# If $X$ is compact and Hausdorff, and $f:X\rightarrow Y$ is continuous, closed, and surjective, prove $Y$ is Hausdorff.

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*}

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.

• Ah, of course. Thank you! – Arbutus Jan 13 at 0:55

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).

$$f$$ is not injective, so you can have $$y_1\in f(X/U_1)$$.