Let $X$ be a compact Hausdorff space. A continuous map $f:X \to X$ defines a bounded linear operator $T\_{f}$ on the Banach space $C(X)=\{\phi:X\to \mathbb{C} \mid \phi\; \text{is continuous}\}$ with $T\_{f}(\phi)=\phi \circ f$.


Put $X=[0,1]$.



> 
> What is an example of a non constant map $f$ such that $T\_{f}$ is a compact operator?
>  What is an example of a map $f$ such that $T\_{f}$ is a Fredholm operator of non-zero index?
> 
> 
> 


For a linked MSE question [see this MSE post](https://math.stackexchange.com/questions/1373804/non-injective-continuous-maps/1373823#1373823).



Continuous maps on compact topological spaces which induce compact (Fredholm) operators
The image of $f$ is an interval $[a,b] \subset [0,1]$.


$T\_f$ induce an isometric inclusion of $C([a,b])$ in $C([0,1])$.


If $T\_f$ was compact then weak convergence of a bounded sequence in $C([a,b])$ would imply norm convergence in $C([0,1])$ which would in turn (as $T\_f$ is isometric) imply norm convergence in $C([a,b])$. which is only possible if $C([a,b])$ is finite dimensional i.e. if $f$ is constant.


(edit: $C([a,b])$ is embeded in $C([0,1])$ by extending the functions by their value on the boundary outside $[a,b]$ and I'm identifying $T\_f$ with its restriction to $C([a,b])$, which is also compact...)


