I'm given an exercise, in a differential geometry class, where I need to detemine wether or not the smooth map between manifolds: \begin{align} f \colon\ &SO(n) \rightarrow SO(n)\\ & A \mapsto A^2 \end{align}
Is homotopic to the identity map for $n\geq 2$.
I tried, for $n=2$, saying that a general element of $SO(2)$ is described by
\begin{equation} \begin{bmatrix} \cos(\theta) & \sin(\theta) \\ -\sin(\theta) & \cos(\theta) \end{bmatrix}\end{equation}
and thus \begin{equation} F(t,\theta):= \begin{bmatrix} \cos(t\theta) & \sin(t\theta) \\ -\sin(t\theta) & \cos(t\theta) \end{bmatrix}\end{equation}
is an homotopy, but I've been told I'm wrong.
Please, could someone tell me why that isn't an homotopy? Maybe computing the degee or, even better, the lefschetz number would help me to prove there's (or there isn't) homotopy, but I can't undestand how to work with $SO(n)$, could someone, please, give some hints? Thank you in advance!
Although it's true that each element of $SO(2)$ can be represented in that form for some $\theta$, it's not possible to do so in such a way that $\theta$ depends continuously on the matrix. (Think about what happens to $\theta$ when you go once around the circle.) Therefore, your formula for $F$ does not define a continuous function from $SO(2)\times [0,1]$ to $SO(2)$.