Min and max elements in lattices of KC ($= T_B$) topological spaces

Question

a space $(X,\tau )$ is said to be minimal KC if $(X,\tau )$ is KC and no topology on X which is strictly smaller than $\tau$ is KC.

A space $(X,\tau)$ is minimal KC iff it is KC and compact.

A space $(X,\tau)$ is minimal KC iff it is maxima compact

Let $X $ and $Y$ be topological space. $ f : X ‎\longrightarrow ‎Y‎ $ is called closed , if for every closed subset $ F \subseteq X$, emage $ f( F) $ will be closed in $Y$.

question: Is it right? Why?

(1):Let ‎$ f : X ‎\longrightarrow ‎Y‎ $ be a continuous map from a maximal compact space $X$ to topological space $Y$. then $Y$ is maximal compact iff $f$ is closed.

(2): Does f need to be surjective? why?

Answer 1

Suppose that $f$ is not closed; then there is a closed $F\subseteq X$ such that $f[F]$ is not closed in $Y$. $F$ is compact, so $f[F]$ is compact, and it follows that $Y$ is not $KC$. Since $Y$ is not $KC$, it cannot be minimal $KC$ and therefore cannot be maximal compact.

Now suppose that $Y$ is not maximal compact; then $Y$ is compact but not minimal $KC$, so $Y$ is not $KC$. Let $K$ be a compact subset of $Y$ that isn’t closed; $f^{-1}[K]$ is a closed subset of $X$, and $X$ is compact, so $f^{-1}[K]$ is compact. Moreover, $X$ is $KC$, so $f^{-1}[K]$ is closed. Finally, $f$ is surjective, so $K=f[f^{-1}[K]]$, and $K$ is therefore the non-closed image of the closed set $f^{-1}[K]$; this shows that $f$ is not closed. Surjectivity is needed to ensure that $K=f[f^{-1}[K]]$: if $f$ were not surjective, we might have $K\supsetneqq f[f^{-1}[K]]$.