[proofplan] The space of differential $k$-forms on a $0$-dimensional manifold is determined by the exterior algebra of its tangent spaces. Since $T_pM = 0$, the alternating multilinear forms on $T_pM$ vanish for every degree $k \ge 1$, while degree $0$ forms are just smooth real-valued functions on the singleton, which are in bijection with $\mathbb{R}$. The [exterior derivative](/theorems/1525) is therefore zero in every degree, and cohomology reduces to the chain groups themselves. Reading off $H^0$ and $H^k$ from this trivial complex gives the claim. [/proofplan] [step:Describe the de Rham complex of $M$ in each degree] Let $M = \{p\}$. For each non-negative integer $k$, the space of smooth differential $k$-forms on $M$ is defined as the space of smooth sections of the bundle $\Lambda^k T^*M \to M$: \begin{align*} \Omega^k(M) := \Gamma(\Lambda^k T^*M). \end{align*} Since $M$ is a single point, a section of any vector bundle $E \to M$ is determined by a single element of the fibre $E_p$. Hence \begin{align*} \Omega^k(M) \cong \Lambda^k T_p^* M. \end{align*} The de Rham complex of $M$ is \begin{align*} 0 \longrightarrow \Omega^0(M) \xrightarrow{\;d^0\;} \Omega^1(M) \xrightarrow{\;d^1\;} \Omega^2(M) \xrightarrow{\;d^2\;} \cdots \end{align*} where $d^k: \Omega^k(M) \to \Omega^{k+1}(M)$ is the [exterior derivative](/theorems/1525) in degree $k$. The de Rham cohomology in degree $k$ is by definition \begin{align*} H^k_{\mathrm{dR}}(M) := \frac{\ker d^k}{\operatorname{im} d^{k-1}}, \end{align*} with the convention $\operatorname{im} d^{-1} := 0$. [guided] We unpack the definitions because every claim below rests on them. A differential $k$-form on a smooth manifold $M$ is a smooth section of the $k$-th exterior power bundle $\Lambda^k T^*M$, whose fibre over $x \in M$ is the [vector space](/page/Vector%20Space) $\Lambda^k T_x^*M$ of alternating $k$-linear functionals on $T_xM$. We denote the space of smooth sections by $\Omega^k(M) := \Gamma(\Lambda^k T^*M)$. When $M = \{p\}$ consists of one point, the manifold is $0$-dimensional, and any vector bundle $E \to M$ has a single fibre $E_p$. A smooth section $s: M \to E$ amounts to a choice of an element $s(p) \in E_p$ (the smoothness condition is vacuous because the source has no positive-dimensional smooth structure to test against). Consequently \begin{align*} \Gamma(E) \xrightarrow{\;\cong\;} E_p, \qquad s \mapsto s(p), \end{align*} is a linear isomorphism. Applying this to $E = \Lambda^k T^*M$ gives \begin{align*} \Omega^k(M) \cong \Lambda^k T_p^* M. \end{align*} The de Rham complex is the cochain complex \begin{align*} 0 \longrightarrow \Omega^0(M) \xrightarrow{\;d^0\;} \Omega^1(M) \xrightarrow{\;d^1\;} \Omega^2(M) \xrightarrow{\;d^2\;} \cdots, \end{align*} whose differentials are the exterior derivatives $d^k$ in each degree. They satisfy $d^{k+1} \circ d^k = 0$. The de Rham cohomology groups are the cohomology of this complex: \begin{align*} H^k_{\mathrm{dR}}(M) := \frac{\ker d^k}{\operatorname{im} d^{k-1}}, \qquad k \ge 0, \end{align*} with $\operatorname{im} d^{-1} := 0$, so that $H^0_{\mathrm{dR}}(M) = \ker d^0$. [/guided] [/step] [step:Compute $\Omega^k(\{p\})$ for $k \ge 1$ from the tangent space at $p$] The tangent space $T_pM$ at $p \in \{p\}$ is the tangent space to a $0$-dimensional manifold, hence \begin{align*} T_p M = 0, \end{align*} the zero [vector space](/page/Vector%20Space). Its dual is also zero: $T_p^*M = 0$. For $k \ge 1$, an element of $\Lambda^k T_p^*M$ is an alternating $k$-multilinear map \begin{align*} \omega: \underbrace{T_pM \times \cdots \times T_pM}_{k \text{ copies}} \to \mathbb{R}. \end{align*} Since $T_pM = 0$, the only input is the $k$-tuple $(0, \dots, 0)$, and multilinearity forces $\omega(0, \dots, 0) = 0$. Therefore \begin{align*} \Lambda^k T_p^*M = 0 \qquad \text{for every } k \ge 1, \end{align*} and combining with Step 1, \begin{align*} \Omega^k(\{p\}) = 0 \qquad \text{for every } k \ge 1. \end{align*} [guided] A $0$-dimensional smooth manifold has, at each point, a tangent space whose dimension