Timeline for Continuous maps on compact topological spaces which induce compact (Fredholm) operators
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12 events
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Dec 3, 2015 at 19:24 | vote | accept | Ali Taghavi | ||
| S Jul 26, 2015 at 6:07 | history | suggested | the_fox | CC BY-SA 3.0 |
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| Jul 26, 2015 at 6:06 | review | Suggested edits | |||
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| Jul 26, 2015 at 4:05 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Jul 26, 2015 at 3:52 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Jul 24, 2015 at 18:53 | comment | added | Ali Taghavi | @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)? | |
| Jul 23, 2015 at 18:58 | comment | added | Bill Johnson | 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. | |
| Jul 23, 2015 at 15:38 | comment | added | Ali Taghavi | @BillJohnson Prof. Johnson Thank you for your very interesting comment. My second question is about X=interval. | |
| Jul 23, 2015 at 15:04 | comment | added | Bill Johnson | 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. | |
| Jul 23, 2015 at 13:27 | answer | added | Simon Henry | timeline score: 6 | |
| Jul 23, 2015 at 13:11 | history | asked | Ali Taghavi | CC BY-SA 3.0 |