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Let $X$ be a compact Hausdorff space. If $f:X\rightarrow Y$ is continuous, closed, and surjective, prove that $Y$ is Hausdorff.

I'm wondering if/where the compactness of $X$ is needed. I have the following 'proof', but I'm rather skeptical as it doesn't depend on the compactness of $X$. Where is the mistake in my proof?

Proof: Let $y_1,y_2\in Y$ be distinct. Then there are $x_1,x_2\in X$ such that $f(x_i)=y_i$. Since $X$ is Hausdorff, there exist disjoint open neighborhoods $U_1, U_2$ of $x_1$ and $x_2$, respectively. As $f$ is closed, $f(X\backslash U_i)\subseteq Y$ are also closed. Thus, if $V_i:=Y\backslash f(X\backslash U_i)$, then $V_1$ and $V_2$ are disjoint open neighborhoods of $y_1$ and $y_2$, respectively, where disjointness follows from \begin{align*} Y\backslash( V_1\cap V_2)&=Y\backslash V_1\cup Y\backslash V_2 \\ &=f(X\backslash U_1)\cup f(X\backslash U_2)\\ &=f(X\backslash U_1\cup X\backslash U_2)\\ &=f(X\backslash (U_1\cap U_2))\\ &=f(X)\\ &=Y. \end{align*}

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3 Answers 3

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The non-injectivity failure in your current proof has already been pointed out. Now instead of disjoint neighbourhoods $U_1,U_2$ of $x_1$ and $x_2$, use disjoint neighbourhoods of the closed sets $F_i = f^{-1}[\{y_i\}]$ ($i=1,2$) of $X$ and the same construction of open sets. This does use compactness (why?) and we do get neighbourhoods of $y_i$ ( this needs a small argument).

A more general theorem:

Suppose that $f: X \to Y$ is surjective, continuous, closed and has compact fibres (all $f^{-1}[\{y\}], y \in Y$ are compact (such a map is called perfect)) then if $X$ is Hausdorff, then so is $Y$.

This captures the essence of the proof and does not require $X$ to be compact (we need the compactness of the fibres, really).

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The mistake is in claiming that $y_1 \in V_1$ and $y_2 \in V_2$. You cannot say this unless $f$ is injective.

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  • $\begingroup$ Ah, of course. Thank you! $\endgroup$
    – Arbutus
    Commented Jan 13, 2019 at 0:55
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$f$ is not injective, so you can have $y_1\in f(X/U_1)$.

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