Kernel of adjoint and cokernel of operator

Question

Let $D$ be a Fredholm operator and $D^\dagger$ is its adjoint. Is the dimension of $ker ~ D^\dagger$ equal to the dimension of $coker ~D$ ? If so, can someone sketch the proof?

Answer 1

I think the answer is yes for the following reason :

Let $D : X \rightarrow Y$ is a Fredholm operator on vector bundles with fiber metric $< , >$.

If $f \in coker ~ D$, we have $0= < Ds, f > = <s, D^\dagger f>$ for any $s \in X$. This means that $D^\dagger f=0$, and $f \in ker ~ D^\dagger$.

If $g \in ker ~ D^\dagger$, we have $0=< D^\dagger g, t > = < g, D t >$ for any $t \in Y$. This means that $g \in coker ~ D$.