[proofplan] We work with compactly supported de Rham cohomology $H^*_c$ and observe that for $M$ compact one has $H^k_c(M) = H^k_{\mathrm{dR}}(M)$ for every $k$. We therefore prove the more general statement that, for every open subset $U \subseteq M$ admitting a finite good cover, the wedge-and-integrate pairing $H^k_{\mathrm{dR}}(U) \times H^{n-k}_c(U) \to \mathbb{R}$ is non-degenerate. The strategy is induction on the cardinality of the good cover. The base case is $U$ diffeomorphic to $\mathbb{R}^n$, where the Poincaré lemma reduces the duality to integration $H^n_c(\mathbb{R}^n) \cong \mathbb{R}$. The inductive step compares the [Mayer–Vietoris](/theorems/1533) long exact sequence for $H^*_{\mathrm{dR}}$ with the [Mayer–Vietoris for Compactly Supported Cohomology](/theorems/2289) via a ladder of duality maps, and the [Five Lemma](/theorems/1938) propagates non-degeneracy across the induction. Specialising to $U = M$ — which admits a finite good cover because $M$ is compact — yields the theorem. [/proofplan] [step:Reduce to a statement about the compactly supported pairing on open subsets of $M$] Let $M$ be as in the hypothesis. For an open subset $U \subseteq M$ and integer $k \geq 0$, define \begin{align*} \Omega^k_c(U) &:= \{\omega \in \Omega^k(U) : \operatorname{supp} \omega \text{ is a compact subset of } U\}, \end{align*} the space of compactly supported smooth $k$-forms on $U$. The [exterior derivative](/theorems/1525) restricts to $d: \Omega^k_c(U) \to \Omega^{k+1}_c(U)$, and the resulting cohomology is the **compactly supported de Rham cohomology** $H^k_c(U)$. Since $M$ is itself compact, every smooth form on $M$ has compact support, hence $\Omega^k_c(M) = \Omega^k(M)$ and consequently \begin{align*} H^k_c(M) = H^k_{\mathrm{dR}}(M) \quad \text{for every } k \geq 0. \end{align*} For an open $U \subseteq M$ inheriting the orientation of $M$, define the **wedge-and-integrate pairing** \begin{align*} P_U^k : H^k_{\mathrm{dR}}(U) \times H^{n-k}_c(U) &\to \mathbb{R}, \\ ([\alpha], [\beta]) &\mapsto \int_U \alpha \wedge \beta. \end{align*} When $U = M$ this coincides with $\langle \cdot, \cdot \rangle_{\mathrm{PD}}$. The theorem will follow once we show $P_M^k$ is a non-degenerate pairing between finite-dimensional spaces. We will prove, more generally, that $P_U^k$ is non-degenerate and that both sides are finite-dimensional whenever $U$ admits a finite good cover (defined in a later step). [/step] [step:Verify well-definedness of the pairing via Stokes' Theorem] Let $\alpha \in \Omega^k(U)$ and $\beta \in \Omega^{n-k}_c(U)$ with $d\alpha = 0$ and $d\beta = 0$. The form $\alpha \wedge \beta \in \Omega^n_c(U)$ has compact support because $\operatorname{supp}(\alpha \wedge \beta) \subseteq \operatorname{supp} \beta$, so $\int_U \alpha \wedge \beta$ is finite by the standard theory of [Integration of Differential Forms](/theorems/1529). We verify the pairing descends to cohomology classes. Suppose $\alpha = \alpha_0 + d\eta$ for some $\eta \in \Omega^{k-1}(U)$. Since $d\beta = 0$, the Leibniz rule for the [exterior derivative](/theorems/1525) gives \begin{align*} d(\eta \wedge \beta) = d\eta \wedge \beta + (-1)^{k-1}\eta \wedge d\beta = d\eta \wedge \beta. \end{align*} The form $\eta \wedge \beta$ has support inside $\operatorname{supp} \beta$, which is compact in $U$. Applying [Stokes' Theorem](/theorems/1530) to the compactly supported form $\eta \wedge \beta$ on the boundaryless manifold $U$: \begin{align*} \int_U d\eta \wedge \beta = \int_U d(\eta \wedge \beta) = 