let $L:V \to U,$ and $M:U \to W$ be linear mappings if $M\circ L$ is one-to-one then $L$ is one-to-one

Question

a problem asks me to prove or disprove the following statement let $L:V \to U,$ and $M:U \to W$ be linear mappings if $M\circ L$ is one-to-one then $L$ is one-to-one

This is my attempt at proving the statement let $\vec{x} \in ker(L)$ then $M\circ L(\vec{x})=M(L(\vec{x}))=M(\vec{0})=0$ thus $\vec{x} \in ker(M\circ L)$

Answer 1

Unless I'm mistaken, there's a much simpler way to do this. Just take some $x, x'\in V$ that are not equal. Suppose that $L$ is not injective. Then $M \circ L(x)\neq M\circ L(x')$. Clearly, due to $M$ being well-defined, it follows that $L(x)\neq L(x' )$, and due to $M\circ L$ being injective, this holds for all pairs of points $x, x'\in V\Rightarrow L$ is injective. In fact, it's more than this and it holds for more than linear maps; you can check out Eric Towers' answer.

Answer 2

It's true that $\ker (M\circ L)\supseteq \ker L$, as you showed correctly. In general it is not true that $\ker (M\circ L)=\ker L$, which is a surprisingly common characteristic of sentences that are introduced by the phrase "Similarly, I can show that...".