[proofplan] We prove by induction on $n$ that any reflection $r: S^n \to S^n$ across a hyperplane through the origin has degree $-1$. The base case $n = 1$ is verified directly using the Mayer--Vietoris computation for $S^1$. The inductive step uses naturality of the Mayer--Vietoris sequence: by choosing the poles to lie in the reflecting hyperplane, the boundary isomorphism $\partial: H_n(S^n) \xrightarrow{\sim} H_{n-1}(S^{n-1})$ intertwines $r_*$ on $S^n$ with $r_*$ on $S^{n-1}$, reducing the problem to lower dimension. [/proofplan]

[step:Establish the Mayer--Vietoris setup with poles in the reflecting hyperplane]Let $r: S^n \to S^n$ be the reflection across a hyperplane $V$ through the origin. Since any two reflections across hyperplanes through the origin are conjugate by an orthogonal transformation (which is a homeomorphism of $S^n$), it suffices to prove the result for a single choice of hyperplane. We may therefore choose coordinates so that $r(x_1, x_2, \ldots, x_{n+1}) = (-x_1, x_2, \ldots, x_{n+1})$. Choose the north pole $N = (0, \ldots, 0, 1)$ and south pole $S = (0, \ldots, 0, -1)$. Both lie in the hyperplane $\{x_1 = 0\}$, so $r(N) = N$ and $r(S) = S$. Define $A = S^n \setminus \{N\}$ and $B = S^n \setminus \{S\}$. Both $A$ and $B$ are invariant under $r$, and both are homeomorphic to $\mathbb{R}^n$ via stereographic projection, hence contractible. Their intersection $A \cap B = S^n \setminus \{N, S\}$ deformation-retracts onto the equatorial sphere $S^{n-1} = S^n \cap \{x_{n+1} = 0\}$, and $r$ restricts to a reflection on this equatorial $S^{n-1}$.[/step]

[guided]The key geometric idea is to place the poles in the mirror so that the decomposition $S^n = A \cup B$ is compatible with the reflection. If we chose poles away from the mirror, $r$ would swap the two halves $A$ and $B$, and we could not use naturality to reduce dimensions. With $N$ and $S$ in the hyperplane $\{x_1 = 0\}$, the reflection $r(x_1, x_2, \ldots, x_{n+1}) = (-x_1, x_2, \ldots, x_{n+1})$ fixes both poles, so $r(A) = A$ and $r(B) = B$. This means $r$ induces self-maps on $A$, $B$, and $A \cap B$, and the entire Mayer--Vietoris sequence is natural with respect to $r$. The intersection $A \cap B = S^n \setminus \{N, S\}$ is the sphere with two antipodal points removed. The deformation retraction onto the equator $S^{n-1} = \{x \in S^n : x_{n+1} = 0\}$ is given by normalising the projection $(x_1, \ldots, x_n, x_{n+1}) \mapsto \frac{(x_1, \ldots, x_n, 0)}{|(x_1, \ldots, x_n)|}$. Under this retraction, $r$ restricts to $(x_1, x_2, \ldots, x_n) \mapsto (-x_1, x_2, \ldots, x_n)$ on $S^{n-1}$, which is again a reflection across a hyperplane through the origin.[/guided]

[step:Use naturality of the Mayer--Vietoris boundary to reduce to $S^{n-1}$]Since $H_k(A) = H_k(B) = 0$ for $k \geq 1$ (as $A$ and $B$ are contractible), the Mayer--Vietoris sequence for $S^n = A \cup B$ gives an isomorphism \begin{align*} \partial: H_n(S^n) \xrightarrow{\sim} H_{n-1}(A \cap B) \cong H_{n-1}(S^{n-1}) \end{align*} for all $n \geq 2$. By naturality of Mayer--Vietoris with respect to the map $r$ (which preserves the cover $\{A, B\}$), the following diagram commutes: \begin{align*} \begin{array}{ccc} H_n(S^n) & \xrightarrow{\partial, \sim} & H_{n-1}(S^{n-1}) \\ \downarrow r_* & & \downarrow r_* \\ H_n(S^n) & \xrightarrow{\partial, \sim} & H_{n-1}(S^{n-1}) \end{array} \end{align*} Since $\partial$ is an isomorphism, $r_*$ on $H_n(S^n)$ equals $r_*$ on $H_{n-1}(S^{n-1})$ (after conjugating by the isomorphism $\partial$). In particular, $\deg(r: S^n \to S^n) = \deg(r|_{S^{n-1}}: S^{n-1} \to S^{n-1})$.[/step]

