Direct proof of "K is projective iff C(K) has the Hahn-Banach property" ?

Question

An object $X$ of a given category is called projective if for each morphism $f : X \rightarrow Z$, and each epimorphism $ g : Y \twoheadrightarrow Z$, there is a morphism $h : X \rightarrow Y$ such that $f=gh$.

An ordered vector space $X$ is said to have the Hahn-Banach extension property if for each real vector space $Y$, each subspace $Z$ of $Y$, each sublinear operator $V : Y \rightarrow X$, and each linear operator $T : Z \rightarrow X$ satisfying $T \leq V_{|Z}$ pointwise, there is a linear operator $\hat T : Y \rightarrow X$ with $\hat T_{|Z}=T$ and $\hat T \leq V$.

A theorem of Gleason asserts that a compact (Hausdorff) space $K$ is projective in the category $\mathbf{CHaus}$ of compact Hausdorff spaces iff $K$ is extremally disconnected. By results of Goodner and Nachbin, the ordered vector space $C(K)$ has the Hahn-Banach extension property iff $K$ is extremally disconnected. It is thus known that: $K$ is projective in $\mathbf{CHaus}$ iff $C(K)$ has the Hahn-Banach extension property $(\star)$.

But : the Hahn-Banach extension property is a kind of projectivity with reversed arrows (reflecting the contravariance of the functor which sends maps $f : X \rightarrow Y$ to $u \in C(Y) \mapsto u \circ f \in C(X)$ ), so that it is reasonable to expect a "direct proof" of $\star$. It is relatively easy to show the direction "$\rightarrow$" in the equivalence (see below), and I would be very happy if the other direction also had a "direct proof" ... Thanks in advance.

Proof of the direction "$\rightarrow$" in $\star$ : Let $K$ be a projective object of $\mathbf{CHaus}$, and $(Y, Z, V, T)$ as in the definition of the Hahn-Banach property. Following Rainwater, let $D$ be the set $K$ endowed with the discrete topology and $r : \beta D \twoheadrightarrow K$ be the (unique) continuous extension of the canonical map $D \rightarrow K$ to the Stone-Cech compactification $\beta D$ of $D$. By projectivity, there is some $i : K \rightarrow \beta D$ with $id_{K} = r \circ i$. Without loss of generality $i$ is an inclusion map. Then $r$ is a retraction of $\beta D$ onto $K$. The maps $r$ and $i$ induce (by the functor described above) maps $\tilde r : C(K) \rightarrow C(\beta D)$ and $\tilde i : C(\beta D) \rightarrow C(K)$. As a consequence of the usual Hahn-Banach theorem, the space $C(\beta D) \simeq \ell^\infty(D)$ has the Hahn-Banach extension property ; applying this to $(Y,Z,\tilde r \circ V, \tilde r \circ T)$ yields an operator $\hat T'$ with $\hat T'_{|Z}=\tilde r \circ T$ and $\hat T'\leq \tilde r \circ V$. The desired operator is $\hat T = \tilde i \circ \hat T'$.

Answer 1

As indicated by Theo Buehler above, Fremlin proves what I want and much more in his book. However, the proof given in this reference can be simplified a lot in my setting, so that I can answer my own question by giving a relatively simple proof :

Basically, Fremlin's proof begins by finding a projective resolution of $K$ inside the dual ball of $C(K)$. But the existence of a projective resolution in $\mathbf{CHaus}$ has a short proof (see this article), so I use it freely below.

Proof of the direction "$\leftarrow$" in $\star$ : Let $X$ be a projective resolution of $K$ in $\mathbf{CHaus}$, and let $p : X \twoheadrightarrow K$ be the corresponding map, which satisfies $p(S) \neq K$ whenever $S$ is a proper closed subspace of $X$. Then $p$ induce an isometry $\tilde p : f \in C(K) \rightarrow f \circ p \in C(X)$. Since $\mathrm{Im} \; \tilde p$ is isometric to $C(K)$, which has the Hahn-Banach extension property, we can extend the identity on $\mathrm{Im} \; \tilde p$ to a norm-one operator $T : C(X) \rightarrow \mathrm{Im} \; \tilde p$.

• Let $h$ be in the unit ball of $C(X)$, and set $S=\overline{ \lbrace h \neq 0 \rbrace } \cup \lbrace Th = 0 \rbrace$. If $S$ is a proper subspace of $X$ then some $p(x) \in K$ doesn't belong to $p(S)$. Let $f \in C(K,\[ 0,1 \])$ be such that $f \circ p(x) = 1$ and $f_{|p(S)}=0$. Then $\lVert \tilde p (f) \pm h \rVert \leq 1$, so that $\lVert \tilde p (f) \pm Th \rVert = \lVert T( \tilde p (f) \pm h )\rVert \leq 1$. But for an appropiate choice of sign, the value of $\tilde p (f) \pm Th$ at $x$ exceeds $1$, a contradiction. Thus $S=X$.

• Let $x_1$ and $x_2$ be distinct points in $X$, hence separated by disjoint closed sets $F_1$ and $F_2$ in $X$. Let $h \in C(X,\[ 0,1 \])$ be such that $h_{|F_1}=0$ and $h_{|F_2}=1$. Then $\overline{ \lbrace h \neq 0 \rbrace } \subset {}^c( \mathring F_1)$, so that by the point above $x_1 \in \mathring F_1 \subset \lbrace Th = 0 \rbrace$, and $Th(x_1)=0$. Similarly, $Th(x_2)=1$. Since $Th \in \mathrm{Im} \; \tilde p$, this shows that $p(x_1) \neq p(x_2)$.

We have shown that $p$ is injective, so that $p$ is a homeomorphism $X \simeq K$. Thus $K$ is projective.

PS: Thanks for the Latex,Theo.

Answer 2

Maybe I made a stupid mistake, but I think something along the following lines should work:

Let $1_K$ be the standard order unit of $C(K)$. There is a canonical identification of $K$ with the subset of $\Phi \subset \mathbb{R}^{C(K)}$ (with the product topology) consisting of Riesz homomorphisms $\varphi \colon C(K) \to \mathbb{R}$ such that $f(1_K) = 1$ (the identification of $K$ with $\Phi$ is given by sending $k \in K$ to evaluation at $k$).

From this we get a functorial map from unit-preserving Riesz homomorphisms $\varphi\colon C(K) \to C(L)$ to continuous functions $f^{\ast} \colon L \to K$ by precomposition $f^\ast(\varphi) = \varphi\circ f$.

Consider the inclusion $i: C(K) \to \ell^{\infty}(K) = C(\beta K_\delta)$ (it corresponds to Rainwater's map $i^\ast\colon\beta K_{\delta} \to K$ where $K_{\delta} = D$ is $K$ with the discrete topology). The map $V\colon \ell^\infty(K)\to C(K), h \mapsto \lVert h\rVert_{\infty} 1_{K}$ is sublinear and it dominates the identity $C(K) \to C(K)$. Since $C(K)$ has the Hahn-Banach property, the identity extends to a left-inverse $l$ of $i$.

However, $l^\ast \colon K \to \beta K_\delta$ is a continuous function such that $f^\ast l^\ast = (lf)^{\ast} = 1_{K}$ whence $K$ is a retract of the projective compact space $\beta K_\delta$, so $K$ is projective, too.