Algebraic Topology – Existence of Stone-?ech Compactification

algebraic-topologycompactification

i have a question regarding the Stone–Čech compactification of some topological space.

On Wikipedia Page, it says "A form of the axiom of choice is required to prove that every topological space has a Stone–Čech compactification", but in a lot of texts regarding this, its always stated that a topological space must at least be Tychonoff, in order to have any compactification. Whats up with that?

Can anyone confirm or give an explanation on how one can see that any topological Space $X$ can have a Stone–Čech compactification? (Or is Wikipedia just spewing nonsense in this case?)

Best Answer

The wikipedia article is misleading. The definition is

The Stone–Čech compactification of the topological space $X$ is a compact Hausdorff space $βX$ together with a continuous map $i_X : X → βX$ that has the following universal property: any continuous map $f : X → K$, where $K$ is a compact Hausdorff space, extends uniquely to a continuous map $βf : βX → K$, i.e. $(βf)i_X = f$.

In the diagram below, however, $i_X$ is depicted as an arrow $$X \hookrightarrow \beta X$$ which suggests that $i_X$ is an embedding as a subspace. This is not required in the definition, although it uses the phrase "$f$ extends to $\beta f$" (normally a map is extended from a subspace to the the total space).

Later Wikipedia correctly states

  • The map from $X$ to its image in $βX$ is a homeomorphism if and only if $X$ is Tychonoff.

So what is going on here?

The map $i_X : X → βX$ always has the above universal property (for arbitrary $X$), but for non-Tychonoff $X$ it is not a compactification of $X$.

Here is the definition of a compactification:

An embedding of a topological space $X$ as a dense subset of a compact space is called a compactification of $X$.

Therefore it is not really adequate to call $i_X : X \to \beta X$ the Stone–Čech compactification for arbitrary $X$. Perhaps one should better call it the Stone–Čech map.

Remark.

Using the Stone–Čech map, we can prove the following well-known

Theorem. Each topological space $X$ has a Tychonoffication. This is a map $p_X : X \to \tau X$ to a Tychonoff space $\tau X$ that has the following universal property: any continuous map $f : X → T$, where $T$ is a Tychonoff space, admits a unique continuous map $\tau f : \tau X → T$ such that $\tau f \circ p_X = f$.

Proof. Let $\tau X = i_X(X) \subset \beta X$ and $p_X: X \xrightarrow{i_X} \tau X$. Let $\iota_X \tau X \to \beta X$ denote inclusion. Given any continuous map $f : X \to T$ to a Tychonoff space, we know that there exists a unique continuous map $\phi : X \to \beta T$ such that $\phi \circ i_X = i_T \circ f$. We have $$\phi(\tau X) = \phi(i_X(X)) = i_T(f(X)) \subset i_T(T) = \tau T .$$ Since $\iota_T$ is an embedding, we get a continuous map $$\tau f = \iota_T^{-1} \circ \phi \circ \iota_X : \tau X \xrightarrow{\phi} \tau T \xrightarrow{i_T^{-1}} T .$$ Then $$\tau f \circ p_X = \iota_T^{-1} \circ \phi \circ \iota_X \circ p_X = \iota_T^{-1} \circ \phi \circ i_X = i_T^{-1} \circ i_T \circ f = f .$$

Let $\psi : \tau X \to T$ be any map such that $\psi \circ p_X = f$. We get

$$i_T \circ \psi \circ p_X = i_T \circ f = \phi \circ i_X = i_T \circ i_T^{-1} \circ \phi \circ \iota_X \circ p_X = i_T \circ \tau f \circ p_X .$$

Since $i_T$ is injective, we get $\psi \circ p_X = \tau f \circ p_X$. Since $p_X$ is surjective, we get $\psi = \tau f$. $\phantom{xx}\square$

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