Show that the result in the previous exercise can fail without the hypothesis that $V$ is finite-dimensional.

Suppose $STU = I$ and suppose $V$ is infinite-dimensional. We basically want a counterexample to 3.69 since that's the core of the argument in 3d/11.

Consider the differentiation map $D \in \mathcal L(\mathcal P(\mathbf F))$. It's surjective but not injective, thus it's a counterexample to 3.69.

Let $S = D$, it's impossible for $T$ to be invertible with $T^{-1}= US$ because $TUS \ne I$ since constants get mapped to zero by $S$. This shows the previous example fails when $S$ is not injective.

Another failure case is where $U$ is injective but not surjective, since then we could an element not in the range of $U$ which again would show $TUS \ne I$. An example of a $U$ like this would be the integration map $Up = \int_0^x p(t)dt$ over $\mathcal P(\mathbf F)$ (note $\int_0^x p(t)dt \in \mathcal P(\mathbf F)$.)