Derivative of the antipodal map of $S^n$ in the context of smooth manifolds. I was reading a solution of a question here (see item $c$ on page $7$), but I don't understand why the antipodal map of $S^n$ has derivative equal to minus the identity on $T_pS^n$ for $p \in S^n$. This is clear if we see $S^n \subset \mathbb{R}^{n+1}$, then it's a simple computation of analysis in Euclidian spaces, but how to prove this in the context of smooth manifolds, where we don't have an ambient space to $S^n$ necessarily?
 A: There are several equivalent ways to define a smooth manifold, the tangent bundle, differentials and so on, beginning with set theory, classical structures studied in analysis on euclidean spaces etc. So to precisely address your question you first have to provide your favourite definitions then probably it will be automatically answered. Let us have in mind that all this different definitions have to mimic the analysis and the geometry of regular curves and surfaces in $\mathbb{R}^3$ when the usual machinery are chosen. For this to count as an answer I will choose mine.
Let $ \ n \in \mathbb{N} \setminus \{ 0 \} \ $ and $ \ S^n = \big\{ u \in \mathbb{R}^{n+1} : \Vert u \Vert = 1 \big\} \, $, where $ \ \Vert \cdot \Vert : \mathbb{R}^{n+1} \to \mathbb{R} \ $ is the usual euclidean norm. This norm induces the usual/euclidean distance function on $ \, \mathbb{R}^{n+1} \, $ and this function, in turn, induces the usual/euclidean topology of $ \, \mathbb{R}^{n+1} \, $. Let $\, \tau \, $ be the subspace topology on $ \, S^n \, $, that is, a subset $ \ X \subset S^n \ $ is open in $ \, S^n \, $ (in symbols $ \ X \in \tau \ $) if and only if (iff) there exists an open set $ \, Y \, $ in the usual topology of $ \, \mathbb{R}^{n+1} \, $ such that $ \ X = Y \cap S^n \ $. In this way $ \, (S^n, \tau) \, $ is the topological subspace of $ \, \mathbb{R}^{n+1} \, $ (with the usual/euclidean topology) we are interested in.
Now we put a smooth structure on $ \, (S^n, \tau) \, $. We say that a function $ \, x \, $ is a chart/coordinate system of $ \, (S^n, \tau) \, $ iff $ \ \varnothing \neq \text{dom}(x) \in \tau \ $, $ \, \text{im}(x) \subset \mathbb{R}^n \ $ and $ \, x \, $ is a topological embedding of $ \, \text{dom}(x) \, $ on $ \, \mathbb{R}^n \, $, that is, $x : \text{dom}(x) \to \mathbb{R}^n \ $ is an open, injective and continuous function, where the topology of the set $ \, \text{dom}(x) \, $ is the subspace induced topology of $ \, \tau \, $. In other words, $ \, x : \text{dom}(x) \to \text{im}(x) \ $ is a homeomorphism (onto its image), where the topology of the set $ \, \text{im}(x) \, $ is the subspace induced topology of the usual/euclidean topology of $ \, \mathbb{R}^n \, $. We say that two charts $ \, x \, $ and $ \, y \, $ are smoothly compatible iff $$\big[ x|_{\text{dom}(y)} \big]^{-1} \circ \big[ y|_{\text{dom}(x)} \big] : x[\text{dom}(y)] \to y[\text{dom}(x)]$$ is a smooth diffeomorphism between the open sets $ \ x[\text{dom}(y)] \ $ and $ \ y[\text{dom}(x)] \ $ of $ \, \mathbb{R}^n \, $. We define a smooth (maximal) atlas for $ \, (S^n , \tau) \, $ as a collection $ \, \mathcal{A} \, $ of charts of $ \, (S^n , \tau) \, $ such that $ \ \displaystyle \bigcup_{x \in \mathcal{A}} \text{dom}(x) = S^n \ $ and, for all $ \ x \in \mathcal{A} \ $ and all charts $ \, y \, $ of $ \, (S^n , \tau) \, $, we have that $ \ y \in \mathcal{A} \ $ if and only if $ \, x \, $ and $ \, y \, $ are smoothly compatible. We say that $ \, \mathcal{A} \, $ is a smooth structure for $ \, (S^n, \tau) \, $ and that $ \, (S^n , \tau , \mathcal{A}) \, $ is a smooth $n$-manifold. The usual atlas for $ \, (S^n, \tau) \, $ is the one which contains the stereographic projections and we will fix exactly this to be our $ \, \mathcal{A} \, $.