0. \end{align*} Hence $\int_U \alpha \wedge \beta = \int_U \alpha_0 \wedge \beta$. The symmetric argument with $\beta = \beta_0 + d\zeta$ for compactly supported $\zeta \in \Omega^{n-k-1}_c(U)$ shows independence on the choice of representative of $[\beta]$ as well. Thus $P_U^k$ is well-defined. [/step] [step:Establish the base case $U \cong \mathbb{R}^n$ for the duality] Let $U \subseteq M$ be open and assume there exists an orientation-preserving diffeomorphism $\varphi: U \to \mathbb{R}^n$. We compute both sides of $P_U^k$. For the de Rham cohomology of $\mathbb{R}^n$, the Poincaré lemma gives \begin{align*} H^k_{\mathrm{dR}}(\mathbb{R}^n) = \begin{cases} \mathbb{R}, & k = 0, \\ 0, & k \geq 1. \end{cases} \end{align*} For the compactly supported cohomology of $\mathbb{R}^n$, the [Top Cohomology of Orientable Manifolds](/theorems/1531) applied to the connected oriented $n$-manifold $\mathbb{R}^n$ asserts that integration gives an isomorphism \begin{align*} \int_{\mathbb{R}^n}: H^n_c(\mathbb{R}^n) \xrightarrow{\;\;\cong\;\;} \mathbb{R}, \qquad [\omega] \mapsto \int_{\mathbb{R}^n} \omega, \end{align*} and the standard computation gives $H^j_c(\mathbb{R}^n) = 0$ for $j \neq n$. Thus, for $k \neq 0$, both vector spaces paired by $P_{\mathbb{R}^n}^k$ are zero, so the pairing is non-degenerate. In degree $k=0$: \begin{align*} P_{\mathbb{R}^n}^0: H^0_{\mathrm{dR}}(\mathbb{R}^n) \times H^n_c(\mathbb{R}^n) &\to \mathbb{R}, \\ (c, [\omega]) &\mapsto c \int_{\mathbb{R}^n} \omega. \end{align*} This is non-degenerate because $H^0_{\mathrm{dR}}(\mathbb{R}^n) = \mathbb{R} \cdot 1$ and the integration map is an isomorphism. Both spaces are one-dimensional. Pulling back along $\varphi$, the same holds for $U$. The case $k=0$ is the heart of the duality: it says integration is the dual pairing between the constants and the top compactly supported forms. Why does the orientation matter? Because integration of an $n$-form on an oriented manifold is well-defined only after a choice of orientation; reversing the orientation flips the sign of $\int$. The non-degeneracy of $P_{\mathbb{R}^n}^0$ amounts to the fact that for any non-zero $[\omega] \in H^n_c(\mathbb{R}^n)$, $\int_{\mathbb{R}^n} \omega \neq 0$, which is precisely the content of [Top Cohomology of Orientable Manifolds](/theorems/1531). All other degrees of $P_{\mathbb{R}^n}^k$ are vacuously non-degenerate because both sides are zero. The Poincaré lemma is invoked here in its standard form: a closed form on a contractible (in particular star-shaped) open subset of $\mathbb{R}^n$ is exact. The vanishing of $H^j_c(\mathbb{R}^n)$ for $j<n$ is a more delicate fact, since compactly supported forms cannot be primitivised by integration over an unbounded ray; the standard proof passes through the compactly supported Poincaré lemma on $\mathbb{R}^n$ (citing a result not yet in the wiki: Compactly Supported Poincaré Lemma for $\mathbb{R}^n$). [/step] [step:Define a good cover and reduce to a statement about finite good covers] A **good cover** of an open subset $U \subseteq M$ is an open cover $\mathcal{U} = \{U_\alpha\}_{\alpha \in I}$ of $U$ such that every non-empty finite intersection $U_{\alpha_1} \cap \cdots \cap U_{\alpha_r}$ is diffeomorphic to $\mathbb{R}^n$. [claim:Every compact smooth manifold admits a finite good cover.] [proof] Equip $M$ with a Riemannian metric $g$ (possible by a [partition of unity](/page/Partition%20of%20Unity) argument; see [Existence of Smooth Partitions of Unity](/theorems/57)). For each $p \in