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Let $f: X\rightarrow Y$ be a continuous, injective, surjective.

Question 1, if $f$ is open or closed, then does $f^{-1}$ continuous?

Question 2, if $f$ is open or closed, then does $f^{-1}$ open or closed?

Thanks ahead.

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2 Answers 2

up vote 4 down vote accepted

Hint: A function $h:A\to B$ is continuous if and only if $h^{-1}U$ is open (closed) in $A$ for every open (closed) $U$ in $B$.

  1. Note that $(f^{-1})^{-1}U=fU$ for every $U\subseteq X$.

  2. Use the fact that $f$ is continuous to derive the conclusion. The assumption of $f$ being also open or closed is not relevant.

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1 second faster! –  P.. Feb 22 '13 at 9:29
@Pambos. Yeah I saw that :) –  Thomas E. Feb 22 '13 at 9:29
@ThomasE.If $f$ continuous, injective, surjective + open (or closed), then $X=Y$? –  Paul Feb 22 '13 at 9:33
@John. Those conditions imply that $X$ and $Y$ are homeomorphic, but not necessarily that $X=Y$. For e.g. the intervals $(0,1)$ and $(0,2)$ are homeomorphic by the function $x\mapsto 2x$, which is continuous; open; surjective; and injective; but $(0,1)\neq (0,2)$. –  Thomas E. Feb 22 '13 at 9:36

Hint: A function $g:X\rightarrow Y$ is continuous $\iff \forall U\subseteq Y$ open $g^{-1}(U)\subseteq X $ is open $ \iff \forall K\subseteq Y$ closed $g^{-1}(K)\subseteq X$ is closed.

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