[proofplan] We prove by induction on the complexity of compact sets that every compact $A$ in an $R$-oriented $d$-manifold $M$ is "good" — meaning (1) there exists a unique class $\mu_A \in H_d(M \mid A; R)$ restricting to each local generator $\mu_x$, and (2) $H_i(M \mid A; R) = 0$ for $i > d$. The induction has three stages: first, convex compact subsets of $\mathbb{R}^d$ are good (by a direct homotopy equivalence argument); second, arbitrary compact subsets of $\mathbb{R}^d$ are good (by a Mayer-Vietoris induction on finite covers by convex compact sets); third, arbitrary compact subsets of $M$ are good (by a Mayer-Vietoris induction on a finite cover by Euclidean chart domains). [/proofplan] [step:Show that convex compact subsets of $\mathbb{R}^d$ are good] Let $A \subset \mathbb{R}^d$ be compact and convex, and fix $x_0 \in A$. The inclusion $\mathbb{R}^d \setminus A \hookrightarrow \mathbb{R}^d \setminus \{x_0\}$ is a homotopy equivalence: a homotopy inverse is provided by the map that retracts $\mathbb{R}^d \setminus \{x_0\}$ to $\mathbb{R}^d \setminus A$ by pushing points of $A \setminus \{x_0\}$ radially outward. More precisely, for $y \in \mathbb{R}^d \setminus \{x_0\}$, if $y \notin A$ then $y$ is fixed; if $y \in A \setminus \{x_0\}$, the ray from $x_0$ through $y$ exits $A$ at a unique boundary point (by convexity), and the retraction sends $y$ to that boundary point. The resulting homotopy is continuous because $A$ is convex. Applying the [Five Lemma](/theorems/1938) to the long exact sequences of the pairs $(M, M \setminus A)$ and $(M, M \setminus \{x_0\})$ connected by the inclusion of pairs, every vertical map on the $H_i(M)$ terms is the identity, and the maps $H_i(\mathbb{R}^d \setminus A) \to H_i(\mathbb{R}^d \setminus \{x_0\})$ are isomorphisms (by the homotopy equivalence above). The Five Lemma gives \begin{align*} H_i(\mathbb{R}^d \mid A; R) \xrightarrow{\;\sim\;} H_i(\mathbb{R}^d \mid x_0; R) \quad \text{for all } i. \end{align*} By the [Local Homology of Manifolds](/theorems/2253), $H_i(\mathbb{R}^d \mid x_0; R) \cong R$ for $i = d$ and $0$ otherwise. The generator $\mu_{x_0} \in H_d(\mathbb{R}^d \mid x_0; R)$ pulls back to a unique class $\mu_A \in H_d(\mathbb{R}^d \mid A; R)$, and for any $x \in A$ the same homotopy equivalence argument shows that the restriction $H_d(\mathbb{R}^d \mid A; R) \to H_d(\mathbb{R}^d \mid x; R)$ is an isomorphism sending $\mu_A$ to $\mu_x$. Therefore convex compact sets are good. [guided] The convexity of $A$ is used in two essential ways. First, it ensures that the inclusion $\mathbb{R}^d \setminus A \hookrightarrow \mathbb{R}^d \setminus \{x_0\}$ is a homotopy equivalence — this would fail for non-convex sets with "holes." Second, it guarantees that for any two points $x, y \in A$, the homotopy equivalences $\mathbb{R}^d \setminus A \simeq \mathbb{R}^d \setminus \{x\}$ and $\mathbb{R}^d \setminus A \simeq \mathbb{R}^d \setminus \{y\}$ are compatible, so the class $\mu_A$ restricts correctly to every local generator. Why does the retraction work? For any $y \neq x_0$, the ray $\{x_0 + t(y - x_0) : t > 0\}$ exits $A$ because $A$ is bounded. Convexity ensures the ray intersects $\partial A$ in exactly one point $p(y)$, and the map $y \mapsto p(y)$ is continuous (by the