[proofplan] We compute $H_n(M, M \setminus \{x\})$ by excising the complement of a Euclidean chart neighbourhood, reducing to $H_n(\mathbb{R}^d, \mathbb{R}^d \setminus \{0\})$. We then compute the latter using the long exact sequence of the pair, together with the facts that $\mathbb{R}^d$ is contractible and $\mathbb{R}^d \setminus \{0\} \simeq S^{d-1}$. [/proofplan] [step:Excise the complement of a chart neighbourhood to reduce to $\mathbb{R}^d$] Since $M$ is a $d$-dimensional manifold, there exists an open neighbourhood $U$ of $x$ and a homeomorphism $\varphi: U \xrightarrow{\sim} \mathbb{R}^d$ with $\varphi(x) = 0$. Set $Z = M \setminus U$. We verify the excision hypothesis: $Z$ is a closed subset of $M$ with $Z \subset M \setminus \{x\}$, and $M \setminus \{x\}$ is open in $M$, so $\overline{Z} = Z \subset \operatorname{int}(M \setminus \{x\}) = M \setminus \{x\}$. By the [Excision Theorem](/theorems/???), we obtain \begin{align*} H_n(M, M \setminus \{x\}) \cong H_n(M \setminus Z, (M \setminus \{x\}) \setminus Z) = H_n(U, U \setminus \{x\}). \end{align*} Since $\varphi: U \xrightarrow{\sim} \mathbb{R}^d$ is a homeomorphism sending $x$ to $0$, this gives $H_n(U, U \setminus \{x\}) \cong H_n(\mathbb{R}^d, \mathbb{R}^d \setminus \{0\})$. [guided] The strategy is to "zoom in" to a small Euclidean chart around $x$, discarding the rest of the manifold. Excision allows us to remove the closed set $Z = M \setminus U$ from both the space and the subspace without changing relative homology, provided $Z$ is contained in the interior of the subspace $M \setminus \{x\}$. Why is the excision hypothesis satisfied? The subspace is $M \setminus \{x\}$, which is open in $M$ (since $\{x\}$ is closed in a Hausdorff space, and manifolds are Hausdorff by definition). Therefore $\operatorname{int}(M \setminus \{x\}) = M \setminus \{x\}$. The set $Z = M \setminus U$ is closed (complement of open $U$), and $Z \subset M \setminus \{x\}$ (since $x \in U$). So $\overline{Z} = Z \subset M \setminus \{x\} = \operatorname{int}(M \setminus \{x\})$, and excision applies. After excision, we are left with the pair $(U, U \setminus \{x\})$. The homeomorphism $\varphi: U \to \mathbb{R}^d$ induces isomorphisms on all relative homology groups, so the problem reduces to computing $H_n(\mathbb{R}^d, \mathbb{R}^d \setminus \{0\})$. [/guided] [/step] [step:Compute $H_n(\mathbb{R}^d, \mathbb{R}^d \setminus \{0\})$ via the long exact sequence of the pair] The long exact sequence of the pair $(\mathbb{R}^d, \mathbb{R}^d \setminus \{0\})$ reads \begin{align*} \cdots \to H_n(\mathbb{R}^d) \to H_n(\mathbb{R}^d, \mathbb{R}^d \setminus \{0\}) \xrightarrow{\partial} H_{n-1}(\mathbb{R}^d \setminus \{0\}) \to H_{n-1}(\mathbb{R}^d) \to \cdots \end{align*} Since $\mathbb{R}^d$ is contractible, $H_n(\mathbb{R}^d) = 0$ for all $n \geq 1$ and $H_0(\mathbb{R}^d) \cong \mathbb{Z}$. Therefore, for $n \geq 2$, the sequence gives isomorphisms \begin{align*} \partial: H_n(\mathbb{R}^d, \mathbb{R}^d \setminus \{0\}) \xrightarrow{\sim} H_{n-1}(\mathbb{R}^d \setminus \{0\}). \end{align*} The inclusion $S^{d-1} \hookrightarrow \mathbb{R}^d \setminus \{0\}$ is a homotopy equivalence (with homotopy inverse given by radial projection $y \mapsto y / |y|$), so $H_{n-1}(\mathbb{R}^d \setminus \{0\}) \cong H_{n-1}(S^{d-1})$. By the [Homology of