$L_1, L_2$ are two linear maps with $L_1 : \mathbb{R}^3 \to \mathbb{R}^2,\; L_2 :\mathbb{R}^2 \to \mathbb{R}^3$ and be $L_2 \circ L_1 : \mathbb{R}^3\to \mathbb{R}^3$ their concatenation.
Can $L_2\circ L_1$ be injective?
Can $L_2\circ L_1$ be surjective?
I found out that $L_2\circ L_1$ can be injective and surjective if and only if $L_2,\;L_2$ is bijective. That means, that
$\forall y\in\mathbb{R}^2$ exists only one $x\in\mathbb{R}^3:L_1(x)=y$ and
$\forall y'\in\mathbb{R}^3$ exists only one $x'\in\mathbb{R}^2:L_2(x')=y'$
Let
$L_2(y)=x',\;L_1(x)=y$ with $x,x'\in\mathbb{R}^3,\;y\in \mathbb{R}^2$ and $L_2\circ L_1=L_2(L_1(x))$
Then $L_2\circ L_1=L_2(L_1(x))=L_2(y)=x'$ and because $L_1$ bijective, it reaches all $y\in \mathbb{R}^2$ and because $L_2$ bijective, we can reach all $x'$. Therefore, $L_2 \circ L_1$ is bijective.
Is this proof correct or am I missing something?
EDIT: $L_2\circ L_1$: First, we need to prove, that $L_1$ injective:
In order to be injective, $\dim(\operatorname{ker}(L_1))=1$ must hold.
$\dim(\mathbb{R}^3)=\dim(\operatorname{im}(L_1))+\dim(\operatorname{ker}(L_1))$.
Because $\dim(\mathbb{R}^2)=2 \implies \dim(\operatorname{im}(L_1))\leq2\implies \dim(\operatorname{ker}(L_1))\geq1$ $\implies \exists x_1,x_2\in L_1:L_1(x_1)=L_1(x_2)$ with $ x_1\neq x_2$ $\implies$ $L_1$ not injective and $L_2\circ L_1$ not injective
Second $L_2\circ L_1$ surjective? Need to prove, that $L_2$ surjective:
$\dim(\mathbb{R}^3)=\dim(\operatorname{im}(L_2))+\dim(\operatorname{ker}(L_2))$
$\dim(R^2)=2$ and $\dim(R^3)=3 \implies \dim(im(L_2))\leq 3$ Therefore, $\dim(ker(L_2))\geq -1$
What means, that $\dim(ker(L_2))=-1$?
A linear map $$T:\mathbb{R}^3 \to \mathbb{R}^2,$$
can never be injective. Best case scenario it's a projection into a the plane.
A linear map
$$T:\mathbb{R}^2 \to \mathbb{R}^3,$$
can never be surjective. Best case: you're embedding a plane (through the origin) into $3$-D space.