convexity and compactness of $A$). The homotopy $H(y, s) = (1-s)y + s \cdot p(y)$ for $y \in A \setminus \{x_0\}$, extended by the identity on $\mathbb{R}^d \setminus A$, provides the deformation retraction. [/guided] [/step] [step:Extend to arbitrary compact subsets of $\mathbb{R}^d$ by Mayer-Vietoris induction] Let $A \subset \mathbb{R}^d$ be an arbitrary compact set. Since $A$ is compact, it can be covered by finitely many open convex sets $V_1, \ldots, V_m$ (e.g., small open balls). Set $A_j := A \cap \overline{V}_j$, which is compact and convex (as the intersection of a compact set with a closed convex set). Then $A = A_1 \cup \cdots \cup A_m$. We induct on $m$. The base case $m = 1$ is handled by the previous step. For the inductive step, write $A = A' \cup A_m$ where $A' = A_1 \cup \cdots \cup A_{m-1}$, and assume both $A'$ and $A_m$ are good (and so is $A' \cap A_m$, since it is a compact subset of $\mathbb{R}^d$ covered by $m - 1$ convex compact sets). The Mayer-Vietoris sequence for the pair $(M, M \setminus A)$ gives \begin{align*} \cdots \to H_{i+1}(M \mid A' \cap A_m; R) \to H_i(M \mid A'; R) \oplus H_i(M \mid A_m; R) \to H_i(M \mid A; R) \to H_i(M \mid A' \cap A_m; R) \to \cdots \end{align*} **Vanishing for $i > d$.** Since $A'$, $A_m$, and $A' \cap A_m$ are all good, $H_i = 0$ for $i > d$ in the first, second, and fourth terms. The exact sequence forces $H_i(M \mid A; R) = 0$ for $i > d$. **Existence and uniqueness of $\mu_A$ for $i = d$.** The portion of the sequence at $i = d$ reads \begin{align*} 0 = H_{d+1}(M \mid A' \cap A_m; R) \to H_d(M \mid A'; R) \oplus H_d(M \mid A_m; R) \xrightarrow{\phi} H_d(M \mid A; R) \to H_d(M \mid A' \cap A_m; R). \end{align*} The map $\phi$ sends $(\alpha, \beta) \mapsto i_*(\alpha) + j_*(\beta)$ where $i, j$ are the inclusions. Consider the classes $\mu_{A'} \in H_d(M \mid A'; R)$ and $\mu_{A_m} \in H_d(M \mid A_m; R)$ provided by the inductive hypothesis. Both restrict to $\mu_x$ at each $x$ in their respective supports. The element $\mu_{A'} - \mu_{A_m}$ maps to zero under the map to $H_d(M \mid A' \cap A_m; R)$ because both classes restrict to $\mu_x$ for every $x \in A' \cap A_m$, and $\mu_{A' \cap A_m}$ is the unique such class, so both restrictions equal $\mu_{A' \cap A_m}$. By exactness, $\phi(\mu_{A'}, \mu_{A_m})$ is a well-defined class $\mu_A \in H_d(M \mid A; R)$ that restricts to $\mu_x$ for every $x \in A$. For uniqueness: suppose $\mu$ and $\mu'$ both restrict to $\mu_x$ for every $x \in A$. Then $\mu - \mu'$ restricts to zero at every point, so its images in $H_d(M \mid A'; R)$ and $H_d(M \mid A_m; R)$ are both zero (by uniqueness of $\mu_{A'}$ and $\mu_{A_m}$). Since $\phi$ has trivial kernel (from the vanishing of $H_{d+1}(M \mid A' \cap A_m; R) = 0$ and the exact sequence), $\mu = \mu'$. [guided] The Mayer-Vietoris induction is a standard technique for extending results from simple compact sets (convex) to general compact sets. The key point is that the intersection $A' \cap A_m$ is again covered by fewer convex sets than $A$, so the inductive hypothesis applies to it as well. The uniqueness argument is worth emphasizing. We need not just existence of $\mu_A$ but uniqueness — two fundamental classes with the same local restrictions must agree. This uses the