Given $T(x,y,z) = (xy, yz, xz)$, determine whether $T$ is one to one and/or onto.
So far, I have come up with a contradiction that proves $T$ is not one to one:
$T(-1, 1, -1) = (-1, -1, 1)$ and, $T(1, -1, 1) = (-1, -1, 1)$. However, I can't figure out whether or not $T$ is onto... I think it is, but I don't know the reaosn why or how to prove it.
Can I say:
Taking arbitrary elements a,b,c $\in \mathbb R$, $T(a,b,c) = (ab, bc, ac)$, and therefore any element $x,y,z \in \mathbb R$ can be generated by $x=ab$, $y = bc$, and $z=ac$?
Assuming both the domain and codomain are $\mathbb{R}^3$.
Suppose on the contrary that it is onto, then $(-1,-1,-1)$ has a preimage.
Suppose $$-1 =ab, -1=bc, -1=ac$$
then we have $a = -\frac{1}b, c= -\frac{1}{b}$
and we have $ac= \frac1{b^2}>0$ which contracts $ac=-1<0$.