Let $ \ k \in \mathbb{N} \setminus \{ 0 \} \ $ and $ \, f \, $ be a continuous function such that $ \ \text{dom}(f) \in \tau \ $ and $ \ \text{im}(f) \subset \mathbb{R}^k \ $. We say that $ \, f \, $ is smooth with respect to the smooth structure $ \, (S^n , \tau , \mathcal{A}) \, $ iff, for all $ \ x \in \mathcal{A} \ $, $$\big[ x|_{\text{dom}(f)} \big]^{-1} \circ \big[ f|_{\text{dom}(x)} \big] : x[\text{dom}(f)] \to \mathbb{R}^k$$ is smooth as a function of euclidean spaces. Let $ \, M \, $, $ \, \mu \, $ and $ \, \mathcal{B} \, $ be sets and $ \ m \in \mathbb{N} \setminus \{ 0 \} \ $ such that $ \, \mu \, $ is a topology for $ \, M \, $ and $ \, \mathcal{B} \, $ is a smooth maximal atlas for the topological space $ \, (M, \mu) \, $ so that $ \, (M, \mu , \mathcal{B} ) \, $ is a smooth $m$-manifold. Let $ \, f \, $ be a continuous function such that $ \ \text{dom}(f) \in \tau \ $ and $ \ \text{im}(f) \subset M \ $. We say that $ \, f \, $ is smooth with respect to (wrt) the smooth structures $ \, (S^n , \tau , \mathcal{A}) \, $ and $ \, (M, \mu , \mathcal{B} ) \, $ iff, for all $ \ x \in \mathcal{A} \ $ and all $ \ y \in \mathcal{B} \ $, $$\big[ x|_{\text{dom}(f)} \big]^{-1} \circ \big[ f|_{\text{dom}(x)} \big] \circ \Big[ y|_{f[\text{dom}(x)]} \Big] : x[\text{dom}(f)] \to \mathbb{R}^m$$ is smooth as a function of euclidean spaces.
Fix $ \ p \in S^n \ $ and let $$C^{\infty}_p (S^n) = \Big\{ f \subset S^n \times \mathbb{R} \ \ \textbf{ : } \ \ p \in \text{dom}(f) \in \tau \ \, \text{ and } \ \, f \ \, \text{ is smooth wrt } \ (S^n , \tau , \mathcal{A} ) \, \Big\} \ \ . $$ Define a binary relation $ \, \sim_p \, $ on $ \ C^{\infty}_p (S^n) \times C^{\infty}_p (S^n) \ $ making, $\forall f,g \in C^{\infty}_p (S^n) \, $, $ \, f \sim_p g \ $ if and only if there exists $ \ X \in \tau \ $ such that $ \ p \in X \ $ and $ \ f|_X = g|_X \ \, $. This is an equivalence relation and its equivalence classes are called the germs of $ \, (S^n , \tau , \mathcal{A} )$-smooth functions at $p \, $. It is straightforward that, for all $ \ \ell \in \mathbb{N} \setminus \{ 0 \} \ $, for any non-zero finite number of $ \, \sim_p$-related functions $f_1, f_2 , ..., f_{\ell} \in C^{\infty}_p (S^n) \, $, there exists the (unique) largest set $ \, X_{p;f_1, f_2, ..., f_{\ell}} \, $ such that $ \ p \in X_{p;f_1, f_2, ..., f_{\ell}} \in \tau \ $, $ \, X_{p;f_1, f_2, ..., f_{\ell}} \subset \text{dom}(f_1) \cap \text{dom}(f_2) \cap ... \cap \text{dom}(f_{\ell}) \ $ and $ \ f_1|_{X_{p;f_1, f_2, ..., f_{\ell}}} = f_2|_{X_{p;f_1, f_2, ..., f_{\ell}}} = ... = f_{\ell}|_{X_{p;f_1, f_2, ..., f_{\ell}}} \ \ $. Define a scalar multiplication $ \ c \cdot [f]_p = [ c \cdot f ]_p \ $, $\forall c \in \mathbb{R} \, $, $ \, \forall [f]_p \in C^{\infty}_p (S^n) / \! \sim_p \ $, and an addition $ \ [h]_p + [g]_p = \left[ h|_{\text{dom}(g)} + g|_{\text{dom}(h)} \right]_p \ \, $, $ \, \forall [h]_p , [g]_p \in C^{\infty}_p (S^n) / \! \sim_p \ $, and we have that $ \, \big( C^{\infty}_p (S^n) / \! \sim_p , \mathbb{R} , + , \cdot \big) \, $ is a $\mathbb{R}$-vector space. Define a multiplication of germs $ \ [h]_p \cdot [g]_p = \left[ h|_{\text{dom}(g)} \cdot g|_{\text{dom}(h)} \right]_p \ \, $, $ \, \forall [h]_p , [g]_p \in C^{\infty}_p (S^n) / \! \sim_p \ $, and we have that $ \, \big( C^{\infty}_p (S^n) / \! \sim_p , \mathbb{R} , + , \cdot , \cdot \big) \, $ is an unital associative and commutative $\mathbb{R}$-algebra. Let $ \, T_p S^n \, $ be the set of all linear functionals $ \ v: C^{\infty}_p (S^n) / \! \sim_p \, \, \to \mathbb{R} \ $ such that $$v([h]_p \cdot [g]_p) = h(p) \cdot v([g]_p) + g(p) \cdot v([h]_p) \ \ , $$ $\forall [h]_p , [g]_p \in C^{\infty}_p (S^n) / \! \sim_p \ $, and we have that $ \, T_p S^n \, $ is a $\mathbb{R}$-vector subspace of the dual space $ \, \big( C^{\infty}_p (S^n) / \! \sim_p \big)^* \, $ and its dimension is $ \, \dim_{\mathbb{R}} ( T_p S^n ) = n \ $. This construction uses derivations and it is my favourite model of the tangent space.
Let $ \ f: (S^n , \tau , \mathcal{A}) \to (M, \mu , \mathcal{B} ) \ $ be a smooth function wrt the smooth structures, where $ \ p \in \text{dom}(f) \in \tau \ $ and $ \, \text{dom}(f) \, $ is not necessarily the whole $ \, S^n \, $. The differential (or pushforward) of $ \, f \, $ at $ \, p \, $ is the linear map $ \ d_p f : T_p S^n \to T_{f(p)}M \ $ such that $ \ \big[ (d_p f) (v) \big] \big( [h]_{f(p)} \big) = v \big( \big[ h|_{\text{im}(f)} \circ f \big]_p \big) \ $, $\forall v \in T_pS^n \, $, $\forall [h]_{f(p)} \in C^{\infty}_{f(p)} (M) / \! \sim_{f(p)} \ $.
