[proofplan] We decompose $S^1$ into two overlapping open arcs $A$ and $B$, each contractible (hence homotopy equivalent to a point), whose intersection $A \cap B$ consists of two disjoint contractible components. The Mayer-Vietoris sequence then reduces the computation to an analysis of the connecting homomorphism $\partial: H_1(S^1) \to H_0(A \cap B)$ and the inclusion-induced map $H_0(A \cap B) \to H_0(A) \oplus H_0(B)$. The kernel-cokernel analysis yields $H_1(S^1) \cong \mathbb{Z}$, $H_0(S^1) \cong \mathbb{Z}$, and $H_k(S^1) = 0$ for $k \ge 2$. [/proofplan] [step:Construct the Mayer-Vietoris cover of $S^1$] Identify $S^1 = \{(x,y) \in \mathbb{R}^2 : x^2 + y^2 = 1\}$. Define the open arcs \begin{align*} A &= \{(\cos\theta, \sin\theta) : -\pi/4 < \theta < 5\pi/4\}, \\ B &= \{(\cos\theta, \sin\theta) : 3\pi/4 < \theta < 9\pi/4\}. \end{align*} Then $A \cup B = S^1$, and each of $A$ and $B$ is homeomorphic to an open interval in $\mathbb{R}$, hence contractible: $A \simeq *$ and $B \simeq *$. The intersection $A \cap B$ consists of two disjoint open arcs, each homeomorphic to an open interval. Denote the two components by $U_p$ (containing the point $p = (0, 1)$) and $U_q$ (containing the point $q = (0, -1)$). Since each component is contractible, $A \cap B \simeq \{p\} \sqcup \{q\}$. [guided] The key constraint is that $A$ and $B$ must be open (required for Mayer-Vietoris), their union must be $S^1$, and each must be contractible. Open arcs that are slightly more than semicircles achieve this. Their intersection then consists of two small arcs near the "east" and "west" points of the circle — or equivalently, near the top and bottom. Either convention works; we choose $p$ and $q$ as the top and bottom points. Since each arc is homeomorphic to $\mathbb{R}$ (via the angle parametrization restricted to an interval of length less than $2\pi$), $A$ and $B$ are contractible. Each component of $A \cap B$ is also an open interval, hence contractible. Therefore the homology of each piece is that of a point, while $A \cap B$ has the homology of a disjoint union of two points. [/guided] [/step] [step:Record the homology of the pieces] Since $A \simeq *$ and $B \simeq *$, the homology of a point gives: \begin{align*} H_k(A) = H_k(B) = \begin{cases} \mathbb{Z} & k = 0, \\ 0 & k \ge 1. \end{cases} \end{align*} Since $A \cap B \simeq \{p\} \sqcup \{q\}$, the homology of a two-point discrete space gives: \begin{align*} H_k(A \cap B) = \begin{cases} \mathbb{Z} \oplus \mathbb{Z} & k = 0, \\ 0 & k \ge 1. \end{cases} \end{align*} Here $H_0(A \cap B) \cong \mathbb{Z}^2$ is generated by the classes $[p]$ and $[q]$ of the two $0$-simplices (one in each connected component). [/step] [step:Extract $H_k(S^1) = 0$ for $k \ge 2$ from the Mayer-Vietoris sequence] The Mayer-Vietoris sequence for the cover $S^1 = A \cup B$ reads, for each $k$: \begin{align*} \cdots \to H_k(A \cap B) \xrightarrow{\Phi_k} H_k(A) \oplus H_k(B) \xrightarrow{\Psi_k} H_k(S^1) \xrightarrow{\partial_k} H_{k-1}(A \cap B) \to \cdots \end{align*} For $k \ge 2$, we have $H_k(A) = H_k(B) = 0$ and $H_{k-1}(A \cap B) = 0$ (since $k - 1 \ge 1$). By exactness, $H_k(S^1) = 0$. [/step] [step:Compute $H_1(S^1) \cong \mathbb{Z}$ via the connecting homomorphism] The relevant portion