[proofplan] We argue by induction on $n$. Cover $S^n$ by the two open sets $U = S^n \setminus \{p_S\}$ and $V = S^n \setminus \{p_N\}$ obtained by deleting the south and north poles. Stereographic projection identifies each of $U, V$ with $\mathbb R^n$, so both are smoothly contractible; the intersection $U \cap V$ is diffeomorphic to $S^{n-1} \times \mathbb R$ and therefore smoothly deformation retracts onto $S^{n-1}$. Feeding this data into the [Mayer–Vietoris Theorem](/theorems/2238) for de Rham cohomology yields, for every $k \ge 1$, isomorphisms or short exact sequences relating $H^k_{\mathrm{dR}}(S^n)$ to $H^{k-1}_{\mathrm{dR}}(S^{n-1})$. The base case $n = 1$ is handled by direct computation from the same sequence using that $S^0$ consists of two points; the inductive step then propagates the result to all higher dimensions. [/proofplan] [step:Construct the stereographic cover and compute the cohomology of its pieces] Write $p_N = (0,\dots,0,1) \in S^n$ and $p_S = (0,\dots,0,-1) \in S^n$ for the north and south poles. Define open subsets \begin{align*} U &:= S^n \setminus \{p_S\}, \\ V &:= S^n \setminus \{p_N\}. \end{align*} Then $U \cup V = S^n$, and the intersection is \begin{align*} U \cap V = S^n \setminus \{p_N, p_S\}. \end{align*} Stereographic projection from $p_S$ provides a diffeomorphism \begin{align*} \sigma_N : U &\to \mathbb R^n, \\ (x_1, \dots, x_{n+1}) &\mapsto \frac{1}{1 + x_{n+1}}(x_1, \dots, x_n), \end{align*} with inverse map \begin{align*} \tau_N : \mathbb R^n &\to U, \\ y &\mapsto \left(\frac{2y}{1 + |y|^2}, \frac{1 - |y|^2}{1 + |y|^2}\right). \end{align*} For $y \in \mathbb R^n$, the point $\tau_N(y)$ lies on $S^n$ because \begin{align*} \left|\frac{2y}{1 + |y|^2}\right|^2 + \left(\frac{1 - |y|^2}{1 + |y|^2}\right)^2 = 1, \end{align*} and its last coordinate is never $-1$; hence $\tau_N(y) \in U$. Direct substitution gives $\sigma_N \circ \tau_N = \operatorname{id}_{\mathbb R^n}$ and $\tau_N \circ \sigma_N = \operatorname{id}_U$. Stereographic projection from $p_N$ provides the diffeomorphism \begin{align*} \sigma_S : V &\to \mathbb R^n, \\ (x_1, \dots, x_{n+1}) &\mapsto \frac{1}{1 - x_{n+1}}(x_1, \dots, x_n), \end{align*} with inverse map \begin{align*} \tau_S : \mathbb R^n &\to V, \\ y &\mapsto \left(\frac{2y}{1 + |y|^2}, \frac{|y|^2 - 1}{1 + |y|^2}\right). \end{align*} For $y \in \mathbb R^n$, the point $\tau_S(y)$ lies on $S^n$ and its last coordinate is never $1$, so $\tau_S(y) \in V$; direct substitution gives $\sigma_S \circ \tau_S = \operatorname{id}_{\mathbb R^n}$ and $\tau_S \circ \sigma_S = \operatorname{id}_V$. Both $U$ and $V$ are therefore diffeomorphic to $\mathbb R^n$, which is smoothly contractible to the origin via the straight-line homotopy \begin{align*} H : \mathbb R^n \times [0,1] &\to \mathbb R^n, \\ (y,t) &\mapsto (1-t)y. \end{align*} Composing $H$ with $\sigma_N,\tau_N$ and with $\sigma_S,\tau_S$ gives smooth contractions of $U$ and $V$, respectively, to points. Next, define \begin{align*} \Psi : S^{n-1} \times (-1, 1) &\to U \cap V, \\ (w, s) &\mapsto \bigl(\sqrt{1 - s^2}\, w, \; s\bigr). \end{align*} This is a diffeomorphism: its smooth inverse sends $(x_1, \dots, x_{n+1}) \in U \cap V$ to $\bigl((x_1, \dots, x_n)/\sqrt{1 - x_{n+1}^2},\, x_{n+1}\bigr)$. The map $R : (U \cap V) \times [0,1] \to U \cap V$ defined by \begin{align*} R\bigl(\Psi(w, s), t\bigr) = \Psi\bigl(w, (1-t)s\bigr) \end{align*} is