If $T\in L(X,X)$ where $X$ is a Banach space and $L(X,X)$ is denoted as the space of bounded linear maps, and $\|I-T\|<1$ where $I$ is the identity operator, then $T$ is invertible?
Here invertible means $T$ is bijective and $T^{-1}$ is bounded. I can prove $T$ is injective but fail to show $T$ is surjective. In fact, we can find $T^{-1}=\sum_0^\infty(I-T)^n$, which can help prove the surjection. But without knowing this, how to prove? Thanks in advance.
Proof: $X^{n+1}-I=(X-I)(I+X+X^2+...+X^n)$. Passing the limit, where we know the right side converges due to the fact that $X$ is a Banach space, together with the fact that the series is absolutely convergent (due to comparison with geometric series), and we know the left side converges to $-I$ since $\Vert X^n \Vert \rightarrow 0$, we have that $$-I=(X-I) \sum X^n$$ $$\implies I=(I-X) \sum X^n.$$
The above proof can also be easily adapted to show that $ I=(\sum X^n) (I-X) $, and we have the result. $\blacksquare$
Now to your proposition. Since $\Vert I-T \Vert <1$, we have that $I-(I-T)=T$ is invertible. The result now follows completely from the open mapping theorem.
OBS: I did not know what you meant by "without knowing this", and neither exactly what "this" was. If my answer does not address your question, please clarify.