Let $\tilde{X}$ be the set of Hausdroff Compactifications of a space $X$.
I define a relation as follows
Let $\tilde{X_1}$ $\leq$ $\tilde{X_2}$ iff there exists a continuous onto map from $\tilde{X_2}$ to $\tilde{X_1}$ which leaves the points of $X$ fixed.
I am trying to show it is antisymmetric i.e if $\tilde{X_2}$ is related to $\tilde{X_1}$ and $\tilde{X_1}$ is related to $\tilde{X_2}$ then $\tilde{X_1}$ = $\tilde{X_2}$.
Please help in this.
Also, if $X$ is locally compact and hausdroff then I am trying to show that $X_\infty$ i.e the one point compactification of $X$ is the minimal element ie. if $\tilde{X}$ is any Hausdroff Compactification of $X$ then there exists a continuous onto map from $\tilde{X}$ to $X_\infty$ which leaves the points of $X$ fixed.
Kindly explain how to justify?
For antisymmetry, assume that we have continuous maps $f: \tilde{X}_2 \to \tilde{X}_1$ and $g: \tilde{X}_1 \to \tilde{X}_2$ that fix $X$. We prove that the compositions $gf$ and $fg$ are the respective identities. By symmetry we just prove this for $gf: \tilde{X}_2 \to \tilde{X}_2$. By assumption, $gf$ agrees with $\operatorname{id}_{\tilde{X}_2}$ on the dense subset $X \subset \tilde{X}_2$. Since $X$ is dense in $\tilde{X}_2$ and since $\operatorname{id}_{\tilde{X}_2}$ and $gf$ are continuous, this implies that $gf = \operatorname{id}_{\tilde{X}_2}$. Similarly $fg = \operatorname{id}_{\tilde{X}_1}$, proving that $\tilde{X}_1 \cong \tilde{X}_2$.
I am not quite sure how to prove that $X_\infty$ is minimal without additional assumptions on the Hausdorff compactifications of $X$ that are allowed. However, the following argument works if we assume that we only consider Hausdorff compactifications $\tilde{X}$ of $X$ such that $X$ is open in $\tilde{X}$.
We claim that for any such Hausdorff compactification $\tilde{X}$, the canonical map $p:\tilde{X} \to X_\infty = X \cup \{\infty\}$ given by \begin{aligned} x \mapsto \begin{cases} X & x \in X; \\ \infty & x \notin X;\end{cases} \end{aligned} is a continuous map. An open subset of $X_\infty$ is either given by $U \subseteq X$ open in $X$, or by $X\setminus C \cup \{\infty\}$ for $C \subseteq X$ closed and compact. We have $$ p^{-1}(X \setminus C \cup \{\infty\}) = \tilde{X} \setminus C, $$ which is open since $C \subseteq \tilde{X}$ is compact and thus closed (as $\tilde{X}$ is Hausdorff). We have $$ p^{-1}(U) = U \subseteq X \subset \tilde{X}, $$ and since $X \subset \tilde{X}$ is open by assumption it follows that $U \subset \tilde{X}$ is also open.