Composition of Fredholm Operators

Question

If $ST$ is a Fredholm operator, then show that $T$ is Fredholm if and only if $S$ is Fredholm.

Answer 1

$\DeclareMathOperator{\coker}{coker}\DeclareMathOperator{\index}{ind}$Using the definition that a Fredholm operator is an operator $F \colon X \to Y$ for which both $\ker{F}$ and $\coker{F} = Y/F(X)$ are finite-dimensional this really is an exercise in linear (or homological) algebra. Notice that the requirement that that $F(X)$ be closed in $Y$ is automatic: Is the closedness of the image of a Fredholm operator implied by the finiteness of the codimension of its image?

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For every composition $ST$ of linear maps $S$ and $T$ there is an exact sequence of vector spaces $$ 0 \to \ker{T} \to \ker{ST} \to \ker{S} \to \coker{T} \to \coker{ST} \to \coker{S} \to 0. $$ Since $ST$ is Fredholm, $\ker{ST}$ and $\coker{ST}$ are finite-dimensional. Therefore $\ker{T}$ and $\coker{S}$ are finite-dimensional, being a subspace and a quotient of a finite-dimensional space.

We are given the additional information that either $\ker{S}$ or $\coker{T}$ is finite-dimensional, depending on whether $S$ or $T$ is Fredholm. Thus, we know that five out of six of the vector spaces in the exact sequence are finite-dimensional, hence so is the sixth.

In other words, we can read off this sequence:

If two out of the three operators $S$, $T$ and $ST$ are Fredholm then so is the third.

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Incidentally, this also gives a quick proof of the additivity of the Fredholm index: recall that the index of a Fredholm operator is $\index{F} = \dim{\ker{F}} - \dim{\coker{F}}$.

From the above sequence and the fact that an exact complex has zero Euler characteristic we get $$ 0 = \dim{\ker{T}} - \dim\ker{ST} + \dim{\ker{S}} - \dim{\coker{T}}+\dim{\coker{ST}} -\dim{\coker{S}} $$ which is immediately rearranged to $$ \index ST = \index S + \index T. $$

Answer 2

What you want to use, if possible, is the definition that $A \in B(H)$ is Fredholm if and only if there exists some $B \in B(H)$ such that $AB = I \bmod K(H)$ and $BA = I \bmod K(H)$. If you don't mind a little bit more machinery, recalling that $K(H)$ is a closed, two-sided $\ast$-ideal in the $C^\ast$-algebra $B(H)$, we can equivalently say that $A$ is Fredholm if and only if the image $[A]$ of $A$ in the Calkin algebra $Q(H) := B(H)/K(H)$ is invertible (i.e., with inverse $[A]^{-1} = [B]$).

With all this in mind, then, consider why the following observation is true:

Let $R$ be a unital ring, let $a$, $b \in R$, and suppose that $ab$ is invertible. Then $a$ is invertible if and only if $b$ is invertible.

EDIT: In the more general case where $A \in B(H_1,H_2)$, we have that $A$ is Fredholm if and only if there exists some $B \in B(H_2,H_1)$ such that $AB = I_{H_2} \bmod K(H_2)$ and $BA = I_{H_1} \bmod K(H_1)$. Given that $K(H_i)$ is a two-sided ideal of $B(H_i)$ for $i=1,2$, the essential elementary observation is the following:

Let $t : E \to F$ and $s : F \to G$, and suppose that $st : E \to G$ is bijective. Then $s$ is bijective if and only if $t$ is bijective.

Of course, you can't apply this literally, but rather have to think about invertibility modulo compacts on $H_1$ and on $H_2$.