Index map defines a bijection to $\mathbb{Z}$? In the book "Spin Geometry" by Lawson and Michelsohn, page 201, proposition 7.1(chapter III), it asserts that the mapping which assign a Fredholm operator from one Hilbert space to another its index ($\dim\ker-\dim\text{coker} $) defines a bijection from the set of connected components of these Fredholm operator to $\mathbb{Z}$. There the authors use a series of lemmas to prove this assertion. However, just think the most simple example--both Hilbert spaces are of finite dimension, with dimensions $m$ and $n$ respectively. Then the index is constant $m-n$ as we all know. From the proof by the authors, we can only say that the map $\mathtt{ind}$ is injective, and can not speak more. Thus the authors made an mistake there. My question is, if we assume both the Hilbert spaces are of infinite dimension (and both separable), then is this very mapping being bijection? Or there is some reasonable way to remedy this defect?

If $H$ has infinite dimension, the shift operator under a base is Fredholm with index $-1$. As $ind$ is a groups homomorphism between $\text{Fredholms}/\text{Compact}$ and $\mathbb{Z}$, and $-1$ is a generator of $\mathbb{Z}$, this proves that $ind$ is indeed surjective.