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I don't know how solve this exercise: "If $f: X \rightarrow Y $ is an identification and the connected components of X are open show that the connected components of Y are open" My attempt: Consider $C$ connected components and consider $f^{-1}(C)$ how can I show that $f^{-1}(C)$ is union of connected components? From here $C$ is open because $f$ is an identification.

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    What is "identification"? A homeomorphism? An open map? – 2017-01-25
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    Y has the quotient topology – 2017-01-25
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    And $f$ is a quotient map? – 2017-01-25
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    f is a map continuos and surjective and Y has the quotient topology – 2017-01-25

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Let $C$ be a connected component of $Y$. Consider $x\in f^{-1}(C)$ and let $B_x$ be a connected component of $x$ in $X$. Since continous functions map connected sets to connected sets, then $f(B_x)\subseteq C$. Thus $B_x\subseteq f^{-1}(C)$. Since $x$ was picked arbitrarly then

$$f^{-1}(C)=\bigcup_{x\in f^{-1}(C)} B_x$$

and since each $B_x$ is open by assumpion, then $f^{-1}(C)$ is open and so is $C$.

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    Thanks http://math.stackexchange.com/q/2113300/227073any ideas also for this question? – 2017-01-25