Invariance of the Fredholm index under finite-dimensional perturbations Let's call a linear map $f : V \to W$ between vector spaces over some field Fredholm if $\ker(f)$ and $\mathrm{coker}(f)$ are finite-dimensional. (Equivalently, it represents an isomorphism in the abelian quotient category (vector spaces)/(finite-dim. vector spaces).) We can then define the Fredholm index by
$$\mathrm{ind}(f) := \dim(\ker(f))-\dim(\mathrm{coker}(f)).$$
Is it true that $\mathrm{ind}(f+g)=\mathrm{ind}(f)$ if $g : V \to W$ has finite-dimensional range?
I have tried to write down exact sequences and use that the Euler characteristic vanishes on such sequences, but this didn't work out. Another idea would be the following: We may assume wlog that $\mathrm{im}(g)$ is $1$-dimensional. Then write $g = w \otimes \alpha$ for some $\alpha \in V^*$ and $w \in W \setminus \{0\}$. Now two cases appear: 1) $w \in \mathrm{im}(f)$. 2) $w \notin \mathrm{im}(f)$. I have tried to compute $\dim(\ker(f+g))$ and $\dim(\mathrm{coker}(f+g))$ in each case, but didn't succeed.
PS: I have borrowed the terminology for linear operators between Hilbert spaces (Fredholm operator). I don't know if this is standard. Notice that the proof in the Wikipedia article that the index is invariant under compact perturbations is analytic and therefore probably cannot be used here.
 A: The result perhaps can be proved inductively showing that $\text{index}\,( T- \delta) = \text{index}\,T$ for $\delta$ of rank $1$. A useful particular case is showing by hand that $\text{index}( I - \delta) = 0$, starting with rank of $\delta=1$. The different cases presented suggests that there could be a uniform proof. We sketch one below. Break it into several steps.


*

*Show the additivity of the index 


Given $$U\xrightarrow[]{S}V \xrightarrow[]{T} W$$
with $S$, $T$ Fredholm, then $T S$ is Fredholm and we have 
$$\text{index}{\,TS} = \text{index}{\,S}+ \text{index}{\,T}$$
This follows right away from the long exact sequence:
$$0 \to \ker \,S\to \ker \,TS\to \ker\, T \to \text{coker}\,S \to \text{coker}\,T S \to \text{coker}\,T \to 0 $$
by using the fact that the Euler characteristic of the sequence is $0$.


*Show that for every $\pi \colon V\to V$ of finite rank, $I- \pi$ is Fredholm of index $0$


$$\text{index} \, (I-\pi)=0$$
This follows from the existence of a natural isomorphism:
$$\frac{\text{im} (I - \pi)}{\ker \pi} \simeq \frac{\text{im} ( \pi)}{\ker(I- \pi)}$$
(proof is left for the reader)


*If $\pi\colon V \to V$ is of finite rank, and $T\colon U \to V$, then $T$ is Fredholm if and only if $(I-\pi) T$ is, and if so,


$$\text{index}\,T = \text{index}(I -\pi)\, T$$
This follows easily from $1.$ and $2.$


*Let $\delta \colon U \to V$ of finite rank and $T \colon U \to V $ Fredholm. Then $T - \delta$ is Fredholm and 


$$\text{index}\,( T -\delta) = \text{index}\, T$$
Indeed, there exists $\pi \colon V \to V$ of finite rank so that $\pi \delta = \delta$, for instance the projection of $V$ onto the image of $\delta$, considered as a map from $V$ to $V$. Therefore $(I-\pi) \delta = 0$. We conclude  $(I -\pi) T = (I-\pi) (T-\delta)$ and from $1.$ and $2.$ we  get:
$$\text{index}\,(T-\delta) = \text{index}\, (I-\pi) (T-\delta) = \text{index}\, (I-\pi) T= \text{index}\, T$$
