Compact onto linear operator from Banach space imply space is finite-dimensional

Question

I am trying to prove this question, which from the book Foundations of Modern Analysis by Avner Friedman pp189 5.1.7. the question is below.

Let T be a compact linear operator from a Banach space X onto itself. If $\mathbf{T^{-1}}$ is a bounded operator, then X is finite-dimensional.

I think this question is not right, because if T is a compact operator then T is a bounded operator and due to the open-mapping theorem, $\mathbf{T^{-1}}$ must be bounded. It seems that this question just tells us that T is one-one and onto.

Answer 1

Two key facts here.

$(1)$ The compact operators form an ideal.

$(2)$ The identity operator on a space $X$ is compact if and only if $X$ has finite-dimension.

By your assumption, $T \circ T^{-1}= 1_X$ and $T$ is compact. Because of $(1)$, we see that $1_X$ is compact. By $(2)$, $X$ has finite-dimension.