My linear algebra textbook by Friedberg states the following Theorem:
Theorem: Let $V$ and $W$ be vector spaces, and let $T: V → W$ be linear. Then $T$ is one-to-one if and only if $N(T) =$ {$0$}.
In one of the exercises, it asks whether the following is true or false:
$T$ is one-to-one if and only if the only vector $x$ such that $T(x) = 0$ is $x = 0$ .
The correct answer is that this statement is false. However, I don't understand how these two are not equivalent.
In that exercise Friedberg assume $T$ is just a function, so this statement is false , for example, take $f(x)=x^2$ on $\Bbb{R}$. Here the only number $x$ such that $f(x)=0$ is $x=0$ but $f$ is not one-one!
So the mentioned result is true only when $T$ is linear (of course , Friedberg assumes $V$ and $W$ are finite dimensional)