Fredholm Alternative for Compact Operators - Why isn't it so simple? Let $\mathfrak{X}$ be a Banach space, and $K\in \mathcal{K}(\mathfrak{X})$ be a compact operator. Then given $\lambda\in\mathbb{C}\backslash\{0\}$, we can show that

*

*$\operatorname{ker}(\lambda I - K)$ is finite dimensional

*$\operatorname{ran}(\lambda I - K)$ is closed
Now, if $\mathfrak{M}\subseteq\mathfrak{X}$ is a finite-dimensional subspace, then $\mathfrak{M}$ is complemented in $\mathfrak{X}$, so there exists a closed subspace $\mathfrak{N}\subseteq\mathfrak{X}$ such that $\mathfrak{X} = \mathfrak{M}\oplus\mathfrak{N}$. Let $\mathfrak{R}$ be $\operatorname{ker}(\lambda I - K)$'s complement in $\mathfrak{X}$, and define
$$
T : \mathfrak{R}\to \mathfrak{X},\qquad y\mapsto(\lambda I - K)y
$$
or in other words $T = (\lambda I - K)|_{\mathfrak{R}}$. It's easy to see that $T$ is injective, and surjective onto it's range, which is closed. Thus $T$ induces an isomorphism $\mathfrak{R}\cong\operatorname{ran}(\lambda I - K)$, and so
$$
\mathfrak{X}\cong\operatorname{ker}(\lambda I - K)\oplus\operatorname{ran}(\lambda I - K)
$$
From this, hypothetically we should be able to obtain the Fredholm alternative: that $\lambda I - K$ is injective if and only if it is surjective.
Actually, I suppose all this proves is that if $\lambda I - K$ is injective, then it must be surjective. However, it's totally possible that $\mathfrak{X}$ is isomorphic to the direct sum of the finite subspace $\operatorname{ker}(\lambda I - K)$ and the isomorphic copy of $\mathfrak{X}$ that is $\operatorname{ran}(\lambda I - K)$, so we can't be sure that $\lambda I - K$ is injective.
However, we can derive an additional factoid:
$$
\dim(\mathfrak{X}/\operatorname{ran}(\lambda I - K)) = \dim\operatorname{ker}(\lambda I - K^*)
$$
where $K^*\in\mathcal{K}(\mathfrak{X}^*)$ denotes the Banach space adjoint (which is also compact). I forget if this is true or not, but I seem to recall that if an operator is surjective, then it's adjoint is injective, so this would be enough to prove the other direction of the Fredholm alternative (assuming such a result exists, which I'll have to verify).
Most Banach space theory texts go through the motion of showing that $\lambda I - K$ has both finite ascent and finite descent (the towers $\{\operatorname{ran}(\lambda I - K)^n\}_{n\in\mathbb{N}}$ and $\{\operatorname{ker}(\lambda I - K)^n\}_{n\in\mathbb{N}}$ both stabilize), and from this derive the Fredholm alternative.
Question: is this extra rigmarole about ascent/descent necessary for deriving the Fredholm alternative? Is there anything wrong with my deduction above, other than the missing "fact" that a surjective operator has injective adjoint?
Edit: Unless I'm just having an off day (I'm prone to those), it seems trivially true that if $T\in\mathcal{B}(\mathfrak{X}, \mathfrak{Y})$ is surjective then $T^*\in\mathcal{B}(\mathfrak{Y}^*, \mathfrak{X}^*)$ is injective. Indeed, supposing $T^*f = T^*g$, then $\forall x\in\mathfrak{X}$, $f(Tx) = g(Tx)$, but since $Tx$ ranges over $\mathfrak{Y}$, this implies $f(y) = g(y)$ for all $y\in\mathfrak{Y}$, so $f = g$. Thus, don't we have the Fredholm alternative?
 A: Okay I think I've identified the problem, and it's a little pedantic, but a crucial detail. Let's be clear about the facts. First:
$$
\mathfrak{X} = \operatorname{ker}(\lambda I - K)\oplus\mathfrak{R}
$$
This is an equality, these spaces are the same. Second:
$$
\mathfrak{X} \cong \operatorname{ker}(\lambda I - K)\oplus\operatorname{ran}(\lambda I - K)
$$
That is, $\mathfrak{X}$ is merely isomorphic (topologically, but not isometrically so) to the direct sum.
This, unfortunately, tells us little about injectivity/surjectivity of $\lambda I - K$. Sure, if $\lambda I - K$ is injective, then $\mathfrak{X}\cong\operatorname{ran}(\lambda I - K)$, but lots of Banach spaces are isomorphic to some of their own proper subspaces. This tells us nothing about surjectivity.
We also obtain nothing from the other route. Since
$$\dim(\mathfrak{X}/\operatorname{ran}(\lambda I - K)) = \dim\operatorname{ker}(\lambda I - K^*)$$
if $\lambda I - K^*$ is injective, then $\mathfrak{X}/\operatorname{ran}(\lambda I - K) = 0$, so $\lambda I - K$ must be surjective. But, all we've shown is that if $\lambda I - K$ is surjective, then... $\lambda I - K$ is surjective.
I'll leave this answer unaccepted for now, because I'm still curious if someone can shed light on why we need ascent/descent for the opposite end of the Fredholm alternative. However, as it stands, the Fredholm alternative is not as simple as it seems.
