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Suppose $f:X\rightarrow Y$ is 1-1 and continuous. Is $f^{-1}:f(X)\rightarrow X$ continuous too? If not, can you explain it?

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Consider taking a line segment and bringing the ends together to form a circle. That's continuous, but the inverse, which tears the circle open, is not. –  Gerry Myerson Nov 17 '12 at 8:42
    
@GerryMyerson But if it is a circle, is it still 1-1? It seems would map an x to two different point for a circle –  Mathematics Nov 17 '12 at 8:44
    
Gerry's comment is explained in detail here: math.stackexchange.com/questions/122013/… –  Julian Kuelshammer Nov 17 '12 at 8:44
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Gerry needs to take a half-open line segment, e.g. $(0, 1]$. –  Qiaochu Yuan Nov 17 '12 at 8:47
    
@QiaochuYuan why taking $(0,1]$ implies the $f^{-1}$ is discontinuous? –  Mathematics Nov 17 '12 at 9:03

1 Answer 1

up vote 2 down vote accepted

Qiaochu proposed the half open one so that we have the required injectivity (if we included 0 we would map both 0 and 1 to the starting point on our circle).

The inverse (taking the circle to the line) is discontinuous because $f(X)\setminus \{ f(1/2) \}$ is connected (it is simply the circle with one point missing) yet it's image under $f^{-1}$ is $ (0,1/2) \cup (1/2,1]$ which is not connected.

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do you mean that besides $1\over 2$ one can also select many other point say $1\over 3$ and inverse of $f$ also not continuous? –  Mathematics Nov 17 '12 at 9:23
    
@Mathematics Yes we could have picked $1/3$ and the proof would be similar, but we could not have picked $1$, since the image of $f(X)\setminus \{ f(1) \}$ under $f^{-1}$ is $(0,1)$ which indeed is connected, which doesn't tell us anything (the image of a connected set may be connected even under discontinuous functions, but the image of a connected set under a continuous function is always connected). –  Katie Dobbs Nov 17 '12 at 9:29

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