a smooth deformation retraction of $U \cap V$ onto the equator $\Psi(S^{n-1} \times \{0\}) \subset U \cap V$, which is itself diffeomorphic to $S^{n-1}$. The contractions above show that $U$ and $V$ are smoothly homotopy equivalent to a point. The map $R$ is smooth because, in the $\Psi$-coordinates, it is the smooth map \begin{align*} \widetilde R : S^{n-1} \times (-1,1) \times [0,1] &\to S^{n-1} \times (-1,1), \\ (w,s,t) &\mapsto (w,(1-t)s), \end{align*} and it restricts to the identity on $S^{n-1} \times \{0\}$. Hence $U \cap V$ is smoothly homotopy equivalent to $S^{n-1}$. By [homotopy invariance of de Rham cohomology](/theorems/3585) together with the Poincaré Lemma stating that $H^k_{\mathrm{dR}}(\mathbb R^n) = 0$ for $k \ge 1$ and $H^0_{\mathrm{dR}}(\mathbb R^n) = \mathbb R$, we obtain \begin{align*} H^k_{\mathrm{dR}}(U) \cong H^k_{\mathrm{dR}}(V) &\cong \begin{cases} \mathbb R, & k = 0, \\ 0, & k \ge 1, \end{cases} \\ H^k_{\mathrm{dR}}(U \cap V) &\cong H^k_{\mathrm{dR}}(S^{n-1}) \quad \text{for all } k \ge 0. \end{align*} [guided] Why this cover? The Mayer–Vietoris machine needs an open cover $\{U, V\}$ of $S^n$ for which the cohomology of each piece, and of the intersection, is already understood. Removing one point of $S^n$ leaves a space diffeomorphic to $\mathbb R^n$ — this is the geometric content of stereographic projection — and removing two antipodal points leaves a cylinder over $S^{n-1}$, which is the smaller-dimensional sphere we will use induction on. So the deleted-pole cover is precisely engineered to feed Mayer–Vietoris. **Construction.** Define $p_N := (0,\dots,0,1)$ and $p_S := (0,\dots,0,-1)$ in $S^n \subset \mathbb R^{n+1}$, and set \begin{align*} U &:= S^n \setminus \{p_S\}, \\ V &:= S^n \setminus \{p_N\}. \end{align*} Every point of $S^n$ lies in at least one of these (only $p_S$ is missing from $U$, only $p_N$ from $V$, and $p_N \ne p_S$), so $U \cup V = S^n$. The intersection is $U \cap V = S^n \setminus \{p_N, p_S\}$. **Stereographic projection identifies $U$ and $V$ with $\mathbb R^n$.** Define \begin{align*} \sigma_N : U &\to \mathbb R^n, \\ (x_1, \dots, x_{n+1}) &\mapsto \frac{1}{1 + x_{n+1}}(x_1, \dots, x_n). \end{align*} The denominator $1 + x_{n+1}$ vanishes only at $p_S$, which has been removed, so $\sigma_N$ is well-defined and smooth. Define \begin{align*} \tau_N : \mathbb R^n &\to U, \\ y &\mapsto \left(\frac{2y}{1 + |y|^2}, \frac{1 - |y|^2}{1 + |y|^2}\right). \end{align*} The displayed formula lands in $S^n$ because \begin{align*} \left|\frac{2y}{1 + |y|^2}\right|^2 + \left(\frac{1 - |y|^2}{1 + |y|^2}\right)^2 = 1, \end{align*} and it avoids $p_S$ because the last coordinate cannot equal $-1$. Thus $\tau_N$ is a well-defined smooth map into $U$. Using $|(x_1,\dots,x_n)|^2 = 1 - x_{n+1}^2$ for $x \in S^n$, direct substitution gives $\sigma_N \circ \tau_N = \operatorname{id}_{\mathbb R^n}$ and $\tau_N \circ \sigma_N = \operatorname{id}_U$, so $\sigma_N$ is a diffeomorphism. For $V$, define \begin{align*} \sigma_S : V &\to \mathbb R^n, \\ (x_1, \dots, x_{n+1}) &\mapsto \frac{1}{1 - x_{n+1}}(x_1, \dots, x_n), \\ \tau_S : \mathbb R^n &\to V, \\ y &\mapsto \left(\frac{2y}{1 + |y|^2}, \frac{|y|^2 - 1}{1 + |y|^2}\right). \end{align*} The denominator $1 - x_{n+1}$ vanishes only at $p_N$, which has been removed. The formula for $\tau_S$ lands in $S^n$ and avoids $p_N$; substituting the