I have two simple linear maps:
$F:\mathbb{R_{≤ 3}[x]} \to \mathbb{R^{2,2}} \ ; \ ax^3+bx^2+cx+d \to \left[ \begin{array}{cc} c-d& b+c \\ a-c & b \\ \end{array}\right],$
$G:\mathbb{R^{2,2}} \to \mathbb{R_{≤ 1}[x]} \ ;\left[ \begin{array}{cc} a& b \\ c & d \\ \end{array}\right] \to \ (a+b+2c)x-2d $.
I have proven that $F$ is invertible:
$\begin{pmatrix}r&s\\ t&u\end{pmatrix}=:\begin{pmatrix}c-d&b+c\\ a-c&b\end{pmatrix}$
So we get: $a = s + t -u$, $b = u$, $c = s - u$ and $d = s-r-u$
$F^{-1}\left(\begin{pmatrix}r&s\\ t&u\end{pmatrix}\right)=\left(s+t-u\right)x^3+ux^2+\left(s-u\right)x+\left(s-r-u\right)$
My question is, why aren't $G$ and $G \circ F$ invertible?
Because $G$ and $G\circ F$ are linear maps from a $4$-dimensional space into a $2$-dimensional space. Such a map is never injective (it follows from the rank-nullity theorem, for instance), and therefore it is never an isomorphism.