Linear Transformation is Fredholm Operator iff Pseudoinverse exists

Theorem
Let $U, V$ be vector spaces.

Let $T: U \to V$ be a linear transformation.

Then $T$ has finite index $T$ has a pseudoinverse.

Sufficient
Suppose that $T$ has a pseudoinverse $S : V \to U$.

That is, both:
 * $D_U := S \circ T - I_U$

and:
 * $D_V := T \circ S - I_V$

are degenerate.

In particular:
 * $\map \ker T \subseteq \map \ker {D_U + I_U}$

and:
 * $\Img {D_V + I_V} \subseteq \Img T$

By Degenerate Linear Operator Plus Identity has Finite Index:
 * $\map \dim {\map \ker T} \le \map \dim {\map \ker {D_U + I_U} } < +\infty$

and:
 * $\map {\mathrm {codim} } {\Img T} \le \map {\mathrm {codim} } {\Img {D_V + I_V} } < +\infty$

Necessary
Suppose that $T$ has finite index.

By Existence of Complementary Subspace, there are subspaces $U' \subseteq U$ and $V' \subseteq V$ such that:
 * $U = \map \ker T \oplus U'$

and:
 * $V = \Img T \oplus V'$

Consider the projections on direct summands:
 * $P : U \to \map \ker T$

and:
 * $Q : V \to V'$

Observe:
 * $\Img P = \map \ker T$

and:
 * $\Img Q = V' \cong V / \Img T$

where the last since the isomorphism follow from First Isomorphism Theorem.

Thus, these are finite-dimensional.

In particular, $P$ and $Q$ are degenerate.

By First Isomorphism Theorem:
 * $U' \cong U / \map \ker T \cong \Img T$

In particular:
 * $T \restriction_{U'} : U' \to \Img T$

is an isomorphism.

Define $S : V \to U$ by:
 * $S u = \begin{cases} \paren {T \restriction_{U'} }^{-1} u &: u \in \Img T \\ 0 &: u \in V' \end{cases}$

Then:
 * $S \circ T = I_U - P$
 * $T \circ S = I_V - Q$