M$, choose a geodesically convex neighbourhood $V_p$ of $p$, i.e., an open neighbourhood such that any two points in $V_p$ are joined by a unique minimizing geodesic contained in $V_p$. The intersection of finitely many geodesically convex sets is again geodesically convex, and a geodesically convex [open set](/page/Open%20Set) is diffeomorphic to $\mathbb{R}^n$ via the exponential map at any of its points. Hence $\{V_p\}_{p \in M}$ is a good cover. By compactness, this open cover has a finite subcover $\{V_{p_1}, \dots, V_{p_N}\}$, which is again a good cover because any finite intersection $V_{p_{i_1}} \cap \cdots \cap V_{p_{i_r}}$ is geodesically convex (citing a result not yet in the wiki: Existence of Geodesically Convex Neighborhoods). [/proof] [/claim] We will prove by induction on $N$ that, for any open $U \subseteq M$ admitting a good cover of cardinality $N$, the pairing $P_U^k$ is non-degenerate for every $k$ and both $H^k_{\mathrm{dR}}(U)$ and $H^{n-k}_c(U)$ are finite-dimensional. Applying this to $M$ itself with the finite good cover produced above yields the theorem. [/step] [step:Set up the Mayer–Vietoris ladder of duality maps] For $V, W \subseteq M$ open with $U = V \cup W$, define the **[Poincaré duality](/theorems/2291) map** at degree $k$ by \begin{align*} \mathrm{PD}_U^k: H^k_{\mathrm{dR}}(U) &\to \left(H^{n-k}_c(U)\right)^*, \\ [\alpha] &\mapsto \left( [\beta] \mapsto \int_U \alpha \wedge \beta \right). \end{align*} By the well-definedness step, $\mathrm{PD}_U^k$ is a well-defined [linear map](/page/Linear%20Map), and $\mathrm{PD}_U^k$ is injective if and only if the pairing $P_U^k$ is non-degenerate in its first variable. Non-degeneracy in the second variable corresponds to surjectivity of $\mathrm{PD}_U^k$ (once both sides are known finite-dimensional and equal-dimensional). The [Mayer–Vietoris](/theorems/1533) long exact sequence for de Rham cohomology gives \begin{align*} \cdots \to H^{k-1}_{\mathrm{dR}}(V \cap W) \xrightarrow{\delta} H^k_{\mathrm{dR}}(U) \to H^k_{\mathrm{dR}}(V) \oplus H^k_{\mathrm{dR}}(W) \to H^k_{\mathrm{dR}}(V \cap W) \to \cdots, \end{align*} where the connecting homomorphism $\delta$ is constructed using a [partition of unity](/page/Partition%20of%20Unity) subordinate to $\{V, W\}$. Dually, the [Mayer–Vietoris for Compactly Supported Cohomology](/theorems/2289) gives, with arrows in the opposite direction: \begin{align*} \cdots \to H^{n-k-1}_c(V \cap W) \to H^{n-k}_c(V) \oplus H^{n-k}_c(W) \to H^{n-k}_c(U) \xrightarrow{\delta_c} H^{n-k+1}_c(V \cap W) \to \cdots. \end{align*} Since all terms are real vector spaces, the contravariant functor $E \mapsto E^*$ is exact on $\mathrm{Vect}_{\mathbb{R}}$: every short exact sequence of real vector spaces dualises to a short exact sequence. Therefore taking the linear dual of the second long exact sequence reverses its arrows and preserves exactness, producing a long exact sequence of dual spaces parallel to the first. The duality maps $\mathrm{PD}^k$ on $V$, $W$, $V \cap W$, and $U$ assemble into a ladder of commutative squares: \begin{align*} \begin{array}{ccccccc} H^{k-1}_{\mathrm{dR}}(V \cap W) & \to & H^k_{\mathrm{dR}}(U) & \to & H^k_{\mathrm{dR}}(V) \oplus H^k_{\mathrm{dR}}(W) & \to & H^k_{\mathrm{dR}}(V \cap W) \\ \downarrow \mathrm{PD} & & \downarrow \mathrm{PD} & & \downarrow \mathrm{PD} \oplus \mathrm{PD} & & \downarrow \mathrm{PD} \\ (H^{n-k+1}_c(V \cap W))^* & \to & (H^{n-k}_c(U))^* & \to & (H^{n-k}_c(V))^* \oplus (H^{n-k}_c(W))^* & \to & (H^{n-k}_c(V \cap