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?
Is the inverse function continuous?
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calculus
analysis
functions
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0@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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0@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