Let $A ∈ SO(n+1)$ and write $τ(A)$ for the induced map $S^n → S^n$. Prove that $τ(A)\simeq id$ .
My attempt:
So $\tau(A)$ is the map $v \mapsto Av$. So have to construct a map $H:S^n \times [0,1] \to S^n$.
Since, $S^n$ is path-connected, thus for a fixed $A \in SO(n+1)$, and given $v \in S^n$ , we always have a map $\gamma_v : [0,1] \to S^n$, with $\gamma_v(0)=v$ and $\gamma_v(1)=Av$ and was tempted to define, $H(v,t):= \gamma_v (t)$, but realized that well-definedness and continuity both are issues!
Tried to modify by picking $\gamma_v$ to be the geodesic joining $v$ and $Av$. But still problem with Continuity and in fact, well-definedness while $v$ and $Av$ are antipodal points! (note that this is always possible since $SO(n+1)$ acts on $S^n$ transitively!)
But got the feeling, that there is some simpler solution looming.
Thanks in Advance for help!
A path $\gamma\colon[0,1]\to SO(n+1)$ such that $\gamma(0)=A$ and $\gamma(1)=I$ will induce a homotopy $\tau\circ\gamma\colon [0,1]\mapsto\text{Maps}(S^n,S^n)$ between $\tau(A)$ and $\tau(I)=id$. So it suffices to show $SO(n+1)$ is path-connected.
Now there are many ways we can do this. One way is to use the fact unitary matrices are diagonalisable over $\mathbb{C}$ to yield $A$ is conjugate to a block-diagonal matrix $$ \begin{bmatrix} I_p\\ &-I_{2q}\\ &&R_{\theta_1}\\ &&&\ddots\\ &&&&R_{\theta_k} \end{bmatrix} $$ where $R_\theta$ is a $2\times 2$ rotation block $\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}$, $\theta_i\notin\pi\mathbb{Z}$. It is easy to construct a path from here to the identity in $SO(n+1)$.