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Let $E,F$ normed linear spaces, let $C$ connected of $E$, $D\subset F$, and $f:C\to D$ such that $f$ is open (i.e. sends open sets in $C$ "which is the same as open sets of $E$ intersected with $C$", to open sets in $D$). Suppose $f$ onto, and $f^{-1}(\{y\})$ connected for all $y\in D$,then $D$ is connected.

Suppose $D$ is disconnected then $D\subseteq O_1 \cup\ O_2$ two disjoint open sets. Let $D_1=\bigcup\{$open sets in $D$ and subsets of $O_1\}$ and $D_2=\bigcup\{$open sets in $D$ and subsets of $O_2\}$ then $D=D_1\cup D_2$. This is a disjoint union. Let $x\in D_1$ and $y\in D_2$. Then there exists $B(x,p_1)\subseteq D_1$ and $B(x,p_2)\subseteq D_2$.

Now write $C$ as the union of open sets in $C$ satisfying the image of each set in there is precisely equal to an open set in $D$ (note: I suppose this is posible since $f$ is onto and $f$ is open function is this true?) In particular there is two disjoint open sets in $C$ called $C_1,C_2$ such that $f(C_i)=B(x,p_i)$ for $i=1,2$ and then $\{f^{-1}(z):z\in B(x,p_i)\}=C_i$ this union of connected sets, note that $C_1\cap C_2=\emptyset$ other case they will have the same image which is a contradiction, since this is posible por all $x\in O_1,\ y\in O_2$ I conclude that $C$ is disconnected, which is a contradiction.

Edit: Since in the comments is there a counterexample, what happen if $C$ is not suppose to be connected, and we need to prove it!! Is that possible?

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    $\begingroup$ Is $f$ also supposed to be continuous? Otherwise the result is false: take $E=F=\mathbb{R}$, $C=[0,\infty)$, $D=\{0,1\}$ and $f$ defined by $f(0)=0$, $f(x)=1$ if $x>0$. Then $f$ is obviously open, $f^{-1}(\{0\})=\{0\}$ and $f^{-1}(\{1\})=(0,\infty)$ are connected, but $D$ is not connected. $\endgroup$ – egreg Jun 22 '15 at 23:22
  • $\begingroup$ On the other hand, if $f$ is continuous, $D$ is obviously connected. So there seems to be some missing hypothesis. $\endgroup$ – egreg Jun 22 '15 at 23:34
  • $\begingroup$ These are all the hypothesis!! Maybe the problem be wrong because your $f$ gives an easy counterexample. Thanks!! $\endgroup$ – Valerin Jun 22 '15 at 23:49
  • $\begingroup$ What happen if we delete the hypothesis of $C$ connected? and then we need to prove it!! Is that posible? $\endgroup$ – Valerin Jun 22 '15 at 23:55
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Under the stated hypotheses, the statement is false.

Consider $E=F=\mathbb{R}$, $C=[0,\infty)$, $D=\{0,1\}$ and $$ f(x)=\begin{cases} 0 & \text{if $x=0$},\\[6px] 1 & \text{if $x>0$}. \end{cases} $$ Then $f\colon C\to D$ is onto and obviously open, but $D$ is not connected. Also $$ f^{-1}(\{0\})=\{0\}, \qquad f^{-1}(\{1\})=(0,\infty) $$ are connected.

If we remove the assumption that $C$ is connected, we can even have $f$ continuous; take $C=\{-1\}\cup(0,\infty)$ and $D$ as before, defining $$ f(x)=\begin{cases} 0 & \text{if $x=-1$},\\[6px] 1 & \text{if $x>0$}. \end{cases} $$

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  • $\begingroup$ Thanks for the observation! This is very useful! $\endgroup$ – Valerin Jun 23 '15 at 8:10
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    $\begingroup$ @LuisValerin I added the other counterexample to my answer. $\endgroup$ – egreg Jun 23 '15 at 8:16

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