Image of Fredholm Operator of Banach Spaces is Closed

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

Let $X, Y$ 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:
 * $\map {\tilde T} {x + \map \ker T} := \map T x$

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:
 * $\set { y_1 + \Img T, \ldots, y_m + \Img T}$

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

We show that $f$ is an isomorphism.

Then the claim will follow, as:
 * $\Img T = \hat T \sqbrk {X \times \set { {\mathbf 0}_{\Bbb F^m} } }$