Image of Fredholm Operator of Banach Spaces is Closed

Theorem
Let $\Bbb F \in \set {\R, \C}$.

Let $\tuple {X, \norm \cdot_X}, \tuple {Y, \norm \cdot_Y}$ be Banach spaces over $\Bbb F$.

Let $T: X \to Y$ be a Fredholm operator.

Let $\Img T$ be the image of $T$.

Then $\Img T$ is closed.

Proof
By Kernel of Bounded Linear Transformation is Closed Linear Subspace, $\map \ker T \subseteq X$ is closed.

Thus, by Characterization of Complete Normed Quotient Vector Spaces, $X / \map \ker T$ is a Banach space.

Define the injective linear transformation $\tilde T : X / \map \ker T \to Y$ by:
 * $\map {\tilde T} {\eqclass x {\map \ker T} } := \map T x$

Then $\Img {\tilde T} = \Img T$.

Replacing $T$ by $\tilde T$ if necessary, we may assume that $T$ is injective.

Since $Y / \Img T$ is finite-dimensional, it admits a finite basis:
 * $\set { \eqclass {y_1} {\Img T}, \ldots, \eqclass {y_m} {\Img T}}$

Consider the direct sum:
 * $X \oplus {\Bbb F^m}$

Define the surjective linear transformation $\hat T : X \oplus {\Bbb F^m} \to Y$ by:
 * $\map {\hat T} {x, k_1, \ldots, k_m } := \map T x + k_1 y_1 + \cdots + k_m y_m$

Observe:

where:
 * $\norm T$ denotes the operator norm of $T$
 * $L := \sqrt {\norm {y_1}_Y^2 + \cdots + \norm {y_m}_Y^2}$

Hence $\hat T$ is bounded.

Furthermore, $\hat T$ is injective.

Indeed, let
 * $\map {\hat T} {x, k_1, \ldots, k_m } = 0$

That is:
 * $k_1 y_1 + \cdots + k_m y_m = - \map T x$

Since:
 * $k_1 \eqclass {y_1} {\Img T} + \cdots + k_m \eqclass {y_m} {\Img T} = 0$

we have:
 * $k_1 = \cdots = k_m = 0$

Then we also have:
 * $\map T x = 0$

As $T$ is injective, it follows that:
 * $x = 0$

So, we have shown the injectivity of $\hat T$.

That is, $\hat T$ is a bijective bounded linear transformation.

By Banach Isomorphism Theorem, $\hat T$ is a linear isomorphism.

Now the claim follows from:
 * $\Img T = \hat T \sqbrk {X \times \set { {\mathbf 0}_{\Bbb F^m} } }$