W))^* \end{array} \end{align*} Commutativity of the squares not involving the connecting homomorphism is immediate from naturality of the wedge-integrate pairing under inclusion of open sets. Commutativity of the squares involving $\delta$ and $\delta_c^*$ requires a direct computation with the partition-of-unity construction of the connecting maps; the sign appearing is $(-1)^k$, which we absorb by replacing $\mathrm{PD}^k$ by $(-1)^{k(k+1)/2} \mathrm{PD}^k$ if necessary (a sign convention; we adopt it tacitly). The construction of the ladder is the technical heart of the proof. Why do the two Mayer–Vietoris sequences run in opposite directions? Because $H^k_{\mathrm{dR}}$ is a contravariant functor on open sets in the natural sense (forms restrict), whereas $H^k_c$ is covariant under open inclusions via extension by zero — a compactly supported form on $V$ extends to a compactly supported form on $U \supseteq V$. Dualising the compactly supported sequence brings the arrows back into the same direction as the de Rham sequence, and the duality maps $\mathrm{PD}$ are the rungs of the ladder. The commutativity of the square involving the connecting homomorphisms is the most delicate point. Briefly: $\delta$ is defined by $\delta[\omega] = [d(\rho_W \omega|_V)]$ for a [partition of unity](/page/Partition%20of%20Unity) $\{\rho_V, \rho_W\}$ subordinate to $\{V, W\}$, while $\delta_c$ on compactly supported forms is defined by an analogous but sign-shifted construction. The verification that \begin{align*} \langle \delta[\alpha], [\beta] \rangle_U = \pm \langle [\alpha], \delta_c[\beta] \rangle_{V \cap W} \end{align*} is a Stokes calculation: expand both sides using the [partition of unity](/page/Partition%20of%20Unity), apply Leibniz, and the boundary terms vanish because $\beta$ is compactly supported. The sign $(-1)^k$ on the right is universal and depends only on $k$; absorbing it into the duality map is a standard normalisation. [/step] [step:Apply the Five Lemma to propagate non-degeneracy across the induction] We induct on the cardinality $N \geq 1$ of a finite good cover of an [open set](/page/Open%20Set) $U \subseteq M$. **Base case $N = 1$**: $U$ is itself diffeomorphic to $\mathbb{R}^n$. By the base-case step, $\mathrm{PD}_U^k$ is an isomorphism for every $k$, and both sides are finite-dimensional (zero except in one degree where they are one-dimensional). **Inductive step**: Suppose the statement holds for every open subset of $M$ admitting a good cover of cardinality $\leq N - 1$, and let $U \subseteq M$ admit a good cover $\{U_1, \dots, U_N\}$. Set \begin{align*} V := U_1 \cup \cdots \cup U_{N-1}, \qquad W := U_N. \end{align*} Then $\{U_1, \dots, U_{N-1}\}$ is a good cover of $V$ of cardinality $N - 1$, so the inductive hypothesis applies to $V$. The set $W$ is diffeomorphic to $\mathbb{R}^n$ (base case). The intersection \begin{align*} V \cap W = (U_1 \cap U_N) \cup \cdots \cup (U_{N-1} \cap U_N) \end{align*} is covered by the $N-1$ open sets $\{U_i \cap U_N\}_{i=1}^{N-1}$. Each $U_i \cap U_N$ is diffeomorphic to $\mathbb{R}^n$ since the original cover is good, and every finite intersection $(U_{i_1} \cap U_N) \cap \cdots \cap (U_{i_r} \cap U_N) = U_{i_1} \cap \cdots \cap U_{i_r} \cap U_N$ is also diffeomorphic to $\mathbb{R}^n$. Hence $\{U_i \cap U_N\}_{i=1}^{N-1}$ is a good cover of $V \cap W$ of cardinality $\leq N-1$, so the inductive hypothesis applies to $V \cap W$ as well. By the inductive hypothesis, $\mathrm{PD}^j$ is an isomorphism on $V$, $W$, and $V \cap W$ for