Prove or disprove: $\ker (TS)=\{0\}$ $\implies$ $\ker (S)=\{0\}$

Question

Prove or disprove: $\ker (TS)=\{0\}$ $\implies$ $\ker (S)=\{0\}$.

$\bullet~$ $\textbf{My attempt:}$ (not necessarily) we will define

$$S:\mathbb{R}^2 \rightarrow \mathbb{R}^2 ~\text{ and }~ T:\mathbb{R}^2 \rightarrow\mathbb{R}^2$$ from the given

$$\ker(TS)=\{0\} \implies T\circ S(\vec v)=\vec 0$$

let's say

$S(v)=Av~$ where $A = [1, 0 ,0 ,0]~$ and $~v = (0,1) ~$ then $Av=\matrix [1, 0 ,0 ,0] \cdot (0,1)$ =$~\vec 0$ $$T\circ S(\vec {0,1} )=T(S(0,1))=T(0)=0 .$$ Thus, $\ker (TS)=\{0\} \not\Rightarrow\ker (S)=\{0\}.$

Answer 1

If $g\circ f$ is injective, then $f$ is necessarily injective. A linear map is injective if and only if its kernel is trivial. This also proves implication in title. For linear maps, it suffices to note that $$\{0\}\subseteq\mathrm{Ker}(S) \subseteq \mathrm{Ker}(TS) = \{0\}.$$