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I need help proving the following.

A function $f:X\to Y$ is an open map if whenever $U$ is an open subset of $X$, then $f(U)$ is an open subset of $Y$. Let $X$ and $Y$ be topological spaces. prove that if $X$ is compact, $Y$ is Hausdorff and connected, and $f\colon X\rightarrow Y$ is a continuous open map, then $f$ is surjective.

Thank you!

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2 Answers 2

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Hint: $f(X)$ is open and closed in $Y$.

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Recall that $Y$ is connected if and only if the only subsets of $Y$ which are both open and closed are $Y$ and $\emptyset$. Recall also that for $X$ compact and $Y$ Hausdorff, if $V$ is a closed subset of $X$, then $f(V)$ is a closed subset of $Y$ - that is, any continuous map $f\colon X\rightarrow Y$ is a closed map.

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  • $\begingroup$ I do understand that. I guess I am just confused on how that means f is surjective. Thank you $\endgroup$
    – kumhmb
    Oct 2, 2013 at 1:47
  • $\begingroup$ $X$ is a closed and open subset of $X$. If $f$ is a closed and open map, then $f(X)$ is a closed and open subset of $Y$. (I can't say much more without writing the punchline of the proof) $\endgroup$
    – Dan Rust
    Oct 2, 2013 at 1:48
  • $\begingroup$ Does it have something to do with an open map being a specific type of quotient map, and a quotient map is surjective by definition? If not, just say no, and I'll work on it more! Thanks! $\endgroup$
    – kumhmb
    Oct 2, 2013 at 1:53
  • $\begingroup$ No. Use the equivalent definition of connected space mentioned in my answer. $\endgroup$
    – Dan Rust
    Oct 2, 2013 at 1:55

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