[proofplan] Write $i: \{*\} \to M$, $* \mapsto p_0$ and $p: M \to \{*\}$ for the unique map. Smooth contractibility gives $H \circ j_1 = \mathrm{id}_M$ and $H \circ j_0 = i \circ p$, where $j_t: M \to M \times [0,1]$, $x \mapsto (x,t)$. We construct an explicit chain homotopy $K: \Omega^k(M) \to \Omega^{k-1}(M)$ between $\mathrm{id}_M^*$ and $(i \circ p)^*$ on the de Rham complex by integrating $H^*\omega$ along the homotopy parameter using the interior product with $\partial_t$, and verify the identity $dK + Kd = \mathrm{id}_M^* - (i\circ p)^*$ via [Cartan's Magic Formula](/theorems/1535). For $k \ge 1$ the right-hand side reduces to $\mathrm{id}_M^*$ because $\Omega^k(\{*\}) = 0$, so every closed $k$-form is exact. For $k = 0$ we use that the smooth path $t \mapsto H(x,t)$ exhibits $M$ as path-connected, and a smooth function with vanishing differential on a connected manifold is constant. [/proofplan] [step:Fix notation for the homotopy and its time slices] Let $\{*\}$ denote the one-point manifold and define \begin{align*} i: \{*\} &\to M, & * &\mapsto p_0, \\ p: M &\to \{*\}, & x &\mapsto *. \end{align*} For each $t \in [0,1]$ define the time-slice inclusion \begin{align*} j_t: M &\to M \times [0,1], & x &\mapsto (x, t). \end{align*} The hypothesis on $H$ asserts the identities of smooth maps $M \to M$ \begin{align*} H \circ j_1 = \mathrm{id}_M, \qquad H \circ j_0 = i \circ p. \end{align*} Let $\partial_t \in \mathfrak{X}(M \times [0,1])$ denote the vector field tangent to the $[0,1]$-factor, given in product coordinates by the lift of the standard coordinate vector field on $[0,1]$. Let $\iota_{\partial_t}: \Omega^k(M \times [0,1]) \to \Omega^{k-1}(M \times [0,1])$ denote the interior product and let $\mathcal{L}_{\partial_t}: \Omega^k(M \times [0,1]) \to \Omega^k(M \times [0,1])$ denote the Lie derivative along $\partial_t$. [/step] [step:Define the chain homotopy operator $K$ by integrating $\iota_{\partial_t} H^*\omega$ along $[0,1]$] For each integer $k \ge 1$ define \begin{align*} K: \Omega^k(M) &\to \Omega^{k-1}(M) \\ \omega &\mapsto \int_0^1 j_t^*\!\left(\iota_{\partial_t} H^*\omega\right) d\mathcal{L}^1(t), \end{align*} where the integrand is, for each fixed $t \in [0,1]$, the $(k-1)$-form $j_t^*(\iota_{\partial_t} H^*\omega) \in \Omega^{k-1}(M)$, and the integral is taken pointwise in $M$: for $x \in M$ and tangent vectors $v_1, \dots, v_{k-1} \in T_x M$, \begin{align*} (K\omega)_x(v_1, \dots, v_{k-1}) := \int_0^1 \left(j_t^*(\iota_{\partial_t} H^*\omega)\right)_x(v_1, \dots, v_{k-1})\, d\mathcal{L}^1(t). \end{align*} Smoothness of the integrand in $(x,t)$ together with compactness of $[0,1]$ allow differentiation under the integral, so $K\omega \in \Omega^{k-1}(M)$. For $k = 0$ set $K = 0$. [guided] We want a chain homotopy: an operator $K$ such that $dK + Kd = \mathrm{id}_M^* - (i\circ p)^*$. The standard construction comes from the fact that the homotopy $H$ identifies $\mathrm{id}_M$ and $i \circ p$ via a smooth one-parameter family of maps. To extract a form of degree $k-1$ on $M$ from a $k$-form on $M$, we must "lose" one degree somewhere; the natural place is the $[0,1]$-factor of the product, which we exhaust by integration. Concretely: lift $\omega$ to $M \times [0,1]$ via $H^*$, contract along the vertical direction $\partial_t$ to drop one degree, restrict to each time slice via $j_t^*$, and integrate the resulting one-parameter family of $(k-1)$-forms on $M$ over $t \in [0,1]$. The integral converges and produces a smooth form because $H^*\omega$ is smooth on the compact-fibered product $M \times [0,1]$ and integration of a smooth $t$-family of $(k-1)$-forms against $d\mathcal{L}^1(t)$ on a compact interval produces a smooth form on $M$ (differentiation under the integral sign). [/guided] [/step] [step:Establish the chain homotopy identity $dK + Kd = \mathrm{id}_M^* - (i\circ p)^*$ via Cartan's Magic Formula] Fix $k \ge 1$ and $\omega \in \Omega^k(M)$. We claim \begin{align*} d(K\omega) + K(d\omega) = \mathrm{id}_M^* \omega - (i\circ p)^* \omega. \end{align*} Since exterior differentiation commutes with pullback and with integration against the parameter $t$ (the integration is over a parameter independent of $M$, and differentiation under the integral sign is justified by the smoothness of $H^*\omega$ on $M \times [0,1]$), \begin{align*} d(K\omega) = \int_0^1 d\bigl(j_t^*(\iota_{\partial_t} H^*\omega)\bigr) d\mathcal{L}^1(t) = \int_0^1 j_t^*\bigl(d\, \iota_{\partial_t} H^*\omega\bigr) d\mathcal{L}^1(t). \end{align*} Similarly, because $d \circ H^* = H^* \circ d$, \begin{align*} K(d\omega) = \int_0^1 j_t^*\bigl(\iota_{\partial_t} H^* d\omega\bigr) d\mathcal{L}^1(t) = \int_0^1 j_t^*\bigl(\iota_{\partial_t}\, d\, H^*\omega\bigr) d\mathcal{L}^1(t). \end{align*} Adding these and applying [Cartan's Magic Formula](/theorems/1535) — which states $\mathcal{L}_X = d\iota_X + \iota_X d$ for any smooth vector field $X$ on any smooth manifold — to $X = \partial_t$ on $M \times [0,1]$, \begin{align*} d(K\omega) + K(d\omega) = \int_0^1 j_t^*\bigl(\mathcal{L}_{\partial_t} H^*\omega\bigr) d\mathcal{L}^1(t). \end{align*} We now identify the integrand with $\frac{d}{dt}\bigl(j_t^* H^*\omega\bigr)$. For any $\eta \in \Omega^k(M \times [0,1])$ and any $x \in M$, $v_1, \dots, v_k \in T_xM$, the time-slice pullback $j_t^*\eta$ satisfies \begin{align*} \frac{d}{dt}\bigl(j_t^*\eta\bigr)_x(v_1, \dots, v_k) = \bigl(j_t^*\mathcal{L}_{\partial_t}\eta\bigr)_x(v_1, \dots, v_k), \end{align*} because the flow of $\partial_t$ on $M \times [0,1]$ is $\Phi_s(x, t) = (x, t+s)$ (defined on the appropriate open subset of $M \times \mathbb{R}$ containing the orbits we use), so $j_t \circ \Phi_s|_{M \times \{0\}} = j_{t+s}$ in a neighbourhood of any $t \in (0,1)$, and the Lie derivative is by definition $\mathcal{L}_{\partial_t}\eta = \frac{d}{ds}\big|_{s=0} \Phi_s^*\eta$, hence $j_t^*\mathcal{L}_{\partial_t}\eta = \frac{d}{ds}\big|_{s=0} j_t^*\Phi_s^*\eta = \frac{d}{ds}\big|_{s=0} j_{t+s}^*\eta$. At the endpoints $t = 0$ and $t = 1$, one-sided derivatives suffice and the formula extends