formulas gives $\sigma_S \circ \tau_S = \operatorname{id}_{\mathbb R^n}$ and $\tau_S \circ \sigma_S = \operatorname{id}_V$. Hence $\sigma_S$ is a diffeomorphism. Finally, define \begin{align*} H : \mathbb R^n \times [0,1] &\to \mathbb R^n, \\ (y,t) &\mapsto (1-t)y. \end{align*} This is a smooth homotopy from $\operatorname{id}_{\mathbb R^n}$ to the constant map $y \mapsto 0$. The maps $(x,t) \mapsto \tau_N(H(\sigma_N(x),t))$ on $U \times [0,1]$ and $(x,t) \mapsto \tau_S(H(\sigma_S(x),t))$ on $V \times [0,1]$ are smooth contractions, so $U$ and $V$ are smoothly contractible. **The intersection is a cylinder over $S^{n-1}$.** Consider \begin{align*} \Psi : S^{n-1} \times (-1, 1) &\to U \cap V, \\ (w, s) &\mapsto \bigl(\sqrt{1-s^2}\,w,\; s\bigr). \end{align*} The image lies in $S^n$ because $|(\sqrt{1-s^2}w, s)|^2 = (1-s^2)|w|^2 + s^2 = 1$, using $|w| = 1$. Since $s \in (-1, 1)$, the last coordinate is strictly between $-1$ and $1$, so the image avoids both poles, landing in $U \cap V$. The map $\Psi$ is smooth because it is built from smooth coordinate functions on $S^{n-1} \times (-1,1)$. Define \begin{align*} \Theta : U \cap V &\to S^{n-1} \times (-1,1), \\ (x_1, \dots, x_{n+1}) &\mapsto \left(\frac{(x_1, \dots, x_n)}{\sqrt{1 - x_{n+1}^2}}, x_{n+1}\right). \end{align*} For $x \in U \cap V$, we have $x_{n+1} \in (-1,1)$, so $1 - x_{n+1}^2 > 0$ and the reciprocal square-root factor is smooth. Also \begin{align*} \left|\frac{(x_1, \dots, x_n)}{\sqrt{1 - x_{n+1}^2}}\right|^2 = \frac{1 - x_{n+1}^2}{1 - x_{n+1}^2} = 1, \end{align*} so $\Theta$ lands in $S^{n-1} \times (-1,1)$. Direct substitution gives \begin{align*} \Theta(\Psi(w,s)) &= (w,s), \\ \Psi(\Theta(x_1, \dots, x_{n+1})) &= (x_1, \dots, x_{n+1}). \end{align*} Thus $\Theta$ is a smooth inverse to $\Psi$, and $\Psi$ is a diffeomorphism. **Deformation retraction onto the equator.** The map \begin{align*} R : (U \cap V) \times [0,1] &\to U \cap V, \\ \bigl(\Psi(w, s), t\bigr) &\mapsto \Psi\bigl(w, (1-t)s\bigr) \end{align*} is smooth, satisfies $R(\cdot, 0) = \operatorname{id}_{U \cap V}$, $R(\Psi(w, s), 1) = \Psi(w, 0)$ for all $(w, s)$, and $R(\Psi(w, 0), t) = \Psi(w, 0)$ for all $t$. So $R$ is a (smooth) deformation retraction of $U \cap V$ onto the submanifold $\Psi(S^{n-1} \times \{0\})$, which is diffeomorphic to $S^{n-1}$. **Cohomology of the pieces.** Two general facts apply. First, the Poincaré Lemma states that $H^0_{\mathrm{dR}}(\mathbb R^n) = \mathbb R$ (locally constant smooth functions on connected $\mathbb R^n$ are constants) and $H^k_{\mathrm{dR}}(\mathbb R^n) = 0$ for $k \ge 1$ — every closed form on a contractible space is exact. Second, [homotopy invariance of de Rham cohomology](/theorems/3585) states that smoothly homotopy equivalent manifolds have isomorphic de Rham cohomology in each degree. Both of these results are foundational and we cite them as not yet in the wiki. Combining: $U \cong \mathbb R^n$ and $V \cong \mathbb R^n$ give \begin{align*} H^k_{\mathrm{dR}}(U) = H^k_{\mathrm{dR}}(V) = \begin{cases} \mathbb R, & k = 0, \\ 0, & k \ge 1. \end{cases} \end{align*} The deformation retraction of $U \cap V$ onto $S^{n-1}$ gives \begin{align*} H^k_{\mathrm{dR}}(U \cap V) \cong H^k_{\mathrm{dR}}(S^{n-1}) \qquad \text{for all } k \ge 0. \end{align*} The right-hand side is exactly what our inductive hypothesis will control. [/guided] [/step] [step:Compute $H^0_{\mathrm{dR}}(S^n)$ from connectedness] For $n \ge 1$ the sphere $S^n$ is path-connected. The kernel of $d : \Omega^0(S^n) \to \Omega^1(S^n)$ consists of smooth functions $f : S^n \to \mathbb R$ with $df = 0$. On a connected manifold, $df = 0$ forces $f$ to be constant, so $\ker(d \,|\, \Omega^0) \cong \mathbb R$. There is no degree $-1$, so the image of $d$ into $\Omega^0$ is zero. Therefore \begin{align*} H^0_{\mathrm{dR}}(S^n) \cong \mathbb R. \end{align*} [/step] [step:Establish the base case $n = 1$ via Mayer–Vietoris] We compute $H^k_{\mathrm{dR}}(S^1)$ for all $k \ge 0$. Degree $0$ is handled above: $H^0_{\mathrm{dR}}(S^1) \cong \mathbb R$. For $n = 1$, the equator $S^{n-1} = S^0$ is the discrete two-point space $\{-1, +1\} \subset \mathbb R$. Smooth functions on $S^0$ are determined by their values at the two points, so $\Omega^0(S^0) \cong \mathbb R^2$, and $\Omega^k(S^0) = 0$ for $k \ge 1$. Hence \begin{align*} H^0_{\mathrm{dR}}(S^0) &\cong \mathbb R^2, \\ H^k_{\mathrm{dR}}(S^0) &= 0 \quad \text{for } k \ge 1. \end{align*} By Step 1, $H^k_{\mathrm{dR}}(U \cap V) \cong H^k_{\mathrm{dR}}(S^0)$ in this case, so $H^0_{\mathrm{dR}}(U \cap V) \cong \mathbb R^2$ and $H^k_{\mathrm{dR}}(U \cap V) = 0$ for $k \ge 1$. The [Mayer–Vietoris Theorem](/theorems/2238) applied to $S^1 = U \cup V$ yields the long exact sequence \begin{align*} 0 \to H^0_{\mathrm{dR}}(S^1) \xrightarrow{(\iota_U^*, \iota_V^*)} H^0_{\mathrm{dR}}(U) \oplus H^0_{\mathrm{dR}}(V) \xrightarrow{j} H^0_{\mathrm{dR}}(U \cap V) \xrightarrow{\delta} H^1_{\mathrm{dR}}(S^1) \to H^1_{\mathrm{dR}}(U) \oplus H^1_{\mathrm{dR}}(V), \end{align*} where $j(\alpha, \beta) = \iota_{U \cap V, U}^* \alpha - \iota_{U \cap V, V}^* \beta$ is the difference of pullbacks under the inclusions $U \cap V \hookrightarrow U$ and $U \cap V \hookrightarrow V$, and $\delta$ is the connecting homomorphism. Substituting the values from Step 1 and the computation of $H^*(S^0)$ above: \begin{align*} 0 \to \mathbb R \xrightarrow{\Delta} \mathbb R \oplus \mathbb R \xrightarrow{j} \mathbb R^2 \xrightarrow{\delta} H^1_{\mathrm{dR}}(S^1) \to 0, \end{align*} where $\Delta$ is the diagonal map $c \mapsto (c, c)$ sending the constant function $c$ on $S^1$ to its restrictions on $U$ and $V$. We compute $j$ explicitly. Both components of $U \cap V$ — the "east" arc $E$ and the "west" arc $W$ — are path-connected, so a class in $H^0_{\mathrm{dR}}(U \cap V) \cong \mathbb R^2$ is specified by its value $(v_E, v_W)$ on the two components. A pair $(a, b) \in H^0_{\mathrm{dR}}(U) \oplus H^0_{\mathrm{dR}}(V)$ represents the pair of constant functions $a$ on $U$ and $b$ on $V$. Restricting and subtracting on each component gives \begin{align*} j(a, b) = (a - b,\; a - b) \in \mathbb R^2. \end{align*} Thus $\operatorname{im}(j)$ is the diagonal $\{(c, c) : c \in \mathbb R\}$, a one-dimensional subspace of $\mathbb R^2$. By exactness at $H^0_{\mathrm{dR}}(U \cap V)$, we have $\ker(\delta) = \operatorname{im}(j)$, a one-dimensional subspace. By exactness at $H^1_{\mathrm{dR}}(S^1)$ and the vanishing of $H^1_{\mathrm{dR}}(U) \oplus H^1_{\mathrm{dR}}(V) = 0$, the map $\delta$ is surjective. Therefore \begin{align*} H^1_{\mathrm{dR}}(S^1) \cong \mathbb R^2 / \operatorname{im}(j) \cong \mathbb R^2 / \{(c, c)\} \cong \mathbb R. \end{align*} For $k \ge 2$, the Mayer–Vietoris sequence reads \begin{align*} H^{k-1}_{\mathrm{dR}}(U \cap V) \to