# $H_{0}(A)$ in chains with zeros?

If we have a non-zero abelian group A and $0\rightarrow{A}\rightarrow0$, am I correct in thinking $H_{0}(A)=A$? If so why because I'm a bit confused and to the image & kernel in this case...

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Are you saying $0\to A \to 0$ is exact? In this case, $A$ is just equal to $0$. – Steven Gubkin Feb 2 '14 at 19:07
I'm actually trying to show that something weird like $0\rightarrow{A}\rightarrow{0}\rightarrow{B}\rightarrow{0}$ where A & B are non-zero abelian groups cannot be exact.... so I was trying to just consider the case $0\rightarrow{A}\rightarrow0$ as I know that for $0\rightarrow\mathbb{Z}\rightarrow0$ we have that $H_{0}(\mathbb{Z})=\mathbb{Z}$ – Rhoswyn Feb 2 '14 at 19:13
$0 \to A \to 0$ is exact means that the image of the first map is the kernel of the second map. The second map has kernel $A$, and the first map has image $0$. So if the image of the first is the kernel of the second, then $A = 0$. – Steven Gubkin Feb 2 '14 at 19:24
If $0 \to A \to 0$ is not exact, then the homology of this sequence is the kernel of the second map mod the image of the first map, which is just $A$. – Steven Gubkin Feb 2 '14 at 19:25
I will try to spell this out a bit nicer in an answer. – Steven Gubkin Feb 2 '14 at 19:25

A chain complex of abelian groups is just a sequence of maps

$...\to A_{-2} \to A_{-1} \to A_{0} \to A_1 \to A_2 \to ...$

with $d_n: A_n \to A_{n+1}$ and $d_{n+1} \circ d_n = 0$ for all $n$.

The homology of the complex is defined by $H_n = ker(d_{n})/im(d_{n-1})$. We say that the complex is exact at $n$ if $H_n = 0$, or in other words if the image of $d_{n-1}$ is the kernel of $d_n$.

So the homology of the complex $0 \to A \to 0$ at $n=0$ is $H_0 = A$, and the complex is exact iff $A = 0$.

Hopefully you have some intuition for why anyone would care about any of this stuff. The canonical example (which you would see in your first multivariable calculus class in a just world) is the de Rham complex. I will assume you have not seen differential forms. Here is an example:

Let $U = \mathbb{R}^2 = {(0,0)}$. Then

$0 \to C^{\infty}(U) \to C^\infty(U) \times C^\infty(U) \to C^\infty(U) \to 0$ is a nice complex, where the first map is given by the gradient operator $\nabla(f) = \begin{bmatrix} \frac{\partial f}{\partial x} \\ \frac{\partial f}{\partial y}\end{bmatrix}$, and the second map is given by the curl operator $curl\left(\begin{bmatrix} P(x,y) \\ Q(x,y)\end{bmatrix}\right) = \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}$.

This is a chain complex because partial derivatives commute (do the computation!).

Now let us compute the homology groups.

$H_0 = \mathbb{R}$ because the kernel of the first map consists of just the locally constant functions. This is really just detecting that $U$ has only one connected component. Observe that if $U$ had $n$ several connected components, this homology group would have been $\mathbb{R}^n$.

$H^1$ is quite tricky. It turns out that $H^1 = \mathbb{R}^1$, and this is detecting the "hole" in the domain $U$. It is generated by the gradient of the "angle function", which is not defined on all of $U$, but the gradients do in fact patch up globally.

$H^2 = 0$. You should be able to do this exercise using what you know from multivariable calculus.

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