Suppose there are linear operators $T$ and $U$ on $V$ such that $TU = I$. Prove that $T$ is invertible and $U = T^{-1}$. Let $T$ be a linear operator on the finite-dimensional space $V$. Suppose there is a linear operator $U$ on $V$ such that $TU = I$. Prove that $T$ is invertible and $U = T^{-1}$. Given an example which shows that this is false when $V$ is not finite-dimensional.
ATTEMPT
Given two functions $f:X\rightarrow Y$ and $g:Y\rightarrow X$, if the function $g\circ f$ is bijective, then $g$ is surjective. Indeed, if $g\circ f$ is bijective, one has
\begin{align*}
X = i(X) = (g\circ f)(X) = g(f(X))\subset g(Y)
\end{align*}
Since $g(Y)\subset X$, one concludes that $g(Y) = X$, whence $g$ is surjective.
In the present case, $f = U$, $g = T$ and $X = Y = V$. Consequently, $T$ is surjective. Thus $T$ is invertible, as desired.
Therefore $T^{1}(TU) = (T^{-1}T)U = U = T^{-1}I = T^{-1}$.
Can someone provide an alternative solution to the first part and help me to solve the second?
 A: An alternative solution to the first part:
Claim : If $\textsf{TU} = I$ then $\textsf{T}$ is surjective. Consequently, $\textsf{T}$ is invertible.
Proof. Take any $v\in V$ and note that we want to find some vector $w\in V$ such that $v = \textsf{T}(w)$. Well, since
$$v = I(v) = (\textsf{TU})(v) = \textsf{T}(\textsf{U}(v))$$
choose $w := \textsf{U}(v)\in V$. $\blacksquare$
And, for the second part take $V$ equal the set of all polynomial functions, $\textsf{T}\in\mathcal{L}(V)$ the differential operator (i.e. the function that maps $f$ into $f'$) and $\textsf{U}\in\mathcal{L}(V)$ the function that takes $f$ into $F$, where
$$F(x) = \int_0^x f(t)dt.$$
The Fundamental Theorem of Calculus tell us that $\textsf{TU} = I$, but it is not true that $\textsf{T}$ is invertible, since $\textsf{T}$ is not one-to-one: $\textsf{T}$ maps every constant polynomial into $0$.  
A: For the second question, you can also consider the right shift and left shift maps on the vector space of sequences of reals converging to 0.
