Components of the sets of idempotents and involutions

Part 1: idempotents.

*Step 1: rank = trace.* If $P^2=P$, then $P$ is conjugate to $E_k=\operatorname{diag}(I_k,0)$ with $k=\operatorname{rank}P$ (Problem 15), so $\operatorname{tr}P=k$. On the set $\mathcal{P}$ of idempotents, $\operatorname{tr}$ is continuous and integer-valued with values in $\{0,\dots,n\}$.

*Step 2: the pieces are clopen.* $\mathcal{P}_k=\mathcal{P}\cap\operatorname{tr}^{-1}(\{k\})$ is closed and open in $\mathcal{P}$ (as the preimage of an isolated value under a continuous map to $\mathbb{Z}$), and $\mathcal{P}=\bigsqcup_{k=0}^n\mathcal{P}_k$, all non-empty.

*Step 3: each piece is connected.* $\mathcal{P}_0=\{0\}$, $\mathcal{P}_n=\{I\}$, and for $0<k<n$, $\mathcal{P}_k$ is path connected (Problem 15).

Hence $\mathcal{P}$ has exactly $n+1$ components.

Part 2: involutions.

*Step 1: diagonalisation.* If $A^2=I$ the minimal polynomial divides $x^2-1=(x-1)(x+1)$, which has simple real roots, so $A$ is diagonalisable over $\mathbb{R}$ with eigenvalues $\pm1$. If $p$ is the multiplicity of $+1$, $A$ is conjugate to $F_p=\operatorname{diag}(I_p,-I_{n-p})$ and $\operatorname{tr}A=2p-n$. So $\operatorname{tr}$ takes the $n+1$ values $-n,-n+2,\dots,n$ on $\mathcal{I}=\{A^2=I\}$, and as above the $n+1$ pieces $\mathcal{I}_p=\{A\in\mathcal{I}:\operatorname{tr}A=2p-n\}$ are clopen in $\mathcal{I}$.

*Step 2: each piece is connected.* $\mathcal{I}_0=\{-I\}$ and $\mathcal{I}_n=\{I\}$. For $0<p<n$, $\mathcal{I}_p=\{QF_pQ^{-1}:Q\in GL_n\}$. The matrix $D=\operatorname{diag}(-1,1,\dots,1)$ is diagonal so it commutes with $F_p$, and $\det D=-1$. As in Problem 15, replacing $Q$ by $QD$ we may take $\det Q>0$. So $\mathcal{I}_p$ is the continuous image of the path connected $GL_n^+$.

Therefore $\mathcal{I}$ has exactly $n+1$ components.

Remark. In both cases the invariant (rank, resp. trace) is *discrete*, which is what splits the sets into pieces; the dimension of $\mathcal{P}_k$ is $2k(n-k)$, and of $\mathcal{I}_p$ also $2p(n-p)$.