[proofplan] We proceed by induction on the type $k$, the minimal number of charts of a good cover. The base case $k = 1$ means $M \cong \mathbb{R}^n$, and the Poincaré lemma computes the cohomology explicitly. For the inductive step we split the cover into a single chart $U_1$ (of type $1$) and the union $V$ of the remaining $k-1$ charts (of type at most $k-1$), apply the Mayer-Vietoris long exact sequence to $M = U_1 \cup V$, and deduce finite dimensionality of $H^i(M)$ from finite dimensionality of $H^i(U_1)$, $H^i(V)$, and $H^i(U_1 \cap V)$ via exactness. [/proofplan] [step:Establish the base case of type $1$ via the Poincaré lemma] If $M$ has type $1$, then by definition $M \cong \mathbb{R}^n$ for some $n$. The [Poincaré Lemma](/theorems/???) computes \begin{align*} H^i_{dR}(\mathbb{R}^n) = \begin{cases} \mathbb{R} & i = 0, \\ 0 & i \ge 1. \end{cases} \end{align*} Since de Rham cohomology is a diffeomorphism invariant (the induced pullback is a cochain isomorphism), $H^i_{dR}(M) \cong H^i_{dR}(\mathbb R^n)$. In every degree the dimension is at most $1$, hence finite. [/step] [step:State the inductive hypothesis and set up the decomposition for type $k$] Fix $k \ge 2$ and assume that $H^i_{dR}(N)$ is finite-dimensional for all $i$ and all smooth manifolds $N$ of type at most $k-1$. Let $M$ be of type $k$, and let $\{U_1, \dots, U_k\}$ be a good open cover of $M$: each $U_j$ and each finite intersection $U_{j_1} \cap \cdots \cap U_{j_m}$ is diffeomorphic to $\mathbb{R}^n$. Set \begin{align*} U := U_1, \qquad V := U_2 \cup \cdots \cup U_k. \end{align*} Both $U$ and $V$ are open, $M = U \cup V$, and \begin{align*} U \cap V = (U_1 \cap U_2) \cup \cdots \cup (U_1 \cap U_k). \end{align*} [guided] We have a finite good cover of size $k$, and we want to apply the induction hypothesis. The natural splitting is to peel off one chart: set $U = U_1$ (type $1$) and $V$ the union of the remaining $k - 1$ charts. The question is whether $V$ and $U \cap V$ have finite type strictly less than $k$ — so we can invoke induction. For $V = U_2 \cup \cdots \cup U_k$: the collection $\{U_2, \dots, U_k\}$ is a cover of $V$ by open sets. Each $U_j$ is diffeomorphic to $\mathbb R^n$. Each finite intersection $U_{j_1} \cap \cdots \cap U_{j_m}$ (with indices in $\{2, \dots, k\}$) is diffeomorphic to $\mathbb R^n$ because the original cover $\{U_1, \dots, U_k\}$ is good. So $V$ has a good cover of size $k - 1$; in particular $V$ has type at most $k - 1$. For $U \cap V = \bigcup_{j=2}^k (U_1 \cap U_j)$: each $U_1 \cap U_j$ is a finite intersection from the good cover, hence diffeomorphic to $\mathbb R^n$. A finite intersection of the form $(U_1 \cap U_{j_1}) \cap \cdots \cap (U_1 \cap U_{j_m}) = U_1 \cap U_{j_1} \cap \cdots \cap U_{j_m}$ is again diffeomorphic to $\mathbb R^n$. So $\{U_1 \cap U_j\}_{j=2}^k$ is a good cover of $U \cap V$ of size $k - 1$. Hence $U \cap V$ also has type at most $k - 1$. [/guided] [/step] [step:Verify that $V$ and $U \cap V$ both have type at most $k - 1$] *Claim for $V$.* The collection $\{U_2, \dots, U_k\}$ is an open cover of $V$ consisting of $k - 1$ sets. For any subcollection of indices $\{j_1, \dots, j_m\} \subseteq \{2, \dots, k\}$, the intersection $U_{j_1} \cap \cdots \cap U_{j_m}$ is a finite intersection from the good cover of $M$, hence diffeomorphic to $\mathbb R^n$. Therefore $V$ has type at most $k - 1$. *Claim for $U \cap V$.