Hensel Topologie II (25.04.29)

Abstract

Talk

Goal: Understand the $ker(h)$ of the map $h$ we defined last time.

Lemma "Hom is left exact"

Given a Short exact sequence of abelian groups and $G$ a group. $0\to A\to B\to C\to 0$ Applying $Hom(\cdot ,G)$ gives us a dual exact sequence. $0\to C^{\ast }\to B^{\ast }\to A^{\ast }$ And is the original sequence splits, then there is a slightly longer exact sequence. $0\to C^{\ast }\to B^{\ast }\to A^{\ast }\to 0$

Proof I will check the proof later.

”$Hom(\cdot ,G)$ is a contravariant functor and the property of the lemma is called left-exactness. Left-exactness of a functor means that the functor applied to a ses gives a ses with the right dropped.”

Remark: $0\to \mathbb{Z}\stackrel{n}{\to }\mathbb{Z}\to \mathbb{Z}/n\to 0$ shows that the splitness of the sequence is necessary. Meaning $Hom$ is in general not exact.

We continue. We want to understand the relationship between $Z,B,C$. Let $Z_n=\text{ker }\partial$ be the cycle group and $B_n=\text{im }\partial$ the boundary group of $C_n$. We write down the ses as last time: $0\to Z_n\stackrel{i}{\to }{C}_n\stackrel{\partial }{\to }{B}_{n-1}\to 0$ Now $Z_{\bullet },C_{\bullet },B_{\bullet }$ can be interpreted as a ses of chain complexes (we choose the easy map $0$ for $B,Z$, the homology is then equal the chain complex).

Since $C_{\bullet}$ is assumed to be a free abelian subgroup, the sequence splits and we can apply the lemma to get the dual ses. Last semester we said that a ses of chain complexes induces a long exact sequence in [[Homology group|Homology]]. The same fact holds for cohomology. In homology we get the les: $$...\to \underbrace{H_{n}(Z_{\bullet})}_{Z_{n}} \overset \pi\to H_{n}(C_\bullet)\overset \partial\to \underbrace{H_{n}(B_{\bullet})}_{B_{n-1}}\overset i\to Z_{n-1}\overset \pi\to ...$$ with the natural connected homomorphism being the inclusion $i$. We therefore suspect that the new connecting homomorphism in cohomology will be the dual $i^{*}:Z_{n}^{*}\to B_{n}^{*}$ of $i$ which is just the restriction of a map $Z_{n}\to G$ to $B_{n}$. The [[Snake lemma]] tells us that the map is natural but we still have to give a construction. We have $\text{ker }0 = Z_{n-1}^{*}$ and $\text{coker }0 = B_{n-1}^{*}$. Taking one element $\varphi:Z_{n-1}\to G \in Z_{n-1}^{*}$ this can be extended to a map $\tilde \varphi:C_{n-1}\to G$ (due to surjectivity of $i^{*}$). We map this to $\phi=\partial^{*} \tilde\varphi = \tilde \varphi \partial : C_{n}\to G$. This can be lifted by a $\tilde \phi$ s.t. $\partial^{*}\tilde \phi = \partial ^{*}\tilde \varphi$. This means that $\tilde \phi$ and $\tilde \varphi$ are equal on $B_{n-1}$, so $\tilde \phi$ is actually the restriction to $B_{n-1}$. To reiterate, we get the [[Long exact sequence]] $$...\overset{\pi^{*}}\to Z_{n-1}^{*} \overset{i^{*}}\to B_{n-1}^{*} \overset{\partial^{*}} \to H^{n}(C_{\bullet}) \overset{\pi^{*}}\to Z_{n}^{*} \to ...$$ With the connecting homomorphism $i^{*}$. From the long exact sequence we now extract a ses again. $$0\to \text{coker }i^{*}_{n-1} \to H^{n}(C_{\bullet})\to \text{ker }i_{n}^{*} \to 0$$ where the last term can be identified with $\text{Hom}(H_{n}(C_{\bullet)}, G)$.

This has $ker(i_n^{\ast }:Z_n^{\ast }\to B_n^{\ast })$ inside. This are all maps that restrict to $0$ in $B_n^{\ast }$, inducing a map on homology $H_n\to G$. This does exactly the thing of $h$, i.e. we have motivated our $h$ slightly more.

Corollary

There is a split SES $0\to Coker(i_{n+1}^{\ast })\to H^n(C_{\bullet },G)\to Hom(H_n(C_{\bullet ),}G)\to 0$

Observe: This Cokernel is measuring st. It measures the “non-right-exactness of Hom applied to a special SES. If $H_{n-1}(C_{\bullet })$ is Free Abelian group then $H^n(C_{\bullet },G)\cong Hom(H_n{C}_{\bullet },G)$” gives exact sequence so the cokernel is $0$. This is a general construction: Whenever you have a left-exact functor, you can construct a right-derived functor..

A short interlude on derived functors:

Definition

$H$ Abelian group. A Free resolution of $H$ is an exact sequence $...\to F_1\to F_0\to H\to 0$

We use infinitely many because some other groups (not abelian) have longer resolutions, and longer abelian resolutions can be useful in calculations.

Lemma

Any abelian group $H$ has a free resolution of length $2$.

This has a nice stupid proof. It uses a free abelian group generated over all elements of $H$.

Corollary

Suppose I have a free resolution of $H$ and $G$ is an Abelian group. Consider the cochain complex obtained by applying Hom. Then this might not be exact. So we can compute cohomology! An the cohomology of this is independent of the free resolution. We call this the first cohomology group $Ex{t}^1(H,G)$. And this Ext Functor is the Derived functor of Hom Functor. This is just Group cohomology!!!! :o

This $Ext$ is the kernel from before (the one we wondered about!) We will use these on problem sets.