[proofplan] We prove $\deg(r) = -1$ by induction on $n$, using the naturality of the Mayer--Vietoris sequence. We choose the Mayer--Vietoris decomposition $S^n = A \cup B$ (complements of north and south poles) so that both $A$ and $B$ are invariant under $r$, which requires placing the poles in the reflecting hyperplane. The Mayer--Vietoris connecting homomorphism $\partial: H_n(S^n) \xrightarrow{\sim} H_{n-1}(A \cap B) \cong H_{n-1}(S^{n-1})$ then commutes with $r_*$, reducing the problem to the $(n-1)$-sphere. The base case $n = 1$ is verified by direct computation: the reflection swaps the two components of $A \cap B$, reversing the generator of $H_1(S^1)$. [/proofplan] [step:Set up the Mayer--Vietoris decomposition so that the reflection preserves both open sets] Let $r: S^n \to S^n$ be a reflection about a hyperplane $P$ through the origin in $\mathbb{R}^{n+1}$. Choose the north and south poles $N, S \in P \cap S^n$ (this is possible since $P$ is an $n$-dimensional subspace of $\mathbb{R}^{n+1}$, so $P \cap S^n$ is an $(n-1)$-sphere and in particular nonempty with at least two antipodal points when $n \geq 1$). Define $A := S^n \setminus \{N\}$ and $B := S^n \setminus \{S\}$. Since $r(N) = N$ and $r(S) = S$ (both poles lie in the reflecting hyperplane $P$, so $r$ fixes them), we have $r(A) = A$ and $r(B) = B$. Furthermore, $r(A \cap B) = A \cap B$. [/step] [step:Use naturality of Mayer--Vietoris to reduce to $S^{n-1}$] The Mayer--Vietoris sequence for the cover $S^n = A \cup B$ gives, for $n \geq 2$, an isomorphism \begin{align*} \partial: H_n(S^n) \xrightarrow{\;\sim\;} H_{n-1}(A \cap B). \end{align*} (This was established in the proof of the [Homology of $S^n$](/theorems/2242): the groups $H_n(A)$, $H_n(B)$, $H_{n-1}(A)$, and $H_{n-1}(B)$ all vanish since $A \simeq B \simeq \mathbb{R}^n$.) The intersection $A \cap B = S^n \setminus \{N, S\}$ deformation retracts onto the equatorial sphere $S^{n-1} = P \cap S^n$. The reflection $r$ restricts to $A \cap B$ and, under the deformation retraction, corresponds to the reflection $r|_{S^{n-1}}: S^{n-1} \to S^{n-1}$ about the hyperplane $P \cap \operatorname{span}(S^{n-1})$ (a codimension-one subspace of the ambient space of $S^{n-1}$). By the naturality of the Mayer--Vietoris sequence with respect to the map $r: (S^n; A, B) \to (S^n; A, B)$ (which is valid since $r$ preserves the decomposition), the following diagram commutes: \begin{align*} \begin{array}{ccc} H_n(S^n) & \xrightarrow{\;\partial\;} & H_{n-1}(A \cap B) \\ \downarrow r_* & & \downarrow r_* \\ H_n(S^n) & \xrightarrow{\;\partial\;} & H_{n-1}(A \cap B) \end{array} \end{align*} Since $\partial$ is an isomorphism, $r_*$ on $H_n(S^n)$ is conjugate to $r_*$ on $H_{n-1}(A \cap B) \cong H_{n-1}(S^{n-1})$. In particular, \begin{align*} \deg(r: S^n \to S^n) = \deg(r|_{S^{n-1}}: S^{n-1} \to S^{n-1}). \end{align*} By induction on $n$, it suffices to verify the base case. [guided] The naturality of the Mayer--Vietoris sequence is the essential ingredient. In general, if $\varphi: X \to X$ is a continuous map that preserves the open cover $X = A \cup B$ (i.e., $\varphi(A) \subset A$ and $\varphi(B) \subset B$), then $\varphi$ induces a map of Mayer--Vietoris sequences that makes the connecting homomorphisms commute with $\varphi_*$. The commutativity of the square \begin{align*} \begin{array}{ccc} H_n(S^n) & \xrightarrow{\;\partial\;} & H_{n-1}(A \cap B) \\ \downarrow r_* & & \downarrow r_* \\ H_n(S^n) & \xrightarrow{\;\partial\;} & H_{n-1}(A \cap B) \end{array} \end{align*} means $\partial \circ r_* = r_* \circ \partial$. Since $\partial$ is an isomorphism (with inverse $\partial^{-1}$), we can write $r_*|_{H_n(S^n)} = \partial^{-1} \circ r_*|_{H_{n-1}(A \cap B)} \circ \partial$. This is a conjugation by the isomorphism $\partial$, so the two copies of $r_*$ act by the same scalar (degree) on the respective copies of $\mathbb{Z}$. The crucial geometric point is that the reflection $r$ on $S^n$ restricts to a reflection on $S^{n-1}$. This is because the reflecting hyperplane $P$ intersects $S^{n-1}$ (the equatorial sphere) in a codimension-one subsphere, and $r|_{S^{n-1}}$ reflects about this subsphere. The induction therefore reduces the computation of $\deg(r)$ from dimension $n$ to dimension $n-1$. [/guided] [/step] [step:Verify the base case $n = 1$ by direct computation] For $n = 1$, the circle $S^1$ is covered by $A = S^1 \setminus \{N\}$ and $B = S^1 \setminus \{S\}$, where $N$ and $S$ are two points in the reflecting line. The intersection $A \cap B = S^1 \setminus \{N, S\}$ consists of two open arcs, with the two arcs lying on opposite sides of the reflecting line. Call these arcs $I_p$ and $I_q$, containing representative points $p$ and $q$ respectively. The Mayer--Vietoris connecting homomorphism $\partial: H_1(S^1) \to H_0(A \cap B)$ is an isomorphism, and $H_0(A \cap B) \cong \mathbb{Z}^2$ is generated by $[p]$ and $[q]$. The generator of $\ker(i_{A*} \oplus i_{B*}) \cong \operatorname{im}(\partial) \cong \mathbb{Z}$ is $[p] - [q]$ (since both $p$ and $q$ map to the same class in $H_0(A)$ and $H_0(B)$). So $\partial$ identifies the generator $\gamma \in H_1(S^1)$ with $[p] - [q]$. The reflection $r$ swaps the two arcs: $r(I_p) = I_q$ and $r(I_q) = I_p$, so $r_*([p]) = [q]$ and $r_*([q]) = [p]$ on $H_0(A \cap B)$. Therefore \begin{align*} r_*([p] - [q]) = [q] - [p] = -([p] - [q]). \end{align*} By the commutative diagram $\partial \circ r_* = r_* \circ \partial$: \begin{align*} \partial(r_*(\gamma)) = r_*(\partial(\gamma)) = r_*([p] - [q]) = -([p] - [q]) = -\partial(\gamma) = \partial(-\gamma). \end{align*} Since $\partial$ is an isomorphism, $r_*(\gamma) = -\gamma$, so $\deg(r) = -1$. [guided] The base case $n = 1$ is where the geometric content of the proof resides. The circle $S^1$ is split into two arcs by the two fixed points $N$ and $S$ of the reflection. These fixed points lie on the reflecting line, and the two arcs are interchanged by $r$. On $H_0$, the reflection permutes the generators: it sends the class of a point in one arc to the class of a point in the other arc. The generator of $H_1(S^1) \cong \mathbb{Z}$ corresponds, via the Mayer--Vietoris connecting homomorphism, to the difference $[p] - [q]$. (Why the difference? Because $\operatorname{im}(\partial) = \ker(i_{A*} \oplus i_{B*})$, and $i_{A*} \oplus i_{B*}$ maps $[p]$ and $[q]$ to the same element $(1, 1)$ in $H_0(A) \oplus H_0(B) \cong \mathbb{Z}^2$. So the kernel consists of multiples of $[p] - [q]$.) Swapping $p$ and $q$ sends $[p] - [q]$ to $[q] - [p] = -([p] - [q])$, which is multiplication by $-1$. The naturality diagram then forces $r_*$ on $H_1(S^1)$ to also be multiplication by $-1$, giving $\deg(r) = -1$. [/guided] [/step] [step:Conclude by induction] The base case gives $\deg(r: S^1 \to S^1) = -1$. The inductive step shows $\deg(r: S^n \to S^n) = \deg(r|_{S^{n-1}}: S^{n-1} \to S^{n-1})$ for all $n \geq 2$. By induction, $\deg(r) = -1$ for all $n \geq 1$. [/step]