every $j$. In the Mayer–Vietoris ladder constructed above, four out of every five rungs are isomorphisms: \begin{align*} \begin{array}{ccccccccc} \cdots & \to & H^{k-1}_{\mathrm{dR}}(V \cap W) & \to & H^k_{\mathrm{dR}}(U) & \to & H^k_{\mathrm{dR}}(V) \oplus H^k_{\mathrm{dR}}(W) & \to & H^k_{\mathrm{dR}}(V \cap W) \to \cdots \\ & & \downarrow \cong & & \downarrow \mathrm{PD}^k & & \downarrow \cong & & \downarrow \cong \end{array} \end{align*} The [Five Lemma](/theorems/1938) applied to this commutative diagram of exact sequences forces $\mathrm{PD}_U^k: H^k_{\mathrm{dR}}(U) \to (H^{n-k}_c(U))^*$ to be an isomorphism as well. Finite-dimensionality of $H^k_{\mathrm{dR}}(U)$ follows from the inductive hypothesis applied to $V$, $W$, $V \cap W$ together with the exactness of the Mayer–Vietoris sequence: $H^k_{\mathrm{dR}}(U)$ is sandwiched between the finite-dimensional spaces $H^{k-1}_{\mathrm{dR}}(V \cap W)$ and $H^k_{\mathrm{dR}}(V) \oplus H^k_{\mathrm{dR}}(W)$, hence is itself finite-dimensional. The same argument applied to the compactly supported sequence shows $H^{n-k}_c(U)$ is finite-dimensional. This completes the induction. The [Five Lemma](/theorems/1938) is the engine. To recall its statement: in a commutative diagram of abelian groups with exact rows \begin{align*} \begin{array}{ccccccccc} A_1 & \to & A_2 & \to & A_3 & \to & A_4 & \to & A_5 \\ \downarrow f_1 & & \downarrow f_2 & & \downarrow f_3 & & \downarrow f_4 & & \downarrow f_5 \\ B_1 & \to & B_2 & \to & B_3 & \to & B_4 & \to & B_5 \end{array} \end{align*} if $f_1, f_2, f_4, f_5$ are isomorphisms, then $f_3$ is an isomorphism. We apply this with $A_3 = H^k_{\mathrm{dR}}(U)$, $B_3 = (H^{n-k}_c(U))^*$, and $A_1, A_2, A_4, A_5, B_1, B_2, B_4, B_5$ the corresponding terms on $V, W, V \cap W$ from the two Mayer–Vietoris sequences. The induction is well-founded because the cardinality of the good cover strictly decreases when passing from $U$ to $V$ and to $V \cap W$. Why does the choice $V = U_1 \cup \cdots \cup U_{N-1}$, $W = U_N$ work specifically? Because both $V$ and $V \cap W$ inherit good covers of strictly smaller cardinality from the good cover of $U$, while $W$ itself is already a base case. This is the standard "peel off one set at a time" induction on Bott–Tu's good cover. [/step] [step:Specialise to $U = M$ and conclude] The previous step established that $\mathrm{PD}_U^k$ is an isomorphism whenever $U$ admits a finite good cover. By the claim in the good-cover step, the compact manifold $M$ itself admits a finite good cover. Apply the inductive result with $U = M$: \begin{align*} \mathrm{PD}_M^k: H^k_{\mathrm{dR}}(M) \xrightarrow{\;\;\cong\;\;} \left( H^{n-k}_c(M) \right)^*. \end{align*} Since $M$ is compact, $H^{n-k}_c(M) = H^{n-k}_{\mathrm{dR}}(M)$ (first step). Hence \begin{align*} \mathrm{PD}_M^k: H^k_{\mathrm{dR}}(M) \xrightarrow{\;\;\cong\;\;} \left( H^{n-k}_{\mathrm{dR}}(M) \right)^*, \quad [\alpha] \mapsto \left( [\beta] \mapsto \int_M \alpha \wedge \beta\right), \end{align*} is an isomorphism, which is the statement of the theorem. Both $H^k_{\mathrm{dR}}(M)$ and $H^{n-k}_{\mathrm{dR}}(M)$ are finite-dimensional by the induction. The existence of an isomorphism $H^k_{\mathrm{dR}}(M) \cong (H^{n-k}_{\mathrm{dR}}(M))^*$ between finite-dimensional real vector spaces implies \begin{align*} b_k(M) = \dim_{\mathbb{R}} H^k_{\mathrm{dR}}(M) = \dim_{\mathbb{R}} \left(H^{n-k}_{\mathrm{dR}}(M)\right)^* = \dim_{\mathbb{R}} H^{n-k}_{\mathrm{dR}}(M) = b_{n-k}(M), \end{align*} completing the proof. [/step]