Homeomorphic but not equivalent compactifications.

Question

I stumbled upon the definition of equivalent compactifications which is:

Two compactifications $Z_1$ and $Z_2$ of the space $X$, are said to be equivalent if there exists a homeomorphism $h:Z_1\rightarrow Z_2$ such that $h(x)=x$ for every $x$ in $X$.

I find it interesting that this definition includes the "$h(x)=x$ for every $x$ in $X$" part as it seems to me very weird to be able to compactify a space $X$ in two different ways, have them be homeomorphic, but have no homeomorphism that fixes every point of $X$. I however know very few examples of compactifications which might explain why I can't produce a counterexample to this.

My question is: Can you give me an example of a space $X$ and two compactifications of $X$ such that they are homeomorphic but not equivalent?

P.S. The book that I'm reading (Munkres) includes Hausdorff in its definition of compactification, which I don't know if it is usual or important here. It goes: If $Y$ is a compact Hausdorff space and $X$ is a proper subspace of $Y$ whose closure equals $Y$, then $Y$ is said to be a compactification of $X$.

Answer 1

Since compactifications also depend on how the embedding is done, we can cheat a little bit (or a lot). Consider the space $X = (0,1) \times \{ 0, 1 \}$; that is, two disjoint copies of the open unit interval.

Two inequivalent but homeomorphic compactifications of $X$ are $$\begin{align} Z_1 &= ( S^1 \times \{ 0 \} ) \cup ( [0,1] \times \{ 1 \} ), \\ Z_2 &= ( [0,1] \times \{ 0 \} ) \cup ( S^1 \times \{ 1 \} ). \end{align}$$ where $S^1$ is the one-point compactification of $(0,1)$, aka the unit circle.

---

For an example using the stronger form of equivlanece from David C. Ullrich's answer, take the original space to be $$X = ( (0,1) \times \{ 0 \} ) \cup ( ( ( 0,1) \cap \mathbb Q ) \times \{ 1 \} )$$ (only use the rationals in the second copy), but the same compactifications $Z_1,Z_2$ as above.

Answer 2

Seems to me the definition should require just that $h|_X$ be a homeomorphism of $X$ instead of requiring it to be the identity on $X$. Examples exist with that revised definition, but I can't think of one.

With the given definition it's easy to give an example. For complex numbers $a$ and $b$ let $(a,b)$ denote the open line segment with endpoints $a$ and $b$. Similarly for $[a,b]$. Let $X=(-1,1)\cup (-i,i)$, with the standard topology.

Let $Y=[-1,1]\cup[-i,i]$, the closure of $X$ in the plane. Let $C_1$ be $Y$ but with the two points $1,i$ identified to a single point, and the two points $-1,-i$ identified to a single point. So $C_1$ is homeomorphic to a figure-eight. Let $C_2$ be $Y$ except with $1$ and $-i$ identified to a point and $-1$ and $i$ identified to a point.

Answer 3

The usual definition of compactifications is actually a bit more general than that: more formally a compactification of a space $X$ is a pair $(i,Y)$, where $Y$ is compact Hausdorff and $i:X \rightarrow Y$ is an embedding (so $i: X \rightarrow i[X] \subseteq Y$ is a homeomorphism) such that $i[X]$ is dense in $Y$.

Two compactifications $(i,Y)$ and $(j,Z)$ are defined to be equivalent compactifications of $X$ iff there exists a homeomorphism $h: Y \rightarrow Z$ such that $h \circ i = j$ on $X$, so a homeomorphism that preserves the way that $X$ embeds into $Y$, essentially.

Some texts require that $X \subseteq Y$ and $X$ be dense, so they identify $X$ with its embedded form $i[X]$, and then the homeomorphism must be the identity (instead of going via the two embeddings) on $X$. But in practice a compactification is often constructed via embeddings (like the Cech-Stone one) or in such a way that there is a natural embedding (identifying fixed ultrafilters with the original points e.g.).

It is indeed possible to have two compactifications where the compact spaces are homeomorphic but the embeddings are not preserved and Ullrich's example in the plane is a good example (where the embeddings are both the identity).