Spheres](/theorems/1945), $H_{n-1}(S^{d-1}) \cong \mathbb{Z}$ if $n - 1 = d - 1$ (i.e., $n = d$) or $n - 1 = 0$ (i.e., $n = 1$), and $H_{n-1}(S^{d-1}) = 0$ otherwise. [guided] The long exact sequence breaks the relative homology computation into two pieces: the homology of the ambient space $\mathbb{R}^d$ and the homology of the deleted space $\mathbb{R}^d \setminus \{0\}$. Since $\mathbb{R}^d$ is contractible, all its positive-dimensional homology vanishes, and the connecting homomorphism $\partial$ becomes an isomorphism in most degrees. The space $\mathbb{R}^d \setminus \{0\}$ deformation-retracts onto $S^{d-1}$ via radial projection: define $r: \mathbb{R}^d \setminus \{0\} \to S^{d-1}$ by $r(y) = y / |y|$, and $\iota: S^{d-1} \hookrightarrow \mathbb{R}^d \setminus \{0\}$ the inclusion. Then $r \circ \iota = \operatorname{id}_{S^{d-1}}$, and $\iota \circ r \simeq \operatorname{id}_{\mathbb{R}^d \setminus \{0\}}$ via the homotopy $H(y, t) = (1-t)y + t \cdot y/|y|$ (which is well-defined since $y \neq 0$). So $H_{n-1}(\mathbb{R}^d \setminus \{0\}) \cong H_{n-1}(S^{d-1})$, and we know $H_k(S^{d-1})$ from the computation of spherical homology: it is $\mathbb{Z}$ in degrees $0$ and $d-1$, and zero elsewhere. [/guided] [/step] [step:Handle the low-degree cases and assemble the final answer] For $n \geq 2$, we have established $H_n(\mathbb{R}^d, \mathbb{R}^d \setminus \{0\}) \cong H_{n-1}(S^{d-1})$. This gives: - $H_d(\mathbb{R}^d, \mathbb{R}^d \setminus \{0\}) \cong H_{d-1}(S^{d-1}) \cong \mathbb{Z}$. - $H_n(\mathbb{R}^d, \mathbb{R}^d \setminus \{0\}) \cong H_{n-1}(S^{d-1}) = 0$ for $n \geq 2$ with $n \neq d$ (since $n - 1 \neq 0$ and $n - 1 \neq d - 1$). For $n = 1$, the relevant portion of the long exact sequence is \begin{align*} H_1(\mathbb{R}^d) \to H_1(\mathbb{R}^d, \mathbb{R}^d \setminus \{0\}) \xrightarrow{\partial} H_0(\mathbb{R}^d \setminus \{0\}) \xrightarrow{i_*} H_0(\mathbb{R}^d). \end{align*} Since $H_1(\mathbb{R}^d) = 0$, the map $\partial$ is injective. Both $\mathbb{R}^d \setminus \{0\}$ (for $d \geq 2$) and $\mathbb{R}^d$ are path-connected, so $H_0(\mathbb{R}^d \setminus \{0\}) \cong \mathbb{Z}$ and $H_0(\mathbb{R}^d) \cong \mathbb{Z}$, and the inclusion-induced map $i_*$ is an isomorphism. Therefore $\partial$ maps into $\ker(i_*) = 0$, giving $H_1(\mathbb{R}^d, \mathbb{R}^d \setminus \{0\}) = 0$ for $d \geq 2$. When $d = 1$, $\mathbb{R}^1 \setminus \{0\}$ has two path components, so $H_0(\mathbb{R}^1 \setminus \{0\}) \cong \mathbb{Z}^2$, and $i_*$ sends both generators to the single generator of $H_0(\mathbb{R}^1) \cong \mathbb{Z}$. The kernel of $i_*$ is $\mathbb{Z}$ (generated by $[p] - [q]$ for points $p$ and $q$ in distinct components), and $\partial$ maps isomorphically onto it, giving $H_1(\mathbb{R}^1, \mathbb{R}^1 \setminus \{0\}) \cong \mathbb{Z}$. This is consistent with $n = d = 1$. For $n = 0$, the sequence $H_0(\mathbb{R}^d) \to H_0(\mathbb{R}^d, \mathbb{R}^d \setminus \{0\})$ and the fact that any $0$-chain in $\mathbb{R}^d$ can be connected to one in $\mathbb{R}^d \setminus \{0\}$ (for $d \geq 1$) gives $H_0(\mathbb{R}^d, \mathbb{R}^d \setminus \{0\}) = 0$. Combining all cases: \begin{align*} H_n(M, M \setminus \{x\}) \cong H_n(\mathbb{R}^d, \mathbb{R}^d \setminus \{0\}) \cong \begin{cases} \mathbb{Z} & n = d, \\ 0 & \text{otherwise.} \end{cases} \end{align*} [/step]