* By distributivity of intersection over union, \begin{align*} U_1 \cap V = U_1 \cap (U_2 \cup \cdots \cup U_k) = \bigcup_{j=2}^k (U_1 \cap U_j). \end{align*} The collection $\{U_1 \cap U_j\}_{j=2}^k$ is an open cover of $U \cap V$ of size $k - 1$. For any subcollection of indices $\{j_1, \dots, j_m\} \subseteq \{2, \dots, k\}$, \begin{align*} (U_1 \cap U_{j_1}) \cap \cdots \cap (U_1 \cap U_{j_m}) = U_1 \cap U_{j_1} \cap \cdots \cap U_{j_m}, \end{align*} which is again a finite intersection from the good cover of $M$, hence diffeomorphic to $\mathbb R^n$. Therefore $U \cap V$ has type at most $k - 1$. By the inductive hypothesis, $H^i_{dR}(V)$ and $H^i_{dR}(U \cap V)$ are finite-dimensional for all $i$. Also $H^i_{dR}(U) = H^i_{dR}(U_1)$ is finite-dimensional by the base case. [/step] [step:Apply the Mayer-Vietoris long exact sequence] By the [Mayer-Vietoris Short Exact Sequence](/theorems/1533) applied to the open cover $M = U \cup V$, we obtain the short exact sequence of cochain complexes \begin{align*} 0 \to \Omega^*(M) \to \Omega^*(U) \oplus \Omega^*(V) \to \Omega^*(U \cap V) \to 0. \end{align*} Passing to cohomology via the zig-zag lemma yields the Mayer-Vietoris long exact sequence in cohomology: \begin{align*} \cdots \to H^{i-1}_{dR}(U \cap V) \xrightarrow{\delta} H^i_{dR}(M) \xrightarrow{\alpha^*} H^i_{dR}(U) \oplus H^i_{dR}(V) \xrightarrow{\beta^*} H^i_{dR}(U \cap V) \to \cdots \end{align*} [guided] A short exact sequence of cochain complexes always induces a long exact sequence in cohomology, by the standard homological-algebra construction known as the zig-zag lemma (or snake lemma). The connecting homomorphism $\delta: H^{i-1}_{dR}(U \cap V) \to H^i_{dR}(M)$ is defined by: represent a class by a closed form $\omega \in \Omega^{i-1}(U \cap V)$, find $(\omega_U, -\omega_V) \in \Omega^{i-1}(U) \oplus \Omega^{i-1}(V)$ mapping to $\omega$ via $\beta$ (using a partition of unity, as in the proof of Mayer-Vietoris), take exterior derivatives; since $d$ is natural, $d\omega_U$ and $d\omega_V$ agree on $U \cap V$ and glue to a closed form on $M$. For our argument only exactness matters, not the concrete form of $\delta$. The piece of the long exact sequence we need is \begin{align*} H^{i-1}_{dR}(U \cap V) \xrightarrow{\delta} H^i_{dR}(M) \xrightarrow{\alpha^*} H^i_{dR}(U) \oplus H^i_{dR}(V), \end{align*} from which we will extract a dimension bound via rank-nullity. [/guided] [/step] [step:Bound the dimension of $H^i_{dR}(M)$ via exactness and rank-nullity] Consider the portion \begin{align*} H^{i-1}_{dR}(U \cap V) \xrightarrow{\delta} H^i_{dR}(M) \xrightarrow{\alpha^*} H^i_{dR}(U) \oplus H^i_{dR}(V). \end{align*} Exactness at $H^i_{dR}(M)$ asserts $\operatorname{Range}(\delta) = \ker(\alpha^*)$. By rank-nullity applied to the linear map $\alpha^*: H^i_{dR}(M) \to H^i_{dR}(U) \oplus H^i_{dR}(V)$, \begin{align*} \dim H^i_{dR}(M) = \dim \ker(\alpha^*) + \dim \operatorname{Range}(\alpha^*). \end{align*} We estimate each term: \begin{align*} \dim \ker(\alpha^*) &= \dim \operatorname{Range}(\delta) \le \dim H^{i-1}_{dR}(U \cap V), \\ \dim \operatorname{Range}(\alpha^*) &\le \dim (H^i_{dR}(U) \oplus H^i_{dR}(V)) = \dim H^i_{dR}(U) + \dim H^i_{dR}(V). \end{align*} Therefore \begin{align*} \dim H^i_{dR}(M) \le \dim H^{i-1}_{dR}(U \cap V) + \dim H^i_{dR}(U) + \dim H^i_{dR}(V). \end{align*} By the previous step, each of $\dim H^{i-1}_{dR}(U \cap V)$, $\dim H^i_{dR}(U)$, $\dim H^i_{dR}(V)$ is finite. Hence $\dim H^i_{dR}(M) < \infty$. [/step] [step:Conclude by induction] By induction on the type $k$: the base case $k = 1$ is established by the Poincaré lemma, and the inductive step from $k - 1$ to $k$ is established by the Mayer-Vietoris bound. Hence for every manifold $M$ of finite type and every degree $i \ge 0$, $H^i_{dR}(M)$ is finite-dimensional. [/step]