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