For a Banach space $X$, and Fredholm operators $T_1, T_2: X \rightarrow X$ with index $0$, I want to decide whether the following statement is true:
The operator $S: Sx = T_1x + T_2x$ is either the zero mapping or a Fredholm operator with index zero.
What I know is that the identity mapping has Fredholm index $0$, and in the finite dimensional case, Fredholm index equals the difference between the dimensions $dim(X) - dim(Y)$ for a Fredholm operator $F: X \rightarrow Y$ . So in the finite dimensional case, the statement is likely to be true.
This is not true. Consider the case where $X=\ell^2$ (or any other classical sequence space), $T_1$ is the identity operator, and $T_2$ is defined by $(T_2x)(n)=x(n)$ if $n=1$ and $(T_2x)(n)=-x(n)$ otherwise. Then $T_1+T_2\neq0$ and $T_1+T_2$ isn't Fredholm.
In fact, if $T_1$ is any Fredholm operator with index $0$, put $T_2=-T_1+K$ for some nonzero compact operator $K$. Then the index of $T_2$ is $0$ as well, but $T_1+T_2=K$ is not Fredholm.
Since you mention the finite-dimensional case, it follows directly from the rank-nullity theorem that any operator on a finite-dimensional space is Fredholm with index zero.