Let $ \ A: S^n \to S^n \ $ be the antipodal map, that is, $ \, A(q)=-q \ \, $, $ \, \forall q \in S^n \ $. In other words, $ \, A = - id_{S^n} \ $. Since $ \ A(p) = -p \neq p \ $, we have $ \ T_p S^n \cap T_{A(p)} S^n = \varnothing \ $. Then in this model of the tangent spaces it is impossible that the pushforward to be a multiple of an identity function, but we can still compute its matrix at some chosen basis. Let $ \, x \, $ be a chart with $ \ p \in \text{dom}(x) \ $ and, for each $ \ \mu \in \{ 1 , 2 , ... , n \} \ $, define $ \ \frac{\partial}{\partial x^{\mu}} \Big|_p  : C^{\infty}_p (S^n) / \! \sim_p \, \, \to \mathbb{R} \ $ such that, $\forall [h]_p \in C^{\infty}_p (S^n) / \! \sim_p \ $, $$\frac{\partial}{\partial x^{\mu}} \bigg|_p \big( [h]_p \big) = \partial_{\mu} (h \circ x^{-1}) \big( x(p) \big) = \lim_{t \to 0} \frac{(h \circ x^{-1}) \big( x(p) + t \cdot e_{\mu} \big) - (h \circ x^{-1}) \big( x(p) \big)}{t} \ \ \ , $$ where $ \ e_{\mu} = \big( \delta_{\mu}^1 , \delta_{\mu}^2 , ... , \delta_{\mu}^n \big) \ $ and $ \, \delta_{\mu}^{\nu} \, $ is the Kronecker delta. This is the $ \, \mu$-th partial derivative of the function $ \, h \circ x^{-1} \, $ at the point $ \, x(p) \, $. It can be proved that $ \ \frac{\partial}{\partial x^{\mu}} \Big|_p \in T_pS^n \ $, $\forall \mu \in \{ 1,2,...,n \} \ $, and that $ \ \mathfrak{B}_{x} = \left\{ \frac{\partial}{\partial x^1} \Big|_p , \frac{\partial}{\partial x^2} \Big|_p , ... , \frac{\partial}{\partial x^n} \Big|_p \right\} \ $ is a basis for $ \, T_pS^n \, $.
Since $ \ p=(p^1,p^2,...,p^n,p^{n+1}) \in S^n \ $, there is some $ \ j \in \{ 1,2,..., n, n+1 \} \ $ such that $ \ p^j \neq 0 \ $. Without loss of generality suppose that $ \ p^j > 0 \ $ (you can work the case $ \ p^j < 0 \ $ as an exercise). Let $ \ V = \big\{ (q^1,...,q^{n+1}) \in S^n : q^j > 0 \big\} \ $, $ \ W = \big\{ (q^1,...,q^{n+1}) \in S^n : q^j < 0 \big\} \ $, $ \ x:V \to \mathbb{R}^n \ $ and $ \ y:W \to \mathbb{R}^n \ $ be such that, $\forall (t^1,...,t^{n+1}) \in V \ $, $ \ x(t^1,...,t^{n+1}) = (t^1,...,t^{j-1},t^{j+1},...,t^{n+1}) \ $, and, $\forall (s^1,...,s^{n+1}) \in W \ $, $ \ y(s^1,...,s^{n+1}) = (s^1,...,s^{j-1},s^{j+1},...,s^{n+1}) \ $. Making a drawing in the case $ \ n=2 \ $ you will see that these functions are projections from the opposite hemispheres to the plane that cut the sphere at the equator. We have that $ \ V,W \in \tau \ $, $ \ x,y \in \mathcal{A} \ $, $ \ p \in V = \text{dom}(x) \ $, $ \ A(p) = -p \in W = \text{dom}(y) \ $, $ \ \text{im}(x) = \text{im}(y) = B^n = \big\{ q \in \mathbb{R}^n : \Vert q \Vert < 1 \big\} \ $ and $ \ x^{-1} : B^n \to V \ $ and $ \ y^{-1} : B^n \to W \ $ are such that, $ \forall (t^1,...,t^n) \in B^n \ $, $$x^{-1} (t^1,...,t^n) = \left( t^1,...,t^{j-1} , \sqrt{1 - \Vert (t^1,...,t^n) \Vert^2} , t^j , t^{j+1},...,t^n \right) \ \ , $$ $$y^{-1} (t^1,...,t^n) = \left( t^1,...,t^{j-1} , - \sqrt{1 - \Vert (t^1,...,t^n) \Vert^2} , t^j , t^{j+1},...,t^n \right) \ \ . $$
