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 ?
 A: 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.
