See Theorem 3.4.3, page 93, of these notes for a detailed proof of the fact that $T: H\to H$ is Fredholm if and only if there exists $Q:H\to H$ such that $QT-1$ is compact. If $Q=(\lambda-T)^{-1}$ is compact then

$$Q T= Q(T-\lambda)+\lambda Q=1+\lambda Q$$

so that

$$ QT-1=\lambda Q =\mbox{compact}. $$

See Theorem 3.4.3, page 93, of these notes for a detailed proof of the fact that $T: H\to H$ is Fredholm if and only if there exists $Q:H\to H$ such that $QT-1$ is compact. If $Q=(T-\lambda)^{-1}$ is compact then

$$Q T= Q(T-\lambda)+\lambda Q=1+\lambda Q$$

so that

$$ QT-1=\lambda Q =\mbox{compact}. $$

See Theorem 3.4.3, page 93, of these notes for a detailed proof of the fact that $T: H\to H$ is Fredholm if and only if there exists $Q:H\to H$ such that $QT-1$ is compact. If $Q=(\lambda-T)^{-1}$ is compact then

$$Q T= K(T-\lambda)+\lambda Q=1+\lambda Q$$

so that

$$ QT-1=\lambda Q =\mbox{compact}. $$

See Theorem 3.4.3, page 93, of these notes for a detailed proof of the fact that $T: H\to H$ is Fredholm if and only if there exists $Q:H\to H$ such that $QT-1$ is compact. If $Q=(\lambda-T)^{-1}$ is compact then

$$Q T= Q(T-\lambda)+\lambda Q=1+\lambda Q$$

so that

$$ QT-1=\lambda Q =\mbox{compact}. $$