Computing $H^\ast(S^1,\underline{\mathbb{Z}})$ I am trying to compute $H^\ast(S^1, \underline{\mathbb{Z}})$. Here is my approach: Let $U=S^1\setminus \{(1,0)\}$ and $V=S^{1}\setminus \{(-1,0)\}$. Then $U\simeq \{\ast\} \simeq V$ and $U\cap V \simeq \{\ast\} \sqcup \{\ast\}$. Since $S^{1}$ is connected, $H^{0}(S^1, \underline{\mathbb{Z}})=\underline{\mathbb{Z}}(S^1)=\mathbb{Z}$. By Mayer-Vietoris, we have $H^{i}(S^1, \underline{\mathbb{Z}})=0$ for $i\geq 2$ as it lies between two zeros. Moreover, $H^{0}(U,\underline{\mathbb{Z}}) \oplus H^{0}(V,\underline{\mathbb{Z}})=\mathbb{Z} \oplus \mathbb{Z}$ and $H^{0}(U\cap V, \underline{\mathbb{Z}})=H^{0}(\{\ast\},\underline{\mathbb{Z}})\oplus H^{0}(\{\ast\},\underline{\mathbb{Z}})=\mathbb{Z}\oplus \mathbb{Z}$, hence we have an exact sequence,
$$0\to \mathbb{Z} \xrightarrow{\alpha} \mathbb{Z} \oplus \mathbb{Z} \xrightarrow{\beta} \mathbb{Z}\oplus \mathbb{Z} \to H^{1}(S^1,\underline{\mathbb{Z}}) \to 0
$$
Thus, $H^{1}(S^1,\underline{\mathbb{Z}}) = \mathbb{Z}^2/\beta(\mathbb{Z}^2)$. Since $\beta(\mathbb{Z}^2)\cong \mathbb{Z}^2/\alpha(\mathbb{Z})$ and $\alpha(\mathbb{Z})$ is free of rank $1$, I feel like $H^1(S^1,\underline{\mathbb{Z}})=\mathbb{Z}$ but I don't think I can really conclude it without computing what $\alpha$ and $\beta$ actually do ($\alpha$ should just map $1$ to $(1,1)$). Is there a way of concluding $H^1(S^1,\underline{\mathbb{Z}})=\mathbb{Z}$ just by looking at the ranks (like it is possible to look at the dimension if this was a sequence of vector spaces)?
 A: Perhaps you could think about what the generators of $H^0$ actually are. For example, if $X$ is path-connected, then $H^0(X)$ has a single generator. This generator is a functional: you "feed in" a 0-cycle (i.e. a bunch of points) and this functional "spits out" a number equal to the number of points that you fed in. If $X$ has multiple path components, then $H^0(X)$ has multiple generators, associated to the various path components. You feed in a 0-cycle, and the generator spits out how many of the points in the 0-cycle lie in the path component that the generator is associated to.
Now let's work out how the morphisms in the Mayer-Vietoris sequence act. Given 0-cycles $\sigma_U$ and $\sigma_V$ in $U$ and $V$ respectively, and given a element $\psi \in H^0(S^1)$, we have
$$ \alpha(\psi)(\sigma_U ) = \psi(i_1(\sigma_U)), \\\alpha(\psi)(\sigma_V) = - \psi(i_2(\sigma_V)),$$
where $i_1: U \to S^1$ and $i_2 : V \to S^1$ are the inclusions. This is by the definition of the Mayer-Vietoris sequence.
Similarly, given a 0-cycle $\sigma_{U \cap V} $ in $U \cap V$, and given an element $\psi_U \in H^0(U )$ and an element $\psi_V \in H^0(V)$,
$$ \beta(\psi_U \oplus \psi_V)(\sigma_{U \cap V}) = \psi_U(j_1(\sigma_{U \cap V})) + \psi_V(j_2(\sigma_{U \cap V})),$$
where $j_1: U \cap V \to U$ and $j_2: U \cap V \to V$ are the natural inclusions.
Combining our description of the generators of the zeroth cohomology groups with our understanding of how the Mayer-Vietoris morphisms act, it shouldn't be too hard to see that
$$ \alpha = \left( \begin{array}{c} 1 \\ -1 \end{array}\right), \ \ \ \beta= \left( \begin{array}{c} 1 & 1 \\ 1 & 1 \end{array}\right).$$
[Alternatively, you can avoid these difficulties by working with reduced cohomology! The Mayer-Vietoris sequence for reduced cohomology gives the long exact sequence
$$ 0 \to 0 \to 0 \to \mathbb Z \to H^1(S^1) \to 0$$
which gives you the answer you want immediately.]