Analogously, $ \ \mathfrak{B}_{y} = \left\{ \frac{\partial}{\partial y^1} \Big|_{-p} , \frac{\partial}{\partial y^2} \Big|_{-p} , ... , \frac{\partial}{\partial y^n} \Big|_{-p} \right\} \ $ is a basis for $ \, T_{-p} S^n \, $. To find the matrix associated with the linear transformation $ \ d_pA : T_pS^n \to T_{-p} S^n \ $ with respect to these bases, we write, for each $ \ \mu \in \{ 1,...,n \} \, $, $$d_p A \left( \frac{\partial}{\partial x^{\mu}} \bigg|_p \right) = \sum_{\nu = 1}^n c_{\mu}^{\nu} \cdot \frac{\partial}{\partial y^{\nu}} \bigg|_{-p} \ \ \ , $$ where $ \ c_{\mu}^{\nu} \in \mathbb{R} \ $ will be found, $ \forall \mu , \nu \in \{ 1,...,n \} \ $. Let $ \ x = (x^1,...,x^n) \ $ and $ \ y = (y^1,...,y^n) \ $ be the decomposition of these charts into their coordinate functions and, for each $ \ \mu \in \{ 1,...,n \} \ $, let $ \ \pi^{\mu}: \mathbb{R}^n \to \mathbb{R} \ $ be the projection onto the $\mu$-th coordinate, that is, $ \ \pi^{\mu} (t^1,...,t^n) = t^{\mu} \ $, $\forall (t^1,...,t^n) \in \mathbb{R}^n \ $. Let $ \ \sigma \in \{ 1,...,n \} \ $. The right hand side (rhs) of the above linear combination computed at the germ $ \ [y^{\sigma}]_{-p} \in C^{\infty}_{-p} (S^n) / \! \sim_{-p} \ $ is \begin{eqnarray*} \sum_{\nu=1}^{n} c_{\mu}^{\nu} \cdot \frac{\partial}{\partial y^{\nu}} \bigg|_{-p} \big( [y^{\sigma}]_{-p} \big) & = & \sum_{\nu=1}^{n} c_{\mu}^{\nu} \cdot \partial_{\nu} \big( y^{\sigma} \circ y^{-1} \big) \big( y(-p) \big) \\ & = & \sum_{\nu=1}^{n} c_{\mu}^{\nu} \cdot \partial_{\nu} \big( \pi^{\sigma}|_{B^n} \circ y \circ y^{-1} \big) \big( y(-p) \big) \\ & = & \sum_{\nu=1}^{n} c_{\mu}^{\nu} \cdot \partial_{\nu} \big( \pi^{\sigma}|_{B^n} \circ id_{B^n} \big) \big( y(-p) \big) \\ & = & \sum_{\nu=1}^{n} c_{\mu}^{\nu} \cdot \partial_{\nu} \big( \pi^{\sigma}|_{B^n} \big) \big( y(-p) \big) \\ & = & \sum_{\nu=1}^{n} c_{\mu}^{\nu} \cdot \big[ \big( \pi^{\sigma}|_{B^n} \big)' \big( y(-p) \big) \big] (e_{\nu}) \\ & = & \sum_{\nu=1}^{n} c_{\mu}^{\nu} \cdot \pi^{\sigma} (e_{\nu}) \\ & = & \sum_{\nu=1}^{n} c_{\mu}^{\nu} \cdot \delta^{\sigma}_{\nu} \\ & = & c_{\mu}^{\sigma} \ \ . \end{eqnarray*} While the left hand side (lhs) computed at the same germ is \begin{eqnarray*} \left[ d_p A \left( \frac{\partial}{\partial x^{\mu}} \bigg|_p \right) \right] \big( [y^{\sigma}]_{-p} \big) & = & \frac{\partial}{\partial x^{\mu}} \bigg|_p \big( [ y^{\sigma} \circ A]_p \big) \\ & = & \partial_{\mu} \big( y^{\sigma} \circ A \circ x^{-1} \big) \big( x(p) \big) \\ & = & \partial_{\mu} \big( \pi^{\sigma}|_{B^n} \circ y \circ A \circ x^{-1} \big) \big( x(p) \big) \\ & = & \big[ \big( \pi^{\sigma}|_{B^n} \circ y \circ A \circ x^{-1} \big)' \big( x(p) \big) \big] (e_{\mu}) \\ & = & \Big\{ \Big[ \big( \pi^{\sigma}|_{B^n} \big)' \Big( \big( y \circ A \circ x^{-1} \big) \big( x(p) \big) \Big) \Big] \circ \big[ \big( y \circ A \circ x^{-1} \big)' \big( x(p) \big) \big] \Big\} (e_{\mu}) \\ & = & \big\{ \pi^{\sigma} \circ \big[ \big( y \circ A \circ x^{-1} \big)' \big( x(p) \big) \big] \big\} (e_{\mu}) \\ & = & \pi^{\sigma} \Big( \big[ \big( y \circ A \circ x^{-1} \big)' \big( x(p) \big) \big] (e_{\mu}) \Big) \\ & = & \pi^{\sigma} \Big( \partial_{\mu} \big( y \circ A \circ x^{-1} \big) \big( x(p) \big) \Big) \ \ . \end{eqnarray*} That is the same as \begin{eqnarray*} \left[ d_p A \left( \frac{\partial}{\partial x^{\mu}} \bigg|_p \right) \right] \big( [y^{\sigma}]_{-p} \big) & = & \frac{\partial}{\partial x^{\mu}} \bigg|_p \big( [ y^{\sigma} \circ A]_p \big) \\ & = & \partial_{\mu} \big( y^{\sigma} \circ A \circ x^{-1} \big) \big( x(p) \big) \\ & = & \partial_{\mu} \big( \pi^{\sigma}|_{B^n} \circ y \circ A \circ x^{-1} \big) \big( x(p) \big) \\ & = & \partial_{\mu} \big( y \circ A \circ x^{-1} \big)^{\sigma} \big( x(p) \big) \ \ . \end{eqnarray*} Thus $ \ c_{\mu}^{\nu} = \partial_{\mu} \big( y \circ A \circ x^{-1} \big)^{\nu} \big( x(p) \big) = \pi^{\nu} \Big( \partial_{\mu} \big( y \circ A \circ x^{-1} \big) \big( x(p) \big) \Big) \ $, $\forall \mu , \nu \in \{ 1,...,n \} \ $. Since the matrix $ \ \big[ d_p A \big]_{\mathfrak{B}_x}^{\mathfrak{B}_y} = \big[ c_{\mu}^{\nu} \big]_{n \times n} \ $, we have that this matrix equals the Jacobian matrix $ \, J_{y \circ A \circ x^{-1}} \big( x(p) \big) \, $ of the function $ \ y \circ A \circ x^{-1} : B^n \to B^n \ $ at the point $ \, x(p) \, $.
For all $ \ q = (q^1,...,q^n) \in B^n \ $ we have \begin{eqnarray*} (y \circ A \circ x^{-1}) (q^1,...,q^n) & = & y \Big( A \big( x^{-1} (q^1,...,q^n) \big) \Big) \\ & = & y \Big( A \big( q^1,...,q^{j-1}, \sqrt{1 - \Vert (q^1,...,q^n) \Vert^2 } , q^j,...,q^n \big) \Big) \\ & = & y \big( - q^1,...,- q^{j-1}, - \sqrt{ 1 - \Vert (q^1,...,q^n) \Vert^2 } , - q^j,...,- q^n \big) \\ & = & ( - q^1,..., - q^{j-1} , - q^j,..., - q^n ) \\ & = & - (q^1,...,q^n) \\ & = & - q \ \ .  \end{eqnarray*} Hence $ \ y \circ A \circ x^{-1} = - id_{B^n} \ $. A quick calculation gives $ \ \pi^{\nu} \Big( \partial_{\mu} \big( y \circ A \circ x^{-1} \big) \big( x(p) \big) \Big) = - \delta_{\mu}^{\nu} \ $, $\forall \mu , \nu \in \{ 1,...,n \} \ $. We are left with
$$\big[ d_p A \big]_{\mathfrak{B}_x}^{\mathfrak{B}_y} =  - Id_n = \left[ \begin{array}{cccc} -1 & 0 & \cdots & 0 \\ 0 & -1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & -1 \end{array} \right]_{n \times n} \ \ \ \ . $$
