= Solution
Use the <matrices> of <1/c/solution>. Their projective classes satisfy $[-M]=[M]$, since $-I_2\in N$. The <group homomorphism> is therefore explicitly
$$
\boxed{\widetilde R(\pm I_2)=[I_2],\quad
\widetilde R(\pm A)=[A],\quad
\widetilde R(\pm B)=[B],\quad
\widetilde R(\pm C)=[C].}
$$
These four classes are distinct: the two diagonal representatives are not scalar multiples of one another, the two off-diagonal representatives are not scalar multiples of one another, and a diagonal representative cannot be a nonzero scalar multiple of an off-diagonal one.
Although $A$ and $B$ anticommute as <matrices>, their projective classes commute:
$$
[A][B]=[AB]=[-BA]=[BA]=[B][A].
$$
Furthermore $[A]^2=[B]^2=[I_2]$ and $[C]^2=[-I_2]=[I_2]$. Thus every nonidentity element of the image has order two. An explicit <group isomorphism> to the <Klein four-group> is
$$
\boxed{\operatorname{Im}\widetilde R\longrightarrow\mathbb Z_2\times\mathbb Z_2,
\qquad [A^aB^b]\longmapsto(a,b),\qquad a,b\in\{0,1\}.}
$$
Indeed reordering products introduces only a sign, which disappears in the <projective linear group>, so this map turns multiplication into addition modulo two in each coordinate. Equivalently, $\ker\widetilde R=G\cap N=\{\pm I_2\}$ and the <first isomorphism theorem> gives
$$
G/\{\pm I_2\}\cong\operatorname{Im}\widetilde R\cong\mathbb Z_2\times\mathbb Z_2.
$$
This <projective image of the square symmetry group> illustrates how a noncommutative <group> can have an <Abelian> projective image: its scalar commutator becomes the identity class.
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