[guided]Naturality of the Mayer--Vietoris sequence is the engine of this proof. What does it say here? The reflection $r$ maps $A$ to $A$, $B$ to $B$, and $A \cap B$ to $A \cap B$, so $r$ is a map of the open cover $\{A, B\}$ to itself. The naturality property states that all squares in the Mayer--Vietoris ladder commute. In particular, the connecting homomorphism $\partial$ commutes with $r_*$. Since $A$ and $B$ are contractible, all terms $H_k(A) \oplus H_k(B)$ vanish for $k \geq 1$. The portion of the sequence around degree $n$ reads \begin{align*} 0 = H_n(A) \oplus H_n(B) \to H_n(S^n) \xrightarrow{\partial} H_{n-1}(A \cap B) \to H_{n-1}(A) \oplus H_{n-1}(B) = 0, \end{align*} so $\partial$ is an isomorphism. Combining $A \cap B \simeq S^{n-1}$ with the commutativity $\partial \circ r_* = r_* \circ \partial$, we get \begin{align*} r_*^{S^n} = \partial^{-1} \circ r_*^{S^{n-1}} \circ \partial, \end{align*} so the degree of $r$ on $S^n$ equals the degree of $r$ on $S^{n-1}$. This reduces the problem by one dimension at each step.[/guided]

[step:Verify the base case $n = 1$: a reflection on $S^1$ has degree $-1$] For $n = 1$, consider $r: S^1 \to S^1$ given by $r(x_1, x_2) = (-x_1, x_2)$, which reflects across the $x_2$-axis. Cover $S^1$ by two open arcs $A$ and $B$, each slightly more than a semicircle, with $A \cap B$ consisting of two disjoint small arcs around the points $p = (0, 1)$ and $q = (0, -1)$ on the $x_2$-axis. Both $A$ and $B$ are contractible, and $A \cap B$ has two path components, so $H_0(A \cap B) \cong \mathbb{Z}^2$, generated by $[p]$ and $[q]$. The Mayer--Vietoris sequence gives an isomorphism \begin{align*} \partial: H_1(S^1) \xrightarrow{\sim} \ker\bigl(H_0(A \cap B) \xrightarrow{(i_{A*}, i_{B*})} H_0(A) \oplus H_0(B)\bigr). \end{align*} Since $A$ and $B$ are each path-connected, both $i_{A*}$ and $i_{B*}$ send $[p]$ and $[q]$ to the same generator, so the kernel is generated by $[p] - [q]$. Thus $\partial$ sends the generator $\gamma$ of $H_1(S^1)$ to $[p] - [q]$ (up to sign; fix the sign by choosing $\gamma$ so that $\partial(\gamma) = [p] - [q]$). The reflection $r$ fixes both $p$ and $q$ (since both lie on the $x_2$-axis), so $r_*([p] - [q]) = [p] - [q]$ on $H_0(A \cap B)$. However, $r$ reverses the orientation of $S^1$: the loop traversing $S^1$ counterclockwise is sent to the loop traversing clockwise. Concretely, parametrise the generator of $H_1(S^1)$ by the singular $1$-simplex $\sigma: [0,1] \to S^1$, $t \mapsto (\cos 2\pi t, \sin 2\pi t)$. Then $r \circ \sigma(t) = (-\cos 2\pi t, \sin 2\pi t) = (\cos(\pi - 2\pi t), \sin(\pi - 2\pi t))$, which traverses $S^1$ in the opposite direction starting from a different point. Since this reversed-orientation loop represents $-[\sigma]$ in $H_1(S^1)$, we have $r_*([\sigma]) = -[\sigma]$, giving $\deg(r) = -1$. [/step]

[step:Complete the induction] By the base case, $\deg(r: S^1 \to S^1) = -1$. The inductive step shows that $\deg(r: S^n \to S^n) = \deg(r: S^{n-1} \to S^{n-1})$ for all $n \geq 2$. Therefore, by induction on $n$, $\deg(r) = -1$ for all $n \geq 1$. [/step]