Let T(x,y)=(x,0)T(x, y) = (x, 0)T(x,y)=(x,0). Is TTT one-to-one?
Yes, because it is linear.
No, because T(0,1)=(0,0)=T(0,0)T(0, 1) = (0, 0) = T(0, 0)T(0,1)=(0,0)=T(0,0).
Yes, because its range is R1\mathbb{R}^1R1.
No, because it is a projection.