Is absolute value map injective? Theorem: A linear transformation is injective if and only if its nullity is trivial
Let $V$ be vector space and $T:V\to \mathbb{R}$
Let $x\in V$
$T(x) = | x |$
$T(x) = 0$ if $x = 0$
By the theorem, $T$ is injective. How?
$T(x) = 2$ if $x = 2, -2 $
(I may have messed up the conditions, for example, maybe it was $T:V\to W$. I don't remember the exact question, sorry.)
 A: The theorem is valid for linear transformations, the absolute value function (also known as norm) may not be a linear transformation.
A: You have chosen your "linear transformation" (in the statement of the theorem) to be the map $x\mapsto\lvert x\rvert,$ but you haven't checked that this is actually a linear transformation.
Indeed, if $T\colon\mathbb{R}\to\mathbb{R}$ is intended, then $x\mapsto\lvert x\rvert$ is not a linear map (for example, $\lvert -1 + 2\rvert = 1 \neq 3 = \lvert -1\rvert +\lvert 2\rvert$), so, although the theorem is true, it simply doesn't apply to your choice of $T,$ because your choice of $T$ doesn't satisfy the given conditions.
[If, as others have astutely pointed out, you take $T\colon\{0\}\to\mathbb{R}$ instead, then you do get a linear map, but this map is obviously injective.]
Try something like $T_{1}\colon\mathbb{R}\to\mathbb{R},x\mapsto2x$ or $T_{2}\colon\mathbb{R}^{2}\to\mathbb{R}^{2},\begin{pmatrix}x \\ y\end{pmatrix}\mapsto\begin{pmatrix}2x \\ x+2y\end{pmatrix}$ instead. In general, if you want examples of linear maps, you should think about multiplication by matrices.
In the case of $T_{2},$ I picked the matrix $\begin{pmatrix} 2 & 0 \\ 1 & 2\end{pmatrix}$ (written under the standard choices of bases). This matrix has determinant $4,$ which is non-zero, so we see that $T_{2}$ must therefore be invertible, and hence injective. I'll leave it to you to check that its kernel is trivial, as the theorem says it should be.
