Index is additive in compositions of Fredholm maps Suppose that given two Fredholm linear maps $f : U\rightarrow V$ and $g : V\rightarrow W$. (These vector spaces are not necesarily finite-dimensional) Then its composition is Fredholm and:
$$
ind(g\circ f) = ind(f) + ind(g)
$$
Notation:
Fredholm maps are those with kernel and cokernel finite dimensional.
Also, index of a linear map $f$ is $ind(f) = dimker(f) - dimcoker(f)$.

The proof of the index theorem is not at all straight forward. You will have to deal a lot with intersections of subspaces and complementary subspaces. Google for "index theorem". I am sure you will find a proof. In order to show that $g\circ f$ is Fredholm, you will have to prove that $\dim\ker(gf) < \infty$ and $\operatorname{codim}\operatorname{im}(gf) < \infty$. I will show here the first of those two. Let $M := \ker(g)$. Then
$$
\ker(gf) = (gf)^{-1}(\{0\}) = f^{-1}(g^{-1}(\{0\})) = f^{-1}(M).
$$
Now, let $W$ be a complementary subspace of $\ker(f)$ in $U$, i.e., $U = \ker(f)\oplus W$. Then
$$
f^{-1}(M) = \{x\in U : f(x)\in M\} = \ker(f)\oplus\{x\in W : f(x)\in M\} = \ker(f)\oplus\left(f|W\right)^{-1}(M).
$$
But $f|W$ is one-to-one. So, $\dim((f|W)^{-1}(M))\le\dim M$. Therefore,
$$
\dim\ker(gf)\le\dim\ker(f) + \dim M = \dim\ker(f) + \dim\ker(g) < \infty.
$$
