Homology calculation

When playing around with a homological calculation I came across a short exact sequence of the form $$0 \to \mathbb Z^{2g} \to H \to \mathbb Z / 2 \to 0$$ My background in algebra is not very strong. Does this sequence already imply the form of $H$?

For a lot of the calculations I am unsure when to switch from geometry to algebra and vice versa, i.e. how much geometrical information about the maps is necessary to pin down the homology group. It would be great if you could point me to some references where examples are worked out in detail.

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 Have you tried working out examples yourself? – Mariano Suárez-Alvarez♦ Apr 10 '12 at 21:10

$\newcommand\ZZ{\mathbb{Z}}$No, that short exact sequence does not determine the structure of $H$.
For example, suppose $g=1$. There are two short exact sequences $$0\to\ZZ^2\xrightarrow{\hskip1ex f_1\hskip1ex}\ZZ^2\oplus\ZZ/2\ZZ\xrightarrow{\hskip1ex g_1\hskip1ex}\ZZ/2\ZZ\to0$$ and $$0\to\ZZ^2\xrightarrow{\hskip1ex f_2\hskip1ex}\ZZ^2\xrightarrow{\hskip1ex g_2\hskip1ex}\ZZ/2\ZZ\to0$$ with \begin{align} &f_1=\begin{pmatrix}1&0&0\\0&1&0\\0&0&0\end{pmatrix}, &&g_1=\begin{pmatrix}0&0&1\end{pmatrix} \\ &f_2=\begin{pmatrix}2&0\\0&1\end{pmatrix}, &&g_2=\begin{pmatrix}1&0\end{pmatrix} \end{align}
@MarianoSuárez-Alvarez What does the notation $\Bbb{Z}^2$ mean, is it $\Bbb{Z} \oplus \Bbb{Z}$? – BenjaLim Apr 11 '12 at 21:24
Indeed. ${}{}{}$ – Mariano Suárez-Alvarez Apr 11 '12 at 21:49