Does one corona set project onto the other?

Question

Let $X$ be a locally compact Hausdorff space. By a compactification of $X$, let us understand a pair $(C,\iota)$ consisting of a compact Hausdorff space $C$ and a topological embedding $\iota : X \hookrightarrow C$ such that $\iota(X)$ is dense in $C$.

Given two compactifications $(C_1,\iota_1)$ and $(C_2,\iota_2)$ of $X$, we can ask whether there exists a continuous mapping $\pi : C_1 \to C_2$ such that $\pi \circ \iota_1 = \iota_2$. If such a $\pi$ exists, it is unique, and necessarily a quotient map. Roughly, one thinks of $C_1$ as being the "larger" of the two compactifications.

My internal picture of how the quotient map $\pi$ works is that that it fixes the copy of $X$ and, meanwhile, collapses down the remaining points a bit. I just realized, however, that I have never ruled out the following possibility:

Question: Is it automatically true that, if $x \in C_1 \setminus \iota_1(X)$, then $\pi(x) \in C_2 \setminus \iota_2(X)$? In other words, must $\pi$ necessarily map the "corona set" of $C_1$ onto the corona set of $C_2$?

Answer 1

I think that your conjecture is true. Suppose in fact $x∈C_1\setminus ι_1(X)$. Being that $X$ is dense and $C_1$ Hausdorff, you can completely identify $x$ with a local base of open sets in $X$ (the intersections of the open neighborhoods with $X$). Not all families of open sets will define a point of $C_1$ of course.

Now suppose that $y = \pi(x)\in X$. Take another preimage $x'\in X, \pi(x') = y$. The preimages of the neighborhoods of $y$ are the neighborhoods of $x'$, thus $x'$ and $x$ are not separated, so $C_1$ is not Hausdorff.

This lacks a few "it's trivial to prove that" steps, but it's a start.

Answer 2

In the comments section of rewritten's answer, I used a net-based argument to fill in the details of rewritten's sketch. Later I went back wrote down the details again, avoiding the use of nets. Thought I'd add that argument here.

Proposition: Suppose $X$ and $Y$ are Hausdorff spaces with dense subspaces $A \subset X$ and $B \subset Y$. Suppose $\pi : X \to Y$ is a continuous mapping whose restriction to $A$ is a homeomorphism of $A$ onto $B$. Then, $x \in X \setminus A$ implies $\pi(x) \in Y \setminus B$.

Proof: Suppose that $x \in X$ is such that $\pi(x) = b \in B$. We know there is a (unique) $a \in A$ with $\pi(a) =b$. We want to argue that $x = a$. This we achieve by showing that $x$ and $a$ cannot be separated by open neighbourhoods. To this end, let $U,V \subset X$ be open sets with $x \in U$ and $a \in V$. Since $V \cap A$ is open in $A$, $\pi(V \cap A)$ is open in $B$. Thus, there is an open set $V' \subset Y$ with $V' \cap B = \pi(V \cap A)$. Since the goal is to show $V \cap U \neq \varnothing$, it does no harm to replace $U$ with a smaller neighbourhood of $x$. Since, $\pi(x) = \pi(a) = b \in V'$, we can replace $U$ with $\pi^{-1}(V') \cap U$ so that now $\pi(U) \subset V'$. Now, take any $a' \in U \cap A$ (recall $A$ is dense in $X$). Since, $\pi(a') \in V' \cap B =\pi(V \cap A)$, we get $a' \in V$, so $V \cap U$ is nonempty.