injectivity provided by the vanishing of $H_{d+1}$ for the intersection, which is part of the inductive hypothesis. Without uniqueness, we could not glue local fundamental classes coherently. The reason we work with compact sets inside $\mathbb{R}^d$ first (rather than directly on the manifold) is that convexity is a Euclidean notion. On a general manifold, we do not have convex sets, but we do have chart domains, which are homeomorphic to $\mathbb{R}^d$ and inside which the Euclidean argument applies. [/guided] [/step] [step:Extend to compact subsets of a general manifold by covering with chart domains] Now let $M$ be a $d$-dimensional $R$-oriented manifold and $A \subseteq M$ compact. Since $M$ is locally homeomorphic to $\mathbb{R}^d$, every point of $A$ has a chart neighbourhood homeomorphic to $\mathbb{R}^d$. By compactness, $A$ is covered by finitely many chart domains $W_1, \ldots, W_k$ with homeomorphisms $\psi_j: W_j \xrightarrow{\sim} \mathbb{R}^d$. We induct on $k$. For the base case $k = 1$: $A \subseteq W_1 \cong \mathbb{R}^d$, so $A$ is a compact subset of $\mathbb{R}^d$ (via $\psi_1$), and the previous step shows $A$ is good. The orientation is used here: the homeomorphism $\psi_1$ transports the $R$-orientation to an orientation of $\mathbb{R}^d$, and the local generators $\mu_x$ are defined by the orientation, ensuring consistency. For the inductive step, write $A = A' \cup A_k$ where $A' = A \cap (W_1 \cup \cdots \cup W_{k-1})$ (more precisely, choose a closed cover: let $\{C_1, \ldots, C_k\}$ be closed sets with $C_j \subseteq W_j$ and $A \subseteq C_1 \cup \cdots \cup C_k$, and set $A' = A \cap (C_1 \cup \cdots \cup C_{k-1})$, $A_k = A \cap C_k$). The inductive hypothesis gives that $A'$, $A_k$, and $A' \cap A_k$ are all good. The same Mayer-Vietoris argument as in the previous step — applied to $A = A' \cup A_k$ — shows that $A$ is good: $H_i(M \mid A; R) = 0$ for $i > d$, and there exists a unique $\mu_A \in H_d(M \mid A; R)$ restricting to $\mu_x$ for every $x \in A$. [/step] [step:Verify the local compatibility condition using the orientation] The $R$-orientation of $M$ provides a continuous choice of generators: for each $x \in M$, a generator $\mu_x \in H_d(M \mid x; R) \cong R$, with the property that for each $x$ there exists an open neighbourhood $U_x$ and a class $\mu_{U_x} \in H_d(M \mid U_x; R)$ restricting to $\mu_y$ for every $y \in U_x$. This local compatibility is precisely the condition used in the Mayer-Vietoris induction: when we glue $\mu_{A'}$ and $\mu_{A_m}$ (or $\mu_{A'}$ and $\mu_{A_k}$), the fact that both restrict to the same local generator $\mu_x$ at each point of the intersection guarantees that the gluing is consistent. The uniqueness of $\mu_A$ follows from the induction: at each stage, the Mayer-Vietoris sequence together with the vanishing of $H_{d+1}$ for the intersection ensures that the fundamental class of the union is uniquely determined by its restrictions to the pieces. Since the pieces have unique fundamental classes by the inductive hypothesis, $\mu_A$ is unique. This completes the proof: for every compact $A \subseteq M$, there exists a unique class $\mu_A \in H_d(M \mid A; R)$ restricting to $\mu_x$ for every $x \in A$, and $H_i(M \mid A; R) = 0$ for $i > d$. [/step]