H^k_{\mathrm{dR}}(S^1) \to H^k_{\mathrm{dR}}(U) \oplus H^k_{\mathrm{dR}}(V). \end{align*} For $k \ge 2$ both flanking groups are zero (the left by $k - 1 \ge 1$ and $H^{\ge 1}(S^0) = 0$; the right by $H^{\ge 1}(\mathbb R) = 0$), so $H^k_{\mathrm{dR}}(S^1) = 0$. This completes the base case. [guided] The base case $n = 1$ is the source of all the non-trivial classes; we must compute it directly because the inductive step relies on knowing $H^*(S^{n-1})$ in advance. The key feature distinguishing $n = 1$ from $n \ge 2$ is that the equator $S^0$ has **two** connected components rather than one, which makes the map $j$ in Mayer–Vietoris fail to be surjective, opening the door for a non-trivial $H^1$ class. **Setting up the data.** From Step 1 we have $H^k_{\mathrm{dR}}(U \cap V) \cong H^k_{\mathrm{dR}}(S^0)$, so we need to know $H^*(S^0)$. The space $S^0 = \{-1, +1\} \subset \mathbb R$ is a zero-dimensional manifold consisting of two isolated points. A smooth $0$-form on $S^0$ is just a function $S^0 \to \mathbb R$, equivalently a pair of [real numbers](/page/Real%20Numbers) $(c_-, c_+)$; thus $\Omega^0(S^0) \cong \mathbb R^2$. For $k \ge 1$ there are no $k$-forms on a $0$-dimensional manifold ($\Lambda^k T^*_p S^0 = 0$ since $T_p S^0 = 0$), so $\Omega^k(S^0) = 0$. The differential $d : \Omega^0(S^0) \to \Omega^1(S^0)$ lands in zero, so its image is zero. Therefore \begin{align*} H^0_{\mathrm{dR}}(S^0) \cong \mathbb R^2, \qquad H^k_{\mathrm{dR}}(S^0) = 0 \text{ for } k \ge 1. \end{align*} **The Mayer–Vietoris sequence.** The [Mayer–Vietoris Theorem](/theorems/2238) for de Rham cohomology states: for any smooth manifold $M$ and open cover $M = U \cup V$, there is a long exact sequence \begin{align*} \cdots \to H^k_{\mathrm{dR}}(M) \xrightarrow{(\iota_U^*, \iota_V^*)} H^k_{\mathrm{dR}}(U) \oplus H^k_{\mathrm{dR}}(V) \xrightarrow{j} H^k_{\mathrm{dR}}(U \cap V) \xrightarrow{\delta} H^{k+1}_{\mathrm{dR}}(M) \to \cdots \end{align*} where $\iota_U : U \hookrightarrow M$ and $\iota_V : V \hookrightarrow M$ are the inclusions, $j(\alpha, \beta) := \iota_{U \cap V, U}^*\alpha - \iota_{U \cap V, V}^*\beta$, and $\delta$ is the connecting homomorphism. We verify the hypothesis: $S^1$ is a smooth manifold and $\{U, V\}$ is an open cover by Step 1, so the theorem applies. Writing out the low-degree segment: \begin{align*} 0 \to H^0_{\mathrm{dR}}(S^1) \to H^0_{\mathrm{dR}}(U) \oplus H^0_{\mathrm{dR}}(V) \xrightarrow{j} H^0_{\mathrm{dR}}(U \cap V) \xrightarrow{\delta} H^1_{\mathrm{dR}}(S^1) \to H^1_{\mathrm{dR}}(U) \oplus H^1_{\mathrm{dR}}(V). \end{align*} Plugging in $H^0(S^1) \cong \mathbb R$ (Step 2), $H^0(U) = H^0(V) = \mathbb R$, $H^0(U \cap V) \cong \mathbb R^2$, $H^1(U) = H^1(V) = 0$ (Step 1): \begin{align*} 0 \to \mathbb R \xrightarrow{\Delta} \mathbb R \oplus \mathbb R \xrightarrow{j} \mathbb R^2 \xrightarrow{\delta} H^1_{\mathrm{dR}}(S^1) \to 0. \end{align*} The first map $\Delta$ sends a global constant $c$ on $S^1$ to its restrictions $(c, c) \in \mathbb R \oplus \mathbb R$ — the diagonal embedding. **Computing $j$ explicitly.