# the continuous image of a compact $T_2$-space

Let $X,Y$ be a topological spaces. The map $f : X \longrightarrow Y$ is said to be closed if for every $F \subset X$ , its image $f[F]$ is also closed in $Y$.

so, is it right to say:

A $KC$-space $Y$ that is the continuous image of a compact $T_2$-space $X$ is a $T_2$-space.?why? does the function need to be onto ?

• If $Y$ is the continuous image of $X$ under the map $f:X \to Y$, then $f$ must by definition be onto. – Alexander Aug 30 '13 at 17:54
• @Alexander: I believe that the OP is asking whether, if we know (i) $f:X\to Y$ is continuous, (ii) $X$ is compact $T_2,$ and (iii) $Y$ is a KC-space, we can conclude that $Y$ is a $T_2$ space. The answer to that question, of course, is "no." Given any non-$T_2$ KC space $Y,$ any constant function $f:X\to Y$ is continuous, regardless of the topology on $X$. – Cameron Buie Aug 30 '13 at 18:04
• @CameronBuie Yes. Another interpretation might be given by the the bold comment that says $Y$ is a continuous image of $X$. Perhaps the author can clarify which statement is intended. – Alexander Aug 30 '13 at 18:15
• @Cameron: The OP intends $f$ to be a continuous surjection. This is Prop. $1$ of Hans-Peter A. Künzi & Dominic van der Zypen, ‘Maximal (sequentially) compact topologies’, arXiv:math/0306082v1 [math.GN], where the part after the first paragraph of my answer is taken to be obvious and therefore not spelled out. – Brian M. Scott Aug 31 '13 at 1:59
• @Brian: I see. I thought perhaps the OP was wondering if the surjective condition was necessary to the proof. – Cameron Buie Aug 31 '13 at 2:21

Suppose that $f:X\to Y$ is a continuous surjection, $X$ is compact and Hausdorff, and $Y$ is a $KC$ space. Let $F$ be any closed subset of $X$. Then $F$ is compact, so $f[F]$ is compact in $Y$ and hence closed in $Y$ (since $Y$ is $KC$). Thus, the map $f$ is closed.
Now let $y_0$ and $y_1$ be distinct points of $Y$. Let $K_0=f^{-1}[\{y_0\}]$ and $K_1=f^{-1}[\{y_1\}]$; $K_0$ and $K_1$ are disjoint closed sets in the compact Hausdorff space $X$, so there are disjoint open sets $U_0$ and $U_1$ in $X$ such that $K_0\subseteq U_0$ and $K_1\subseteq U_1$. Let $F_0=X\setminus U_0$ and $F_1=X\setminus U_1$; then $F_0$ and $F_1$ are closed in $X$, and $F_0\cup F_1=X$. Let $C_0=f[F_0]$ and $C_1=f[F_1]$; $f$ is closed, so $C_0$ and $C_1$ are closed in $Y$, and clearly $C_0\cup C_1=Y$. Let $V_0=Y\setminus C_0$ and $V_1=Y\setminus C_1$; then $V_0$ and $V_1$ are open in $Y$, and $V_0\cap V_1=\varnothing$. I claim that $y_0\in V_0$ and $y_1\in V_1$.
But this is clear. $K_0\subseteq U_0$, so $K_0\cap F_0=\varnothing$; $K_0$ is the entire inverse image of $\{y_0\}$, so it follows that $f(x)\ne y_0$ for each $x\in F_0$ and hence that $y_0\notin f[F_0]=C_0$ and therefore that $y_0\in V_0$. A similar argument shows that $y_1\in V_1$. Thus, $y_0$ and $y_1$ have disjoint open nbhds, and since $y_0$ and $y_1$ were an arbitrary pair of distinct points of $Y$, we’ve shown that $Y$ is Hausdorff.
• I think, you mean $y_0 \not\in f [F_0] = C_0$? $F_0 \cup F_1 = X$? why?and $V_0 \cap V_1 = \emptyset$? – fatemeh Sep 5 '13 at 19:00
• @fatemeh: Yes, that’s $y_0$. $F_0\cup F_1=X$ because $U_0\cap U_1=\varnothing$. $V_0\cap V_1=\varnothing$ because $C_0\cup C_1=Y$. Both of these are just De Morgan’s laws. – Brian M. Scott Sep 5 '13 at 19:06
• why do you use surjection ?I mean, is it right to say $f$ is closed ,where $f$ is $f : X \longrightarrow‎ Y$if, for every close subset $F$ of a topological space $X$, image $f(F)$ will be close in $Y$? – fatemeh Sep 18 '13 at 13:05
• @fatemeh: I used a surjection because the theorem gives you one, and I was proving the theorem: ‘$Y$ is the continuous image of ... $X$’. If $f$ is not a surjection, it’s quite possible that $Y\setminus f[X]$ is Hausdorff. – Brian M. Scott Sep 18 '13 at 19:28