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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.2012-11-17
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    @GerryMyerson But if it is a circle, is it still 1-1? It seems would map an x to two different point for a circle2012-11-17
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    Gerry's comment is explained in detail here: http://math.stackexchange.com/questions/122013/a-bijective-map-that-is-not-a-homeomorphism2012-11-17
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    Gerry needs to take a half-open line segment, e.g. $(0, 1]$.2012-11-17
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    @QiaochuYuan why taking $(0,1]$ implies the $f^{-1}$ is discontinuous?2012-11-17
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    @Mathematics: it implies nothing of the sort; taking $(0, 1]$ is necessary for the inverse of the map Gerry defined to exist. For me to answer the question of why the inverse is discontinuous I'd first have to know which definition of continuity you're familiar with.2012-11-17

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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?2012-11-17
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    @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).2012-11-17