prove or disprove that If $M \circ L$ is one-to-one then $M$ is one-to-one

Question

Im trying to prove or disprove that if $L:V \to U,$ and $M:U \to W$ are linear mappings and if $M\circ L$ is one-to-one then $M$ is one-to-one

I have already proved that $L$ is one-to-one but intuitively to me it seems that $M$ does not have to be one-to-one due to the fact that there is no way of knowing if $L$ is onto

Answer 1

You're right, there's no reason for $M$ to be one-to-one. Let $V=W=\mathbb{R}$, and let $U= \mathbb{R}^2$. Let $L:V \to U$ be the inclusion onto the left factor, and let $M:U \to W$ be the projection $(u,v) \mapsto u$. Then the composition $L \circ M$ is the identity, but $dim(ker(M))=1$

Answer 2

The application $L : \mathbb{R} \rightarrow \mathbb{R}^2$ and $M : \mathbb{R}^2 \rightarrow \mathbb{R}$ defined by $$L(x)=(x,0) \quad \text{ and } \quad M(x,y)=x$$

satisfy that $M \circ L = \mathrm{Id}$ is one-to-one, but $M$ is not.

Answer 3

Let $L:\mathbb{R}^2 \to \mathbb{R}^3$ be given by $$L\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}x\\y\\x+y\end{bmatrix}$$ and Let $M:\mathbb{R}^3 \to \mathbb{R}^2$ be given by $$M\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}x\\y\end{bmatrix}$$ Then Let $M \circ L:\mathbb{R}^2 \to \mathbb{R}^2$ be given by $$M\circ L\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}x\\y\end{bmatrix}$$ is one-one but $M$ is not.