Simplicial cohomology of $ \Bbb{R}\text{P}^2$ I've managed to confuse myself on a simple cohomology calculation.  I'm working with the usual $\Delta$-complex on $X = \mathbf{R}\mathbf{P}^2$ and I've computed the complex as $\newcommand{Z}{\mathbf{Z}}$
$$ 0 \to \Z \oplus \Z \stackrel{\partial^0}{\to} \Z \oplus \Z \oplus \Z \stackrel{\partial^1}{\to} \Z \oplus \Z \to 0 $$
with $\partial^0$ given by $(l, m) \mapsto (-l+m, -l+m, 0)$ and $\partial^1$ by $(l,m,n) \mapsto (l+m-n,  -l+m+n)$.  Then $\mathrm{Ker}(\partial^0) = \left<(1,1)\right> \cong \Z$ and $\mathrm{Im}(\partial^0) = \left<(1,1,0)\right> \cong \Z$.  For $\partial^1$ I got $\mathrm{Ker}(\partial^1) = \left<(1,0,1)\right> \cong \Z$ and I'm pretty confident about everything so far.
Now for $\mathrm{Im}(\partial^1)$ I first got $2\Z \oplus 2\Z$, since $(2, 0)$ and $(0, 2)$ are both in the image while $(1, 0)$ and $(0, 1)$ are not.  I don't see what's wrong with this logic, but it doesn't give the right answer: $H^2(X) \cong \Z \oplus \Z / (2\Z \oplus 2\Z) \cong \Z/2\Z \oplus \Z/2\Z$ while I believe the correct answer has only one copy.
A second approach I tried is the "isomorphism theorem" which says $\mathrm{Im}(\partial^1) \cong \Z \oplus \Z \oplus \Z / \mathrm{Ker}(\partial^1) = (\Z \oplus \Z \oplus \Z) / \Z \cong \Z \oplus \Z$.  But then $H^2(X) \cong \Z \oplus \Z / (\Z \oplus \Z) = 0$ is still wrong.
What's wrong with both of these approaches, and what's the correct one?
EDIT: I just realised that of course $\Z \oplus \Z \cong 2\Z \oplus 2\Z$ so both approaches actually give the same answer for $\mathrm{Im}(\partial^1)$.  More specifically I think it is generated by $\left<(1, 1), (1, -1)\right>$.  So I can only assume I'm computing the quotient $\Z^2/\mathrm{Im}(\partial^1)$ incorrectly.
To be very precise, we have the isomorphism
$$ H^2(X) = \Z \oplus \Z / \mathrm{Im}(\partial^1) \stackrel{\cong}{\to} \Z $$
given by $(m, n) + \left<(1, 1), (1, -1)\right> \mapsto m + n$.  Since $(m,n) \sim (0, m+n)$ this map is injective; and it is obviously surjective because $(n, 0)$ always maps to $n$ for any $n \in \Z$.
This is so weird......
EDIT: Of course, the problem with the above "isomorphism" is that it is not actually a well-defined homomorphism, as it doesn't agree on $(1, 1)$ and $(1, -1)$ (hence we mod out $2\Z$...)

Assuming that you have computed your cochain complex correctly, the problem with your first approach is that it is not true that
$$(A \oplus B)/(C \oplus D) \cong A/B \oplus C/D.$$
Instead to calculate $H^2(X)$ you will need to work with generators and relations. Define $a := (1,0)$ and $b:= (0,1)$, your basis vectors of $\Bbb{Z} \oplus \Bbb{Z}$. A basis for the image of $\partial^1$ is given by $a -b$ and $a + b$. So when you quotient out by $\operatorname{im} \partial^1$, you are effectively saying that 
$$H^2(X) \cong \langle a,b | a +b = a-b= 0\rangle$$
The relations $a + b= 0$ and $a - b = 0$ combine to give $2a = 0$, $a = -b$. This means
$$H^2(X) \cong \langle a | 2a = 0 \rangle \cong \Bbb{Z}/2\Bbb{Z}.$$