** The two components of $U \cap V = S^1 \setminus \{p_N, p_S\}$ are the open east arc $E$ (containing $(1, 0)$) and the open west arc $W$ (containing $(-1, 0)$); both are path-connected. A class in $H^0_{\mathrm{dR}}(U \cap V) \cong \mathbb R^2$ is a locally constant smooth function, i.e., a function constant on $E$ and constant on $W$ — encoded as a pair $(v_E, v_W)$. Take $(a, b) \in H^0(U) \oplus H^0(V) \cong \mathbb R \oplus \mathbb R$, representing the constants $f_U \equiv a$ on $U$ and $f_V \equiv b$ on $V$. The pullback to $U \cap V$ of $f_U$ is the constant $a$ on each of $E$ and $W$; similarly for $f_V$. So \begin{align*} j(a, b) = (\iota^*_{U \cap V, U} f_U - \iota^*_{U \cap V, V} f_V)\big|_{E, W} = (a - b,\; a - b) \in \mathbb R^2. \end{align*} The image $\operatorname{im}(j) = \{(c, c) : c \in \mathbb R\}$ is the diagonal — one-dimensional, **not** all of $\mathbb R^2$. This non-surjectivity is precisely the geometric obstruction: a locally constant function on $U \cap V$ that takes different values on $E$ and $W$ cannot be written as the difference of constants on $U$ and on $V$. **Reading off $H^1$.** Exactness at $H^0_{\mathrm{dR}}(U \cap V)$ says $\ker(\delta) = \operatorname{im}(j)$, which is the diagonal — a one-dimensional subspace of $\mathbb R^2$. Exactness at $H^1_{\mathrm{dR}}(S^1)$ together with $H^1_{\mathrm{dR}}(U) \oplus H^1_{\mathrm{dR}}(V) = 0$ says $\delta$ is surjective. By the [first isomorphism theorem](/theorems/791), \begin{align*} H^1_{\mathrm{dR}}(S^1) \cong H^0_{\mathrm{dR}}(U \cap V) / \ker(\delta) \cong \mathbb R^2 / \{(c, c)\} \cong \mathbb R. \end{align*} **Higher degrees.** For $k \ge 2$, exactness of \begin{align*} H^{k-1}_{\mathrm{dR}}(U \cap V) \to H^k_{\mathrm{dR}}(S^1) \to H^k_{\mathrm{dR}}(U) \oplus H^k_{\mathrm{dR}}(V) \end{align*} with both flanking groups vanishing (the left because $H^{k-1}(S^0) = 0$ for $k - 1 \ge 1$, the right because $H^k(\mathbb R) = 0$ for $k \ge 1$) forces $H^k_{\mathrm{dR}}(S^1) = 0$. This finishes the base case. [/guided] [/step] [step:Carry out the inductive step for $n \ge 2$] Fix $n \ge 2$ and assume, as the inductive hypothesis, that \begin{align*} H^k_{\mathrm{dR}}(S^{n-1}) \cong \begin{cases} \mathbb R, & k = 0, \\ \mathbb R, & k = n - 1, \\ 0, & \text{otherwise}. \end{cases} \end{align*} We apply the [Mayer–Vietoris Theorem](/theorems/2238) to the cover $\{U, V\}$ of $S^n$ from Step 1. Its hypotheses are met: $S^n$ is a smooth manifold and $\{U, V\}$ is an open cover. The long exact sequence reads \begin{align*} \cdots \to H^{k-1}_{\mathrm{dR}}(U) \oplus H^{k-1}_{\mathrm{dR}}(V) \xrightarrow{j} H^{k-1}_{\mathrm{dR}}(U \cap V) \xrightarrow{\delta} H^k_{\mathrm{dR}}(S^n) \to H^k_{\mathrm{dR}}(U) \oplus H^k_{\mathrm{dR}}(V) \to \cdots \end{align*} By Step 1, $H^k_{\mathrm{dR}}(U) \oplus H^k_{\mathrm{dR}}(V) = 0$ for all $k \ge 1$, and $H^k_{\mathrm{dR}}(U \cap V) \cong H^k_{\mathrm{dR}}(S^{n-1})$ for all $k \ge 0$. **Case $k = 1$.** The segment is \begin{align*} 0 \to \mathbb R \xrightarrow{\Delta} \mathbb R \oplus \mathbb R \xrightarrow{j} H^0_{\mathrm{dR}}(U \cap V) \xrightarrow{\delta} H^1_{\mathrm{dR}}(S^n) \to 0, \end{align*} using $H^0(S^n) = \mathbb R$ (Step 2), $H^0(U) = H^0(V) = \mathbb R$, and $H^1(U) \oplus H^1(V) = 0$. Since $n \ge 2$, the sphere $S^{n-1}$ is connected, so $H^0_{\mathrm{dR}}(U \cap V) \cong H^0_{\mathrm{dR}}(S^{n-1}) \cong \mathbb R$ by the inductive hypothesis. The map $j$ takes $(a, b) \in \mathbb R \oplus \mathbb R$ to the locally constant function $a - b$ on the single component of $U \cap V$, so $j(a, b) = a - b \in \mathbb R$. This is surjective. By exactness $\delta = 0$, and since $\delta$ is also surjective onto $H^1_{\mathrm{dR}}(S^n)$, we conclude \begin{align*} H^1_{\mathrm{dR}}(S^n) = 0. \end{align*} **Case $2 \le k \le n - 1$ (vacuous if $n = 2$).