Connected components of the projective linear group

Idea. $\det(\lambda A)=\lambda^n\det A$: the sign of the determinant is either destroyed or preserved by scaling, according to the parity of $n$.

Connected images. Let $q:GL_n\to PGL_n$ be the quotient map. It is continuous and onto, so images of connected sets are connected. We know $GL_n^+$ and $GL_n^-$ are connected (polar decomposition).

$n$ odd. Take $A\in GL_n^-$. Then $-A$ has $\det(-A)=(-1)^n\det A=-\det A>0$, so $-A\in GL_n^+$ and $q(A)=q(-A)\in q(GL_n^+)$. Hence $q(GL_n)=q(GL_n^+)$, a continuous image of the connected set $GL_n^+$: $PGL_n$ is connected.

$n$ even. Now $\det(\lambda A)=\lambda^n\det A$ has the same sign as $\det A$ for every $\lambda\ne0$. So $\mathrm{sgn}\det$ is constant on each scalar class and induces a well-defined continuous map $\overline{\mathrm{sgn}\det}:PGL_n\to\{\pm1\}$. It is onto (e.g. $I$ and $\mathrm{diag}(-1,1,\dots,1)$), so $PGL_n$ is disconnected. Moreover $PGL_n=q(GL_n^+)\cup q(GL_n^-)$ is a union of two connected sets, which are the two (disjoint, since the sign differs) fibres of $\overline{\mathrm{sgn}\det}$. Hence there are exactly two components.

Check. $n=1$: $PGL_1$ is a point. $n=2$: $PGL_2(\mathbb{R})\supset PSL_2(\mathbb{R})$ with index 2 and has two components.

Remark. For odd $n$, $PGL_n(\mathbb{R})\cong SL_n(\mathbb{R})$ (every class has a unique representative of determinant $1$, namely the real $n$-th root of $1/\det A$ times $A$).