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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.

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    $\begingroup$ It is easy to check that $T_f$ is compact iff it has finite rank iff $f[X}$ is a finite set. For the not completely obvious implication, compose $T_f$ with the restriction mapping $R$ from $C(X)$ to $C(f[X])$. The composition $RT_f$ is a quotient map by Tietze, and $C(f[X])$ is infinite dimensional if $f[X}$ is infinite. In particular, if $X$ is connected then $T_f$ is not compact unless $f$ is constant. As for your second question, let $X$ be the range of a sequence of distinct points together with its limit and let $f$ act as a shift. $\endgroup$
    – Bill Johnson
    Commented Jul 23, 2015 at 15:04
  • $\begingroup$ @BillJohnson Prof. Johnson Thank you for your very interesting comment. My second question is about X=interval. $\endgroup$
    – Ali Taghavi
    Commented Jul 23, 2015 at 15:38
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    $\begingroup$ The only way for $T_f$ to be Fredholm when $X=[0,1]$ is for it to be a surjective homeomorphism. If $f$ is not surjective, then $T_f(\phi)=0$ when every $\phi$ is supported off of the interval $f[0,1]$. If $f$ is not $1-1$, then there are infinitely many pairs $(a,b)$ of distinct points s.t. $f(a)=f(b)$, which implies that $T_f$ cannot have finite codimensional range. $\endgroup$
    – Bill Johnson
    Commented Jul 23, 2015 at 18:58
  • $\begingroup$ @BillJohnson Does this 1-1 argument works if we replace the interval by a compact manifold M? If yes, we would obtain an alternative proof for the fact that the spase of homeomorphisms of a compact manifold is an open set(As it is proved in Hircsh Diff. topology). This would be a consequence of openness of fredholm operatores. So is it obvious that for a non 1-1 map f on M there are infinite pairs (a,b) with f(a)=f(b)? $\endgroup$
    – Ali Taghavi
    Commented Jul 24, 2015 at 18:53

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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...)

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