** The Mayer–Vietoris segment is \begin{align*} 0 = H^{k-1}_{\mathrm{dR}}(U) \oplus H^{k-1}_{\mathrm{dR}}(V) \to H^{k-1}_{\mathrm{dR}}(U \cap V) \xrightarrow{\delta} H^k_{\mathrm{dR}}(S^n) \to H^k_{\mathrm{dR}}(U) \oplus H^k_{\mathrm{dR}}(V) = 0, \end{align*} so $\delta$ is an isomorphism: \begin{align*} H^k_{\mathrm{dR}}(S^n) \cong H^{k-1}_{\mathrm{dR}}(U \cap V) \cong H^{k-1}_{\mathrm{dR}}(S^{n-1}). \end{align*} For $2 \le k \le n - 1$, the exponent $k - 1$ satisfies $1 \le k - 1 \le n - 2$, which is neither $0$ nor $n - 1$, so by the inductive hypothesis $H^{k-1}_{\mathrm{dR}}(S^{n-1}) = 0$. Therefore $H^k_{\mathrm{dR}}(S^n) = 0$ for $2 \le k \le n - 1$. **Case $k = n$.** The same isomorphism argument applies: from \begin{align*} 0 \to H^{n-1}_{\mathrm{dR}}(U \cap V) \xrightarrow{\delta} H^n_{\mathrm{dR}}(S^n) \to 0, \end{align*} we get $H^n_{\mathrm{dR}}(S^n) \cong H^{n-1}_{\mathrm{dR}}(S^{n-1}) \cong \mathbb R$ by the inductive hypothesis. **Case $k \ge n + 1$.** The Mayer–Vietoris segment is \begin{align*} 0 \to H^k_{\mathrm{dR}}(S^n) \to 0, \end{align*} once we know $H^{k-1}_{\mathrm{dR}}(U \cap V) \cong H^{k-1}_{\mathrm{dR}}(S^{n-1}) = 0$. For $k \ge n + 1$, the exponent $k - 1 \ge n$ is strictly greater than $n - 1$, so by the inductive hypothesis $H^{k-1}_{\mathrm{dR}}(S^{n-1}) = 0$. Hence $H^k_{\mathrm{dR}}(S^n) = 0$. [guided] We now carry out the inductive step. Fix $n \ge 2$ and assume the theorem holds for $S^{n-1}$: \begin{align*} H^k_{\mathrm{dR}}(S^{n-1}) \cong \begin{cases} \mathbb R, & k = 0, \\ \mathbb R, & k = n - 1, \\ 0, & \text{otherwise}. \end{cases} \end{align*} Our goal is to read off $H^k_{\mathrm{dR}}(S^n)$ for every $k \ge 0$ from the [Mayer–Vietoris Theorem](/theorems/2238) applied to the cover $\{U, V\}$ constructed in Step 1. **Setting up the sequence.** Mayer–Vietoris requires only that $\{U, V\}$ is an open cover of the smooth manifold $S^n$, which we verified in Step 1. Its long exact sequence in de Rham cohomology is \begin{align*} \cdots \to H^{k-1}_{\mathrm{dR}}(U) \oplus H^{k-1}_{\mathrm{dR}}(V) \xrightarrow{j} H^{k-1}_{\mathrm{dR}}(U \cap V) \xrightarrow{\delta} H^k_{\mathrm{dR}}(S^n) \to H^k_{\mathrm{dR}}(U) \oplus H^k_{\mathrm{dR}}(V) \to \cdots \end{align*} By Step 1, $H^k_{\mathrm{dR}}(U) = H^k_{\mathrm{dR}}(V) = 0$ for every $k \ge 1$, so the right-hand "$H^k(U) \oplus H^k(V)$" term vanishes for $k \ge 1$. Similarly $H^{k-1}(U) \oplus H^{k-1}(V) = 0$ for $k \ge 2$. The behaviour of the sequence therefore changes at $k = 1$ (where the left flank is non-zero) versus $k \ge 2$ (where both flanks vanish, making $\delta$ an isomorphism). And $H^k_{\mathrm{dR}}(U \cap V) \cong H^k_{\mathrm{dR}}(S^{n-1})$ for every $k \ge 0$ — this is where the inductive hypothesis enters. **Case $k = 1$.