by continuity since the integrand is smooth in $t$. Applying the [fundamental theorem of calculus](/theorems/632) to the smooth $t$-family of $k$-forms $t \mapsto j_t^* H^*\omega$, \begin{align*} d(K\omega) + K(d\omega) = \int_0^1 \frac{d}{dt}\bigl(j_t^* H^*\omega\bigr) d\mathcal{L}^1(t) = j_1^* H^*\omega - j_0^* H^*\omega. \end{align*} By the contravariance of pullback applied to the identities $H \circ j_1 = \mathrm{id}_M$ and $H \circ j_0 = i \circ p$, \begin{align*} j_1^* H^*\omega = (H \circ j_1)^*\omega = \mathrm{id}_M^*\omega = \omega, \qquad j_0^* H^*\omega = (H \circ j_0)^*\omega = (i\circ p)^*\omega. \end{align*} Combining, \begin{align*} d(K\omega) + K(d\omega) = \omega - (i \circ p)^*\omega. \end{align*} [guided] The heart of the calculation is the identification \begin{align*} j_t^*\bigl(d\, \iota_{\partial_t}\eta + \iota_{\partial_t}\, d\eta\bigr) = j_t^*\mathcal{L}_{\partial_t}\eta = \frac{d}{dt}\bigl(j_t^*\eta\bigr), \end{align*} valid for any $\eta \in \Omega^*(M \times [0,1])$. The first equality is [Cartan's Magic Formula](/theorems/1535), which we apply on the manifold $M \times [0,1]$ with $X = \partial_t$ — a smooth vector field tangent to a complete one-parameter group of translations in the $[0,1]$ factor. The second equality requires unpacking the definition of Lie derivative. The flow of $\partial_t$ on $M \times [0,1]$ moves a point $(x, t)$ to $(x, t+s)$. Hence $\Phi_s^*\eta$ at $(x,t)$ is $\eta$ at $(x, t+s)$ in an obvious sense, and pulling back to time-slice $t = $ const becomes evaluating $\eta$ on a moving time slice. The chain rule then says differentiating the family $t \mapsto j_t^*\eta$ in $t$ is the same as taking $\mathcal{L}_{\partial_t}$ first and then $j_t^*$. Once this identity is in hand, the integrand becomes a total derivative in $t$, and the [fundamental theorem of calculus](/theorems/632) collapses the integral to a difference of endpoint values $j_1^* - j_0^*$. The endpoint values are determined by the homotopy conditions: $H \circ j_1 = \mathrm{id}_M$ contributes $\omega$, while $H \circ j_0 = i \circ p$ contributes $(i\circ p)^*\omega$. This is the standard construction; what is sometimes called the "fibre integration" or "Poincaré chain homotopy". [/guided] [/step] [step:Deduce vanishing of $H^k_{\mathrm{dR}}(M)$ for $k \ge 1$ from $(i\circ p)^* = 0$ in positive degree] Fix $k \ge 1$ and let $\omega \in \Omega^k(M)$ be closed: $d\omega = 0$. By the identity established in the previous step, \begin{align*} \omega = (i\circ p)^*\omega + d(K\omega) + K(d\omega) = (i\circ p)^*\omega + d(K\omega). \end{align*} We show $(i\circ p)^*\omega = 0$. The pullback $(i\circ p)^* = p^* \circ i^*$ factors through $\Omega^k(\{*\})$. The cotangent space $T_*^*\{*\}$ at the unique point of $\{*\}$ is the zero [vector space](/page/Vector%20Space) (since $T_*\{*\} = 0$), and consequently $\Lambda^k T_*^*\{*\} = 0$ for every $k \ge 1$. Thus $\Omega^k(\{*\}) = \Gamma\bigl(\Lambda^k T^*\{*\}\bigr) = 0$, and in particular $i^*\omega = 0$. Therefore $(i\circ p)^*\omega = p^*(i^*\omega) = p^*(0) = 0$ and \begin{align*} \omega = d(K\omega). \end{align*} Thus every closed $k$-form $\omega$ is the [exterior derivative](/theorems/1525) of $K\omega \in \Omega^{k-1}(M)$, so \begin{align*} H^k_{\mathrm{dR}}(M) = \{\omega \in \Omega^k(M) : d\omega = 0\} \,/\, d\bigl(\Omega^{k-1}(M)\bigr) = 0. \end{align*} [/step] [step:Show $M$ is path-connected via the smooth path $t \mapsto H(x, t)$] For any $x \in M$, the map \begin{align*} \gamma_x: [0,1] &\to M, & t &\mapsto H(x, t) \end{align*} is smooth (as the composition of the smooth inclusion $t \mapsto (x, t): [0,1] \to M \times [0,1]$ with the smooth map $H$) and satisfies $\gamma_x(0) = p_0$, $\gamma_x(1) = x$. Hence every point of $M$ is connected to $p_0$ by a continuous path, so $M$ is path-connected, and therefore connected. [/step] [step:Compute $H^0_{\mathrm{dR}}(M) \cong \mathbb{R}$ via constancy of functions with vanishing differential on a connected manifold] By definition $H^0_{\mathrm{dR}}(M) = \{f \in C^\infty(M) : df = 0\}$, since there are no $(-1)$-forms. Let $f \in C^\infty(M)$ satisfy $df = 0$. We show $f$ is constant. Fix $q \in M$ and define \begin{align*} A := \{x \in M : f(x) = f(q)\} \subseteq M. \end{align*} The set $A$ is closed in $M$ because $A = f^{-1}(\{f(q)\})$ and $f$ is continuous and $\{f(q)\} \subset \mathbb{R}$ is closed. We show $A$ is also open. Fix $x \in A$ and choose a connected chart $(U, \varphi)$ at $x$ with $\varphi(U) \subseteq \mathbb{R}^n$ open and connected; here $n = \dim M$. The function $f \circ \varphi^{-1}: \varphi(U) \to \mathbb{R}$ is smooth and its differential vanishes identically because, for every $y \in U$ and every $v \in T_yM$, $df_y(v) = 0$, and the differential of $f \circ \varphi^{-1}$ at $\varphi(y)$ is $df_y \circ (d\varphi^{-1})_{\varphi(y)}$. A smooth function on a connected open subset of $\mathbb{R}^n$ with zero differential everywhere is constant — this is the [mean value theorem](/theorems/186) applied along straight-line segments inside any convex sub-neighbourhood, combined with the connectedness of $\varphi(U)$ (the set of points where $f \circ \varphi^{-1}$ equals its value at $\varphi(x)$ is both open and closed in $\varphi(U)$). Therefore $f$ is constant on $U$, and since $x \in U \subseteq A$, the set $A$ is open. Since $M$ is connected (previous step), $A$ is nonempty (contains $q$), open, and closed, so $A = M$. Hence $f \equiv f(q)$ is constant. Conversely, every constant function $c \in \mathbb{R}$ lies in $H^0_{\mathrm{dR}}(M)$ since $dc = 0$. The map \begin{align*} \mathbb{R} &\to H^0_{\mathrm{dR}}(M), & c &\mapsto c \cdot \mathbb{1}_M \end{align*} (where $\mathbb{1}_M$ is the constant function $1$ on $M$) is therefore an $\mathbb{R}$-linear bijection, i.e., an isomorphism $H^0_{\mathrm{dR}}(M) \cong \mathbb{R}$. [/step] [step:Conclude] Combining the conclusions of the previous steps: $H^k_{\mathrm{dR}}(M) = 0$ for every $k \ge 1$, and $H^0_{\mathrm{dR}}(M) \cong \mathbb{R}$. This is the asserted statement of the theorem. [/step]