For $X$ compact Hausdorff, $f: X \to Y$ surjective, there is at most one topology on $Y$ that makes $f$ continuous and $Y$ Hausdorff. I know the theorem that if a space $X$ is compact Hausdorff then there is no finer or coarser topology that is both compact and Hausdorff. So if we assume some topology $\tau$ exists that makes $f$ compact and $Y$ Hausdorff, there is no coarser or finer topology that satisfies this. How do we prove that there isn't some incomparable topology which causes this to be true?
 A: Suppose a topology $\tau$ makes $f$ continuous and $Y$ Hausdorff.  Then $\tau$ is contained in the quotient topology with respect to $f$, since the quotient topology is the finest topology that makes $f$ continuous.  But the quotient topology is compact and $\tau$ is compact Hausdorff, so this can only happen if the quotient topology is the same as $\tau$.  Thus the only possible such topology $\tau$ is the quotient topology.
A: This is basically rephrasing Eric's answer:
Given $f \colon X \to Y$ continuous and surjective, this map factors as $X \twoheadrightarrow X/f \to Y$, where $X/f$ is the quotient. The map $\bar f \colon X/f \to Y$ is bijective (because $f$ is surjective) and continuous. But $X/f$ is compact and $Y$ is Hausdorff, so automatically $\bar f$ is a homeomorphism. (This is easily seen by considering images of closed sets.)  Thus the topology on $Y$ is unique. 
Note that we didn't need the assumption that $X$ is Hausdorff.
A: Suppose that $\tau$ and $\tau'$ are Hausdorff topologies on $Y$ making $f$ continuous. Let $U\in\tau$; $f^{-1}[U]$ is open in $X$. Let $K=X\setminus f^{-1}[U]$; $K$ is closed in $X$ and therefore compact, so $f[K]$ is compact and therefore closed in $\tau'$. But $f[K]=Y\setminus U$, so $U\in\tau'$, and hence $\tau\subseteq\tau'$. Similarly, $\tau'\subseteq\tau$, so $\tau=\tau'$.
