Order between compactifications

Question

I'm reading a book that defines an order between compactifications of a space like it is shown in the image.

I don't fully understand this definition, however. If the $f$ mapping fixes all points in $ X $, doesn't it mean that $f$ is the identity function?

Answer 1

$f$ is a function whose domain is $dx$, which is strictly larger than $x$. It can fix $x$ without being the identity function everywhere else.

What is really meant is: $c$ embeds $x$ into $cx$, and $d$ embeds $x$ into $dx$. $f$ maps $dx$ to $cx$, and if $f\circ d =c$ then we say $c≤d$.

Answer 2

It seems that your book understands "compact" as compact Hausdorff (otherwise the Lemma would not be true).

Let us consider the open interval $X = (0,1)$. It has compactifications $dX = [0,1]$ with $d =$ inclusion $X \to [0,1]$ and $cX = S^1 = \{ (x,y) \in \mathbb R^2 \mid x^2 + y^2 = 1\}$ with $c : X \to S^1, d(t) = (\cos(2\pi t),\sin(2\pi t))$. Note that we identity $X$ with its image $c(X) \subset S^1 = cX$. The map $f : dX \to cX, f(t) = (\cos(2\pi t),\sin(2\pi t))$, is a continuous surjection which fixes all points of $X$, but is certainly not the identity.

Observe that given compactifications $cX, dX$, there exists at most one continuous $f : dX \to cX$ fixing the points of $X$. If $f_1, f_2$ are two such maps, then $A = \{ x \in dX \mid f_1(x) = f_2(x) \}$ is closed in $dX$ because $dX$ is Hausdorff. But $X \subset A$, hence $dX = \overline{X}^{dX} \subset A$.

We can alternatively define $cX \le dX$ iff there exists a continuous $f : dX \to cX$ which fixes the points of $X$. The difference to the definition in your question is that we do not require $f$ to be onto.

We can now show that $f$ is automatically surjective (recall that we deal with Hausdorff compactifications). In fact $f(dX)$ is a compact subspace of $cX$ which contains $X$, thus a closed subspace containing $X$. Since $X$ is dense in $cX$, we get $cX = \overline{X} \subset f(dX)$.