Consider the linear transformation $\ltdefn{T}{\complex{3}}{M_{22}}$ defined by \begin{equation*} \lt{T}{\colvector{a\\b\\c}}=\begin{bmatrix} a-b&2a+2b+c\\3a+b+c&-2a-6b-2c \end{bmatrix} \end{equation*}
Suppose we were to search for an inverse function $\ltdefn{S}{M_{22}}{\complex{3}}$.
First verify that the $2\times 2$ matrix $A=\begin{bmatrix} 5&3\\8&2 \end{bmatrix} $ is not in the range of $T$. This will amount to finding an input to $T$, $\colvector{a\\b\\c}$, such that \begin{align*} a-b&=5\\ 2a+2b+c&=3\\ 3a+b+c&=8\\ -2a-6b-2c&=2 \end{align*}
As this system of equations is inconsistent, there is no input column vector, and $A\not\in\rng{T}$. How should we define $\lt{S}{A}$? Note that \begin{equation*} \lt{T}{\lt{S}{A}}=\lt{\left(\compose{T}{S}\right)}{A}= \lt{I_{M_{22}}}{A}=A \end{equation*}
So any definition we would provide for $\lt{S}{A}$ must then be a column vector that $T$ sends to $A$ and we would have $A\in\rng{T}$, contrary to the definition of $T$. This is enough to see that there is no function $S$ that will allow us to conclude that $T$ is invertible, since we cannot provide a consistent definition for $\lt{S}{A}$ if we assume $T$ is invertible.
Even though we now know that $T$ is not invertible, let us not leave this example just yet. Check that \begin{align*} \lt{T}{\colvector{1\\-2\\4}}&=\begin{bmatrix}3&2\\5&2\end{bmatrix}=B& \lt{T}{\colvector{0\\-3\\8}}&=\begin{bmatrix}3&2\\5&2\end{bmatrix}=B \end{align*}
How would we define $\lt{S}{B}$? \begin{align*} \lt{S}{B}&=\lt{S}{\lt{T}{\colvector{1\\-2\\4}}} =\lt{\left(\compose{S}{T}\right)}{\colvector{1\\-2\\4}} =\lt{I_{\complex{3}}}{\colvector{1\\-2\\4}}=\colvector{1\\-2\\4} \end{align*} or \begin{align*} \lt{S}{B}&=\lt{S}{\lt{T}{\colvector{0\\-3\\8}}} =\lt{\left(\compose{S}{T}\right)}{\colvector{0\\-3\\8}} =\lt{I_{\complex{3}}}{\colvector{0\\-3\\8}}=\colvector{0\\-3\\8} \end{align*}
Which definition should we provide for $\lt{S}{B}$? Both are necessary. But then $S$ is not a function. So we have a second reason to know that there is no function $S$ that will allow us to conclude that $T$ is invertible. It happens that there are infinitely many column vectors that $S$ would have to take to $B$. Construct the kernel of $T$, \begin{equation*} \krn{T}=\spn{\set{\colvector{-1\\-1\\4}}} \end{equation*}
Now choose either of the two inputs used above for $T$ and add to it a scalar multiple of the basis vector for the kernel of $T$. For example, \begin{equation*} \vect{x}=\colvector{1\\-2\\4}+(-2)\colvector{-1\\-1\\4}=\colvector{3\\0\\-4} \end{equation*} then verify that $\lt{T}{\vect{x}}=B$. Practice creating a few more inputs for $T$ that would be sent to $B$, and see why it is hopeless to think that we could ever provide a reasonable definition for $\lt{S}{B}$! There is a “whole subspace's worth” of values that $\lt{S}{B}$ would have to take on.