equals the dimension of the manifold. Here $\dim M = 0$, so $T_pM$ is a $0$-dimensional real [vector space](/page/Vector%20Space), which means $T_pM = \{0\}$, the zero [vector space](/page/Vector%20Space). We then ask: what are the alternating $k$-multilinear maps on the zero [vector space](/page/Vector%20Space) for $k \ge 1$? By definition, $\omega \in \Lambda^k T_p^*M$ is a function \begin{align*} \omega: (T_pM)^k \to \mathbb{R} \end{align*} that is linear in each argument and changes sign under transposition of arguments. The product $(T_pM)^k = \{(0, \dots, 0)\}$ is a one-point set, and multilinearity in any one argument applied at the zero vector gives \begin{align*} \omega(0, \dots, 0) = \omega(0 \cdot 0, 0, \dots, 0) = 0 \cdot \omega(0, 0, \dots, 0) = 0. \end{align*} Thus the only alternating $k$-form on $T_pM$ is the zero form, and $\Lambda^k T_p^*M = 0$ for $k \ge 1$. By the isomorphism $\Omega^k(M) \cong \Lambda^k T_p^*M$ established in Step 1, this gives $\Omega^k(\{p\}) = 0$ for all $k \ge 1$. (Note: the convention $\Lambda^0 V = \mathbb{R}$ for any [vector space](/page/Vector%20Space) $V$, including $V = 0$, is what saves degree zero; we treat that case separately in the next step.) [/guided] [/step] [step:Identify $\Omega^0(\{p\})$ with $\mathbb{R}$] By convention, $\Lambda^0 V = \mathbb{R}$ for every real [vector space](/page/Vector%20Space) $V$; equivalently, $\Omega^0(M) = C^\infty(M)$ is the algebra of smooth real-valued functions on $M$. For $M = \{p\}$, a smooth function \begin{align*} f: \{p\} &\to \mathbb{R}\\ p &\mapsto f(p) \end{align*} is determined by the single real number $f(p) \in \mathbb{R}$, and every such choice defines a smooth function (smoothness is vacuous). The evaluation map \begin{align*} \operatorname{ev}_p: \Omega^0(\{p\}) &\to \mathbb{R}\\ f &\mapsto f(p) \end{align*} is therefore an $\mathbb{R}$-linear isomorphism, giving $\Omega^0(\{p\}) \cong \mathbb{R}$. [/step] [step:Conclude that every exterior derivative on $\{p\}$ is the zero map] For each integer $k \ge 0$, the [exterior derivative](/theorems/1525) on $M = \{p\}$ is a [linear map](/page/Linear%20Map) \begin{align*} d^k: \Omega^k(\{p\}) \to \Omega^{k+1}(\{p\}). \end{align*} By Step 2, the codomain $\Omega^{k+1}(\{p\}) = 0$ for every $k \ge 0$ (since $k+1 \ge 1$). A [linear map](/page/Linear%20Map) into the zero [vector space](/page/Vector%20Space) is necessarily the zero map, so \begin{align*} d^k = 0 \qquad \text{for every } k \ge 0. \end{align*} [guided] Why is this step easy? Because all of the analytic content of the [exterior derivative](/theorems/1525) — partial derivatives, chart changes, antisymmetrisation — is irrelevant when the target is zero. There is exactly one [linear map](/page/Linear%20Map) from any [vector space](/page/Vector%20Space) to the zero [vector space](/page/Vector%20Space), namely the zero map, so $d^k$ is forced to be zero for $k \ge 0$ without any computation. We do not even need to know the formula for $d$ on a general manifold; we only need the target to be the zero space, which Step 2 has established for every degree $\ge 1$. [/guided] [/step] [step:Read off the cohomology in each degree] From Step 4 the de Rham complex of $\{p\}$ collapses to \begin{align*} 0 \longrightarrow \mathbb{R} \xrightarrow{\;0\;} 0 \xrightarrow{\;0\;} 0 \xrightarrow{\;0\;} \cdots, \end{align*} where the first $\mathbb{R}$ stands for $\Omega^0(\{p\})$ identified via Step 3. **Degree $0$.** Using $\operatorname{im} d^{-1} = 0$ by convention and $\ker d^0 = \Omega^0(\{p\}) = \mathbb{R}$ (since $d^0 = 0$ on a domain $\cong \mathbb{R}$ has full kernel), \begin{align*} H^0_{\mathrm{dR}}(\{p\}) = \frac{\ker d^0}{\operatorname{im} d^{-1}} = \frac{\mathbb{R}}{0} \cong \mathbb{R}. \end{align*} **Degree $k \ge 1$.** Both $\ker d^k$ and $\operatorname{im} d^{k-1}$ are subspaces of $\Omega^k(\{p\}) = 0$, hence both are $0$, and \begin{align*} H^k_{\mathrm{dR}}(\{p\}) = \frac{0}{0} = 0. \end{align*} This establishes the two assertions in the statement and completes the proof. [/step]