What will happen if we remove the hypothesis that $V$ is finite-dimensional in this problem Original problem:
Suppose $V$ is finite-dimensional and $S, T,U ∈L(V)$and $STU = I$. Show that T is invertible and that $inv(T) = US$.
I know that it is because of the hypothesis of finite-dimensional that we can take advantage of the basis or spanning list of vectors. But I'm not quite clear of the power of "Finite-Dimensional" in linear algebra.
Please help me with some examples and insights.
Really really appreciate it.

This is definitely false in infinite dimensions.
Let $V$ be an vector space of countably infinite dimension, and pick a basis $\{v_n\}_{n \in \mathbb{N}}$ for $V$.
Let $T$ be the be the map which sends $v_1 \mapsto 0$ and $v_n \mapsto v_n$ for $n\geq 2$.
Let $U$ be the map which sends $v_n \mapsto v_{n+1}$.
Let $S$ be the map which sends $v_1 \mapsto 0$ and $v_n \mapsto v_{n-1}$ for $n \geq 2$.
Then $STU=I_V$ but $T$ is not invertible since nothing maps to $v_1$ under $T$.