of the Mayer-Vietoris sequence for $k = 1$ is: \begin{align*} H_1(A) \oplus H_1(B) \xrightarrow{\Psi_1} H_1(S^1) \xrightarrow{\partial_1} H_0(A \cap B) \xrightarrow{\Phi_0} H_0(A) \oplus H_0(B). \end{align*} Since $H_1(A) = H_1(B) = 0$, exactness at $H_1(S^1)$ gives $\ker(\partial_1) = \operatorname{im}(\Psi_1) = 0$, so $\partial_1$ is injective. By exactness at $H_0(A \cap B)$, the image of $\partial_1$ equals $\ker(\Phi_0)$. We now compute $\Phi_0$. The map $\Phi_0$ is defined by \begin{align*} \Phi_0: H_0(A \cap B) &\to H_0(A) \oplus H_0(B), \\ [x] &\mapsto (i_{A*}[x], \, i_{B*}[x]), \end{align*} where $i_A: A \cap B \hookrightarrow A$ and $i_B: A \cap B \hookrightarrow B$ are the inclusions. Since $A$ is path-connected, both $i_{A*}[p]$ and $i_{A*}[q]$ map to the unique generator of $H_0(A) \cong \mathbb{Z}$. Similarly for $B$. Writing $H_0(A \cap B) = \mathbb{Z}[p] \oplus \mathbb{Z}[q]$: \begin{align*} \Phi_0(a[p] + b[q]) = (a + b, \, a + b). \end{align*} Therefore $\ker(\Phi_0) = \{a[p] + b[q] : a + b = 0\} = \mathbb{Z} \cdot ([p] - [q])$, which is isomorphic to $\mathbb{Z}$. Since $\partial_1: H_1(S^1) \hookrightarrow \ker(\Phi_0) \cong \mathbb{Z}$ is injective, and $\operatorname{im}(\partial_1) = \ker(\Phi_0) \cong \mathbb{Z}$ by exactness, we conclude $H_1(S^1) \cong \mathbb{Z}$. [guided] Why is the connecting homomorphism the right tool here? Because $H_1(A) = H_1(B) = 0$, the Mayer-Vietoris sequence forces $\partial_1$ to be an injection from $H_1(S^1)$ into $H_0(A \cap B)$. Meanwhile, exactness at $H_0(A \cap B)$ identifies the image of $\partial_1$ as the kernel of $\Phi_0$. So $H_1(S^1) \cong \ker(\Phi_0)$, reducing the problem to a concrete linear algebra computation. The map $\Phi_0$ sends both generators $[p]$ and $[q]$ of $H_0(A \cap B)$ to the same element in each summand because $p$ and $q$ lie in the same path-component of $A$ (and of $B$). The kernel consists of linear combinations $a[p] + b[q]$ with $a + b = 0$, i.e., multiples of $[p] - [q]$. This generator represents a $1$-cycle that traverses the circle once: it enters through one component of $A \cap B$ and exits through the other. [/guided] [/step] [step:Compute $H_0(S^1) \cong \mathbb{Z}$] The Mayer-Vietoris sequence continues: \begin{align*} H_0(A \cap B) \xrightarrow{\Phi_0} H_0(A) \oplus H_0(B) \xrightarrow{\Psi_0} H_0(S^1) \to 0. \end{align*} The map $\Psi_0$ is surjective (the sequence terminates at $H_{-1} = 0$). By exactness at $H_0(A) \oplus H_0(B)$, we have $H_0(S^1) \cong (H_0(A) \oplus H_0(B)) / \operatorname{im}(\Phi_0)$. Since $\operatorname{im}(\Phi_0) = \{(a+b, a+b) : a, b \in \mathbb{Z}\} = \{(n, n) : n \in \mathbb{Z}\}$ (the diagonal subgroup of $\mathbb{Z} \oplus \mathbb{Z}$), the quotient $(\mathbb{Z} \oplus \mathbb{Z}) / \{(n,n)\} \cong \mathbb{Z}$ via $(a, b) \mapsto a - b$. Alternatively, since $S^1$ is path-connected, $H_0(S^1) \cong \mathbb{Z}$ by the [$H_0$ and Path Components](/theorems/2235) theorem. [/step] [step:Assemble the result] Combining the computations: \begin{align*} H_k(S^1) = \begin{cases} \mathbb{Z} & k = 0, 1, \\ 0 & k \ge 2. \end{cases} \end{align*} Since the singular chain groups $C_k(X)$ are zero for $k < 0$, we also have $H_k(S^1) = 0$ for $k < 0$. This completes the computation. [/step]