Computing homology groups, algebra confusion

Question

In the process of computing homology groups, (at least from what I've been taught so far) I come across matrices representing the homomorphisms in the chain complex. For example, I have the matrix: $$A:= \begin{bmatrix} -7 & -6 & -2 & -4 \\ 4&3&2&4\\ 5&3&4&8\\ \end{bmatrix} $$

I compute the kernel and cokernel in the typical ways using elementary row operations. I find that $\ker A=span \{\begin{bmatrix}-2\\2\\1\\0\end{bmatrix},\begin{bmatrix}-4\\4\\0\\1\end{bmatrix}\}$ and $\text{coker}A=\mathbb{Z}^3/span\{\begin{bmatrix}21\\0\\21\end{bmatrix},\begin{bmatrix}0\\3\\9\end{bmatrix}\}$. From here I should identify the ismorphism type of each group, but have no idea how to proceed.

Questions: I don't really understand how we can be taking the quotient in this way.

Is the span of two vectors over $\mathbb{Z}$ just $\mathbb{Z}^2?$

I'm lacking both justification and intuition, so both are appreciated.

Answer 1

The $k$-th homology group is the quotient $\text{Ker}\phi_k/ \text{Im}\phi_{k-1}$. What this means is:

• It does not involve the cokernel of the maps (except when $\phi_k=0$ as Thomas pointed out).
• To compute one homology group you need two maps, not one.
• The formal $\mathbb{Z}$-module spanned by two objects is $\mathbb{Z}^2$ indeed, but here you really need to understand how the modules are embedded into each other.