# When $C(X)$ is an injective $C(X)$-module? Current answer is erroneous

It is an old question if every injective Banach space is isomorphic as Banach space to $C(X)$-space.

I would like to know if the weakened module version of this question is answered. More precisely: For which compact Hausdorff spaces $X$ the module $C(X)$ is an isomorphically injective $C(X)$-module. I know that for $X$ Stonean $C(X)$ is even an isometrically injective $C(X)$-module.

• Is this relative injectivity as in Helemskii's theory, or the stricter version? – Yemon Choi Dec 6 '14 at 22:47
• In other words: are we testing over what Helemskii calls the admissible embeddings, or just the embeddings with closed range? – Yemon Choi Dec 6 '14 at 22:49
• The stricter version, where embeddings have closed range. – Norbert Dec 6 '14 at 22:58
• I think this version is usually called "strict injectivity", at least in Helemskii's book. So just to clarify: you require that whenever $M$ is a closed sub-module of a Banach $C(X)$-module $N$, every bounded $C(X)$-module map $M\to C(X)$ has an extension to a bounded $C(X)$-module map $N\to C(X)$. Is that correct? – Yemon Choi Dec 6 '14 at 23:02
• If you use the absolutely strictest version of injectivity (viewing C(X) as a ring, then saying it is an injective module in the ring-theoretic sense---which is how I interpreted the question initially), then the answer when $X$ is compact is simply that $X$ be finite (since a commutative unital ring with no nilpotents, on being injective as a module, must be von Neumann regular (aka absolutely flat) ...) – David Handelman Dec 7 '14 at 0:08

It's equivalent to that $C(X)$ is isometrically ijective (i.e., $X$ is stonean). We take the definition cited by Yemon in the comment and let $\iota\colon C(X)\hookrightarrow C(Y)$ be a faithful $*$-homomorphism from $C(X)$ into an injective abelian $C^*$-algebra $C(Y)$. Then, there is a $C(X)$-module projection $\Phi$ from $C(Y)$ onto $C(X)$. In particular, $C(X)$ is isomorphically injective, but Huruya'e example (Proc AMS 1984) says isomorphic injectivity need not imply isometric injectivity (though his example is isomorphic as a Banach space to an isometrically injective one). We will exploit the fact that the projection $\Phi$ is a $C(X)$-module map. Let $\Phi^*\colon X \to C(Y)^*$ be the continuous map define by $\langle\Phi^*(x),f\rangle = \Phi(f)(x)$. Here $C(Y)^*$ is equipped with the weak$^*$-topology. Note that $\|\Phi(x)\| \geq 1$ for every $x\in X$, and define a continuous map $\Psi^*\colon X\to C(Y)^*$ by $\Psi^*(x)=|\Phi^*(x)|/\|\Phi^*(x)\|$. This in turn gives rise to a unital positive contractive map $\Psi\colon C(Y) \to C(X)$, which is given by $\Psi(f)(x) = \langle\Psi^*(x),f\rangle$. We claim that $\Psi$ is a projection onto $C(X)$. Let $h\in C(X)_+$ be given arbitrary. Since $\Phi(\iota(h)f)(x)=h(x)\Phi(f)(x)$ for all $f\in C(Y)$, one has $\iota(h)\Phi^*(x)=h(x)\Phi^*(x)$ as an element in $C(Y)^*$ (which is also viewed as a complex Radon measure on Y). Since $\iota(h)\geq0$, this implies that $\iota(h)|\Phi^*(x)|=|\iota(h)\Phi^*(x)|=h(x)|\Phi^*(x)|$ and so $\iota(h)\Psi^*(x)=h(x)\Psi^*(x)$ for every $x\in X$. This means that $\Psi(\iota(h))(x)=h(x)$ for every $x \in X$.
• Unfortunately, I do not understand one step of your proof. Why is $\Psi^*$ continuous? The continuity of $\Phi^*$ is clear, but the operation of total variation doesn't preserve weak convergence of measures so I don't see why $\Psi^*(x)=\frac{|\Phi^*(x)|}{\Vert\Phi^*(x)\Vert}$ must be continuous. Could you explain that step? – Norbert Feb 2 '15 at 20:12