Alternative proof for Fredholm's alternative I am trying to prove the following variant of Fredholm's alternative:

Let $X$ be a Banach space, $T \in GL(X)$ (invertible) and $A \in K(X)$
(compact operator). Prove that $T+A$ is invertible iff $T+A$ is
injective.

The invertible $\implies$ injective is immediate, but the other direction I am quite stuck on. I thought of using the regular form of Fredholm's alternative to prove this, but was unsuccessful. I then thought of using the same steps used in the actual proof for the original Fredholm's theorem, with some alterations, but that would be like a repeat of the original proof which seems redundant.
Is there a simpler way I am missing? (note - I am at an elementary level in Functional Analysis, so preferably this wouldn't require any advanced theorems).
Thanks.
 A: $\newcommand{\codim}{\operatorname{codim}}\newcommand{\im}{\operatorname{im}}$Under the given conditions, it is known that $T+A$ is Fredholm of index $0$, that is, $\dim\ker(T+A)=\codim\im(T+A)$. Suppose $T+A$ is injective: then $\dim\ker(T+A)=0$ hence $\codim\im(T+A)=0$. That is, $\im(T+A)=X$ and $T+A$ thus bijects.
Whether or not this counts as simple depends on whether or not you know this theorem. If you don't, you can find it on page 332 of this text and I will try to explain it.
Essentially, $T+A=T\circ[\mathrm{Id}+T^{-1}\circ A]$ and $T^{-1}\circ A\in\mathcal{K}(X)$ (why?). By Riesz-Schauder, $\mathrm{Id}+T^{-1}\circ A$ is then Fredholm-$0$, and the composition of it with an invertible operator ($T$) is also Fredholm-$0$ (the book leaves this as an exercise, but I have done this exercise and will demonstrate it if you wish). So, we have shown that $T+A$ is Fredholm-$0$.
A: Recall the following standard form of the Fredholm Alternative

Let $X$ be a Banach space, $I$ the identity operator, $\lambda$
a complex number, and $A_0 \in K(X)$
(a compact operator). If $A_0-\lambda I$ is injective,
then $A_0-\lambda I$ is   invertible.

A good source for this is [1], see also [2].
Now let $X$ be a Banach space, $T \in GL(X)$ (invertible) and $A \in K(X)$.
Suppose that $T+A$ is injective. Then $T^{-1}\circ(T+A)=I+T^{-1}\circ A$ is also injective, and $A_0:=T^{-1}\circ A \in K(X)$. By the Fredholm Alternative with $\lambda=-1$, the operator $I+A_0$ is invertible, so
$T\circ(I+A_0)=T+A$ is also invertible.
[1] https://www-users.cse.umn.edu/~garrett/m/fun/fredholm-riesz.pdf
[2]   W. Arveson, A short course on spectral theory, Springer, 2002
