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 $\hat T = \tilde i \circ \hat T'$.

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    $\begingroup$ Your definition of "projective" doesn't actually use the object $X$. $\:$ $\endgroup$
    – user5810
    Nov 1, 2012 at 21:11
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    $\begingroup$ Out of curiosity, which authors were "assisting Rainwater" in this particular paper? $\endgroup$
    – Yemon Choi
    Nov 2, 2012 at 3:01
  • $\begingroup$ @Ricky Demer: Corrected, thanks. @Yemon Choi: I'm not sure to understand the question. $\endgroup$
    – js21
    Nov 2, 2012 at 4:56
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    $\begingroup$ You can find a relatively short proof in Fremlin's volume 3, 363R: essex.ac.uk/maths/people/fremlin/mt.htm. A very good and detailed reference is Albiac-Kalton's book, see also Day's book on normed spaces for a lot of historical references. I wouldn't call the arguments there direct, though. they certainly are much more involved than the implication you prove. mr x's idea is the basic starting point, but it needs to be refined quite abit, as far as I can tell. Finally, Yemon is alluding to the multiple personalities of John Rainwater: en.wikipedia.org/wiki/John_Rainwater $\endgroup$ Nov 2, 2012 at 8:05
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    $\begingroup$ Slightly off-topic, but to satisfy Yemon's curiosity: This is Rainwater's second paper (linked to in the answer), and according to Phelps at.yorku.ca/t/o/p/d/47.htm Rainwater's assistant was Isbell: en.wikipedia.org/wiki/John_R._Isbell $\endgroup$ Nov 2, 2012 at 18:08

2 Answers 2


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.

  • $\begingroup$ Very nice: that seems very close to optimal. I took the liberty to make some minor edits so that braces, etc. are displayed properly. I didn't change the mathematics. $\endgroup$ Nov 2, 2012 at 12:54

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.

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    $\begingroup$ Two minor typos, I think : $\phi$ should be $f$ in the third paragraph, and $f=i$ in the last one. How to ensure that $l$ is indeed a homomorphism (it is a linear unit-preserving map, but it seems multiplicativity is missing) ? $\endgroup$
    – js21
    Nov 2, 2012 at 5:17

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