Based on this question:Spectrum of a projection
Let $X$ be a banach space and $P\in B(X)$ a projection.
Show that $\sigma(P) \subset \{0,1\}$.
I understand the answer that
If $\lambda$ is neither $0$ nor $1$ then from algebraic calculation (using only the fact that $P^2 = P$) we have
$$
\begin{align*}
(\lambda I – P)\left(\frac{1}{\lambda} I – \frac{1}{\lambda(1-\lambda)} P\right ) & = I – \frac{1}{(1-\lambda)} P – P\left(\frac{1}{\lambda} I – \frac{1}{\lambda(1-\lambda)} P\right )\\
& = I – \frac{1}{(1-\lambda)} P – \frac{1}{\lambda} P + \frac{1}{\lambda(1-\lambda)} P \\
& = I,
\end{align*}
$$
A similar computation (or just the observation that $I$ and $P$ commute with $I$ and $P$) shows that the same operator is a left inverse of $\lambda I – P$, so that $\lambda I – P$ is invertible for all $\lambda \not \in \{0, 1\}$.
But my question is that our definition of invertible is as follows
A bounded linear operator $T: E\to F$ between normed spaces $E, F$ is called invertible if $T$ is bijective and $T^{-1}$ is bounded.
I can show that if $T_\lambda:=\lambda I – P$ has left inverse then $T_\lambda$ is injective. Let $S:=\left(\frac{1}{\lambda} I – \frac{1}{\lambda(1-\lambda)} P\right )$. Because for $\forall x, y\in X$, $ST_\lambda(x)=ST_\lambda(y)$ implies that $x=y$.
Also, for any $y\in X$, there exists $x=S(y)$ so that
$$
T_\lambda (S (y))=y
$$
So $T_\lambda$ is surjective. [Question: does it make sense?]
Moreover, for $\|x\|\le 1$,
$$
\|Sx\|=\|\frac{1}{\lambda} x – \frac{1}{\lambda(1-\lambda)} Px \|\le |\lambda^{-1}|\|x\|+|\lambda^{-1}(\lambda-1)^{-1}\|Px\|
$$
then it is bounded.
Best Answer
Let assume $T_\lambda$ is a linear bijective map. Then by open mapping theorem and the fact that $X$ is a Banach space, $T_\lambda^{-1}$ is continuous, thus it is bounded. Therefore it is enough to give $T_\lambda^{-1}$ explicitly, as shown in the link you attached.
What you wrote in the post is the proof that $T$ is invertible linear map then T is bijective. Here by "invertible" I mean there exists a right inverse and left inverse.