Let $f:A \to B$ be a continuous function. I know that if $U$ is an open set in $B$ then $f^{-1}(U)$ is open in $A$. But is the convese true? i.e. Is it true that if $f^{-1}(U)$ is open in $A$ then $U$ is open in $B$?
Thank you.
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$f:\mathbb{R}\to\mathbb{R}:x\mapsto 0$, $U={0}$ is a counterexample. – rogerl Oct 07 '15 at 14:49
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Please let us know what your definition of "continuous" is. In general topology $f^{-1}(U)$ being open for all $U$ open is usually the definition of continuity. – nullUser Oct 07 '15 at 14:51
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So how would you prove that if $g$ is continuous and if $f\circ g$ is continuous, then $f$ is continuous ? I wanted to show that if $U$ is open then $f^{-1}(U)$ is open, but it looks complicate since for $U$ open, if $g^{-1}(f^{-1}(U))$ we don't necessarily have that $f^{-1}(U)$ is open. For your information, $g$ is a quotient map, may be if $g$ is not a quotient map, the result is wrong. – Rick Oct 07 '15 at 15:00
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@AlexS : no, this is not the definition of an open map and there is open maps that doesn't verify this property. Take $\phi : \Bbb R \to \Bbb Z$ defined by $\phi(x) = \lfloor x \rfloor$(with the usual topology). It's an open map, ${0}$ is open, but clearly $\phi^{-1}( {0} ) [0,1[$ is not open. – Tryss Oct 07 '15 at 15:16
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@AlexS An open map is if $U$-open in $Dom(f)$ then $f(U)\subset R(f)$ is open. – Svetoslav Oct 07 '15 at 15:18
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1@Rick : just use the definition of a quotient map : If $U$ is open, then $(f\circ g)^{-1}(U)$ is open. Let $V= f^{-1}(U)$, then $g^{-1}(V)$ is open, and as $g$ is a quotient map, $V$ is open. Hence $f^{-1}(U)$ is open. – Tryss Oct 07 '15 at 15:21
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The statement should be $f$ is continuous iff $f^{-1}(U)$ is relatively open in $Dom(f)$ for all open $U$. Here the definition for relatively open set is http://math.stackexchange.com/questions/296432/what-are-relative-open-sets – Svetoslav Oct 07 '15 at 15:22
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@Tryss: great. Thanks :-) Could you give me an example of function $f$ such that $f\circ g$ is continuous, $g$ is continuous but $f$ is not continuous ? – Rick Oct 07 '15 at 15:32
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Not necessarily. For example if $f:\mathbb R\to\mathbb R$ is the constant zero function $f(x)=0$ and $U=[-1,1]$, then $f^{-1}(U)$ is open, but $U$ itself is not open.
Or for a less extreme example you could take $f=\sin$ and the same $U$.
hmakholm left over Monica
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So how would you prove that if $g$ is continuous and if $f\circ g$ is continuous, then $f$ is continuous ? I wanted to show that if $U$ is open then $f^{-1}(U)$ is open, but it looks complicate since for $U$ open, if $g^{-1}(f^{-1}(U))$ we don't necessarily have that $f^{-1}(U)$ is open. For your information, $g$ is a quotient map, may be if $g$ is not a quotient map, the result is wrong. – Rick Oct 07 '15 at 15:02
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@Rick: I wouldn't prove that, because it's not true. A counterexample would be to let $f$ be any discontinuous function of your choice and let $g$ be a constant function. – hmakholm left over Monica Oct 07 '15 at 15:46