** The exact segment is \begin{align*} 0 \to H^0_{\mathrm{dR}}(S^n) \to H^0_{\mathrm{dR}}(U) \oplus H^0_{\mathrm{dR}}(V) \xrightarrow{j} H^0_{\mathrm{dR}}(U \cap V) \xrightarrow{\delta} H^1_{\mathrm{dR}}(S^n) \to H^1_{\mathrm{dR}}(U) \oplus H^1_{\mathrm{dR}}(V). \end{align*} Substituting $H^0(S^n) \cong \mathbb R$ (Step 2), $H^0(U) = H^0(V) = \mathbb R$, and $H^1(U) = H^1(V) = 0$: \begin{align*} 0 \to \mathbb R \xrightarrow{\Delta} \mathbb R \oplus \mathbb R \xrightarrow{j} H^0_{\mathrm{dR}}(U \cap V) \xrightarrow{\delta} H^1_{\mathrm{dR}}(S^n) \to 0. \end{align*} What is $H^0_{\mathrm{dR}}(U \cap V)$? This is where the hypothesis $n \ge 2$ matters. For $n \ge 2$, the equator $S^{n-1}$ is connected (a positive-dimensional sphere is path-connected); since $U \cap V$ deformation retracts onto $S^{n-1}$, it too is connected. So $H^0_{\mathrm{dR}}(U \cap V) \cong H^0_{\mathrm{dR}}(S^{n-1}) \cong \mathbb R$ — a single copy, not two. (Contrast Step 4, where $n = 1$ gave $H^0(S^0) = \mathbb R^2$, and that extra $\mathbb R$ ultimately became $H^1(S^1)$.) So the sequence is \begin{align*} 0 \to \mathbb R \xrightarrow{\Delta} \mathbb R \oplus \mathbb R \xrightarrow{j} \mathbb R \xrightarrow{\delta} H^1_{\mathrm{dR}}(S^n) \to 0. \end{align*} The map $j$ sends a pair of constants $(a, b)$ to the (locally) constant $a - b$ on the single component of $U \cap V$. This is surjective onto $\mathbb R$. By exactness at the middle $\mathbb R$, $\ker(\delta) = \operatorname{im}(j) = \mathbb R$, so $\delta = 0$. Combined with the next exactness ($\delta$ surjective onto $H^1(S^n)$), we obtain \begin{align*} H^1_{\mathrm{dR}}(S^n) = 0 \quad \text{for } n \ge 2. \end{align*} This is the geometric content of the connectedness leap from $S^0$ to $S^{n-1}$ with $n \ge 2$: when the equator is connected, the Mayer–Vietoris obstruction vanishes and there is no $H^1$ class on $S^n$. **Case $2 \le k \le n - 1$ (empty when $n = 2$).** Now both flanking terms vanish: \begin{align*} 0 = H^{k-1}(U) \oplus H^{k-1}(V) \to H^{k-1}(U \cap V) \xrightarrow{\delta} H^k_{\mathrm{dR}}(S^n) \to H^k(U) \oplus H^k(V) = 0. \end{align*} Exactness immediately gives $\delta$ both injective and surjective, hence \begin{align*} H^k_{\mathrm{dR}}(S^n) \cong H^{k-1}_{\mathrm{dR}}(U \cap V) \cong H^{k-1}_{\mathrm{dR}}(S^{n-1}). \end{align*} For $2 \le k \le n - 1$ we have $1 \le k - 1 \le n - 2$ — strictly between $0$ and $n - 1$ — so by the inductive hypothesis $H^{k-1}(S^{n-1}) = 0$. Therefore $H^k_{\mathrm{dR}}(S^n) = 0$ in this range. **Case $k = n$.** The same isomorphism $H^n(S^n) \cong H^{n-1}(S^{n-1})$ holds, by the identical Mayer–Vietoris argument (both flanks vanish because $n \ge 2$ makes the relevant degrees $\ge 1$). By the inductive hypothesis $H^{n-1}(S^{n-1}) \cong \mathbb R$, so \begin{align*} H^n_{\mathrm{dR}}(S^n) \cong \mathbb R. \end{align*} This is the connecting-homomorphism shift that propagates the single top-degree class from $S^{n-1}$ to $S^n$. **Case $k \ge n + 1$.** Again both flanks vanish and $\delta$ is an isomorphism: $H^k(S^n) \cong H^{k-1}(S^{n-1})$. For $k \ge n + 1$ we have $k - 1 \ge n$, so $k - 1$ exceeds the top degree $n - 1$ of $S^{n-1}$, and the inductive hypothesis gives $H^{k-1}(S^{n-1}) = 0$. Therefore $H^k_{\mathrm{dR}}(S^n) = 0$ for $k \ge n + 1$. [/guided] [/step] [step:Synthesize the inductive conclusion] Combining the four cases of Step 5 with Step 3: \begin{align*} H^k_{\mathrm{dR}}(S^n) \cong \begin{cases} \mathbb R, & k = 0 \quad \text{(Step 3)}, \\ 0, & 1 \le k \le n - 1 \quad \text{(Step 5, cases $k = 1$ and $2 \le k \le n - 1$)}, \\ \mathbb R, & k = n \quad \text{(Step 5, case $k = n$)}, \\ 0, & k \ge n + 1 \quad \text{(Step 5, case $k \ge n + 1$)}. \end{cases} \end{align*} The base case $n = 1$ (Step 4) and the inductive step $n \ge 2$ (Step 5) together establish the claim for every $n \ge 1$. This completes the proof. [/step]