image of boundary operator in homology I'm starting to learn homologies and I'm facing a lot of problems. We have a matrix $A$, let's say:
$$ \begin{pmatrix}
        1 & 2\\
        3 & 4\\
        \end{pmatrix}$$
And there is a chain:
$$\ldots \to 0 \to \mathbb{Z}^2 \to \mathbb{Z}^2 \to 0$$
Where the boundary operator between these nontrivial groups is given by a matrix $A$. How should I proceed find the image of the boundary operator to compute the homology group?
 A: Since $\alpha = (1,0)$ and $\beta = (0,1)$ generate $\Bbb{Z}^2$, let's see where your operator takes them.  Let $T = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$. Then $T\alpha = (1,3)$ and $T\beta = (2,4)$.  Then for arbitrary, $a,b \in \Bbb{Z}$, $T(a \alpha + b \beta) = (a+2b, 3a+4b)$.  As $a,b$ independently range through all of the integers, you get some subset of the $\Bbb{Z}^2$ in your long sequence.  Which one?
We can take linear combinations of $a+2b$ and $3a+4b$ and still have en element in their range.  In fact, as long as we don't apply a singular transformation, we can replace these two vectors with two linear combinations and span the same subset of $\Bbb{Z}^2$.  Subtracting the first from the second twice, we have $\{a + 2b, 3a +4b - (2a + 4b) \} = \{a+2b, a\}$.  Then subtracting the second from the first, $\{2b,a\}$.  Notice that all we've really done is Gaussian elimination.  This should suggest any number of ways to reduce the first pair of vectors to a useful pair.  Now you know the image is $\Bbb{Z} \times 2\Bbb{Z}$.
The kernel of the next map is $\Bbb{Z} \times \Bbb{Z}$ and the image of this map is $\Bbb{Z} \times 2 \Bbb{Z}$, so the quotient is $\Bbb{Z}/2\Bbb{Z}$, as requested.
I note that with practice, you quickly get to where you reduce $T$ by Gaussian elimination in your head and then write down the quotient.  This takes practice, though.
