We compute the simplicial cohomology of some basic cell complexes directly from the definitions, using coefficients in both $\mathbb{Z}$ and $\mathbb{Z}^2$.

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(a) Let $X = T^2$, with the cell complex structure given in Hatcher section 2.1. First work with coefficients in $\mathbb{Z}$. The cochain complex is:

$$0 \leftarrow \Delta^2 \xleftarrow{\delta_1} \Delta^1 \xleftarrow{\delta_0} \Delta^0 \leftarrow 0$$

Specifically,

$$0 \leftarrow \la U^*, L^* \ra \xleftarrow{\delta_1} \la a^*,b^*,c^* \ra \xleftarrow{\delta_0} \la v^* \ra \leftarrow 0$$

where we are viewing, for example $a^*$, as the map assigning $1$ to the edge $a$ and $0$ to the other edges. Clearly then, the group $\Delta^1$ is generated by $a^*$, $b^*$, and $c^*$ (similarly for the others).

We have $(\delta_0 v^*)(e) = v^*(v) - v^*(v)$ for all edges $e$, so $\delta_0 = 0$. For $\delta_1$, we have $(\delta_1 a^*)(U) = a^*(a) + a^*(b) - a^*(c) = 1$, and $(\delta_1 a^*)(L) = a^*(c) - a^*(a) - a^*(b) = -1$. Then, we can express $\delta_1 a^*$ as $\delta_1 a^* = U^* - L^*$. Similarly, we find $\delta_1 b^* = U^* - L^*$ and $\delta_1 c^* = L^* - U^*$. To summarize,

$$\delta_0 = \left( \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right) \xrightarrow{SNF} \left( \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right) \qquad \delta_1 = \left( \begin{array}{ccc} 1 & 1 & -1 \\ -1 & -1 & 1 \end{array} \right) \xrightarrow{SNF} \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right)$$

Determining the homology of the cochain complex is now a matter of algebra determined completely from the matrices (keeping in mind the homology groups are $\ker \delta_n / \im \delta_{n-1}$, instead of $\im \delta_{n+1}$), and we find:

$$H^0 = \mathbb{Z} \qquad H^1 = \mathbb{Z}^2 \qquad H^2 = \mathbb{Z}$$

Of course, this matches the homology of the torus exactly, because there is no torsion.

If we work with coefficients in $\mathbb{Z}_2$, the maps change slightly. We still have $\delta_0 = 0$, but $\delta_1(a^*) = \delta_1(b^*) = \delta_1(c^*) = U^* + L^*$. Then we have

$$\delta_1 = \left( \begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1 & 1 \end{array} \right) \xrightarrow{SNF} \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right)$$

and the homology is:

$$H^0 = \mathbb{Z}_2 \qquad H^1 = \mathbb{Z}_2^2 \qquad H^2 = \mathbb{Z}_2$$

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(b) Let $X = \mathbb{R}P^2$, with the cell complex structure from Hatcher 2.1. First work with coefficients in $\mathbb{Z}$. The cochain complex is:

$$0 \leftarrow \la U^*, L^* \ra \xleftarrow{\delta_1} \la a^*, b^*, c^* \ra \xleftarrow{\delta_0} \la v^*, w^* \ra \leftarrow 0$$

We have $(\delta_0 v^*)(a) = v^*(w) - v^*(v) = -1$, and likewise $(\delta_0 v^*)(b) = -1$. Also $(\delta_0 v^*)(c) = v^*(v) - v^*(v) = 0$. So $\delta_0 v^* = -(a^* + b^*)$. Similarly we find $\delta_0 w^* = a^* + b^*$. For $\delta_1$, we have $(\delta_1 a^*)(U) = a^*(a) - a^*(b) - a^*(c) = 1$ and $(\delta_1 a^*)(L) = a^*(c) + a^*(a) - a^*(b) = -1$. Thus $\delta_1 a^* = U^* - L^*$. Similarly we find $\delta_1 b^* = -(U^* + L^*)$ and $\delta_1 c^* = -U^* + L^*$. Summarizing:

$$\delta_0 = \left( \begin{array}{cc} -1 & 1 \\ -1 & 1 \\ 0 & 0 \end{array} \right) \xrightarrow{SNF} \left( \begin{array}{cc} 1 & 0 \\ 0 & 0 \\ 0 & 0 \end{array} \right) \qquad \delta_1 = \left( \begin{array}{ccc} 1 & -1 & -1 \\ -1 & -1 & 1 \end{array} \right) \xrightarrow{SNF} \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 2 & 0 \end{array} \right)$$

and we arrive at:

$$H^0 = \mathbb{Z} \qquad H^1 = 0 \qquad H^2 = \mathbb{Z}_2$$

If we work with coefficients in $\mathbb{Z}^2$, the maps change as follows:

$$\delta_0 = \left( \begin{array}{cc} 1 & 1 \\ 1 & 1 \\ 0 & 0 \end{array} \right) \xrightarrow{SNF} \left( \begin{array}{cc} 1 & 0 \\ 0 & 0 \\ 0 & 0 \end{array} \right) \qquad \delta_1 = \left( \begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1 & 1 \end{array} \right) \xrightarrow{SNF} \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right)$$

and we arrive at:

$$H^0 = \mathbb{Z}_2 \qquad H^1 = \mathbb{Z}_2 \qquad H^2 = \mathbb{Z}_2$$

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(c) Finally, let $X = K^2$ with the cell structure from 2.1. Again, work with $\mathbb{Z}$ coefficients first. The cochain complex is:

$$0 \leftarrow \la U^*, L^* \ra \xleftarrow{\delta_1} \la a^*,b^*,c^* \ra \xleftarrow{\delta_0} \la v^* \ra \leftarrow 0$$

We have $\delta_0 = 0$, because there is only one vertex (reasoning as in part (a)). We have $(\delta_1 a^*)(U) = a^*(a) + a^*(b) - a^*(c) = 1$, and $(\delta_1 a^*)(L) = a^*(c) + a^*(a) - a^*(b) = 1$. So we find $\delta_1 a^* = U^* + L^*$. For the others, we find $\delta_1 b^* = U^* - L^*$ and $\delta_1 c^* = -U^* + L^*$. So the maps are:

$$\delta_0 = \left( \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right) \xrightarrow{SNF} \left( \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right) \qquad \delta_1 = \left( \begin{array}{ccc} 1 & 1 & -1 \\ 1 & -1 & 1 \end{array} \right) \xrightarrow{SNF} \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 2 & 0 \end{array} \right)$$

and we have:

$$H^0 = \mathbb{Z} \qquad H^1 = \mathbb{Z} \qquad H^2 = \mathbb{Z}_2$$

If we work with coefficients in $\mathbb{Z}_2$, $\delta_1$ changes as follows:

$$\delta_1 = \left( \begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1 & 1 \end{array} \right) \xrightarrow{SNF} \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right)$$

and we arrive at:

$$H^0 = \mathbb{Z}_2 \qquad H^1 = \mathbb{Z}_2^2 \qquad H^2 = \mathbb{Z}_2$$