated to a double complex:

$E^2_{p,q} = H_p^vH_q^h(C) \Rightarrow H_{p+q}(\mathrm{Tot}(C))$
$E^2_{p,q} = H_p^hH_q^v(C) \Rightarrow H_{p+q}(\mathrm{Tot}(C))$

Since we want to deduce something about $c$, we should first apply the $H_p^hH_q^v$-spectral sequence. At the $E^2$-page we get this:

I have indicated two of the $E^2$-differentials here, which hints that indeed, we probably only will need that $a$ is an epimorphism and $e$ is a monomorphism. Note also that it is at this stage that we are using all the conditions on the vertical maps.

At any rate, we get that $E_{2,0}^\infty = \mathrm{coker}(c)$ and $E_{2,1}^\infty = \ker(c)$. Recall that we have a filtration on $H_*(\mathrm{Tot}(C)$ (abbreviate $H_*$), which we write as $F_*H_*$.

So we get $\ker(c)\cong F_2H_3/F_1H_3$ and $\mathrm{coker}(c) \cong F_2H_2/F_1H_2$. Thus, in order to finish the proof, we just need to show that these two objects are zero. This is where we use the other spectral sequence. Now, in this $H_p^vH_q^h$ spectral sequence, we have to be careful because the filtration on $H_*$ is not necessarily the same as the filtration on $H_*$ associated to the first spectral sequence.

This doesn't matter, because if we can show that $H_2 = H_3 = 0$, then we will be done.

### The Next Spectral Sequence

Now, the objective is to prove that $H_2 = H_3 = 0$, and we will use exclusively the $H_p^vH_q^h$-spectral sequence, which uses the force of exactness of rows to accomplish its task. In order to keep the notation clean, I will use also the $E_{*,*}^\infty$ and $F_*H_*$ to denote the respective parts of this spectral sequence, but remember that $F_1H_2$ of this spectral sequence might not be the same as the $F_1H_2$ piece of $H_2$ of the $H_p^hH_q^v$-spectral sequence, but this shouldn't cause too much confusion because we will just attempt to prove that $H_2 = H_3 = 0$, and from now on we won't speak of the other spectral sequence in this section!

So let us begin: I will outline every step, and the reader should be prepared to do the routine calculations with differentials.

$H_3 = 0$
Proof.

1. $E_{0,3}^\infty = 0$ shows that $F_0H_3 = 0$.
2. $E_{1,2}^\infty = 0$ shows that $F_1H_3 = F_0H_3 = 0$
3. $E_{2,1}^\infty = 0$ shows that $F_2H_3 = F_1H3 = 0$
4. $E_{3,0}^\infty = 0$ shows that $F_3H_3 = F_2H_3 = 0$

Hence $H_3 = 0$.

This shows that $\ker(c) = 0$.

$H_2 = 0$
Proof.

1. $E_{0,2}^\infty = 0$ shows that $F_0H_2 = 0$.
2. $E_{1,1}^\infty = 0$ shows that $F_1H_2 = F_0H_2 = 0$
3. $E_{2,0}^\infty = 0$ shows that $F_2H_2 = F_1H_2 = 0$

Hence $H_2 = 0$.

This shows that $\mathrm{coker}(c) = 0$. Thus $c$ is an isomorphism. Great! Is is true now that this proof appears long, but that is only because I put so much detail in it, but with practice the arguments should become clear with not as much verbosity. Until next time!