MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $K$ be an object of $\mathbf{CHaus}$, the category of compact Hausdorff spaces, and $K \xrightarrow{\ \ \sigma \ \ } K$ be an involutory morphism without fixed points. Define $C^{\sigma}(K)$ as the Banach space $\lbrace f \in C(K) | f \circ \sigma = -f \rbrace$.

Now let $p : X \twoheadrightarrow K$ be a projective resolution of $K$ in $\mathbf{CHaus}$. By projectivity there is some morphism $X \xrightarrow{\ \ \rho \ \ } X$ with $p \rho = \sigma p$. One can show that $\rho$ is an involution without fixed points, so that we have a linear isometry $i : f \in C^{\sigma}(K) \mapsto f \circ p \in C^{\rho}(X)$.

My question is : is it always true that $C^{\sigma}(K) \xrightarrow{\ \ i \ \ } C^{\rho}(X)$ is an injective hull of $C^{\sigma}(K)$ (in the category of normed linear spaces with linear contractions as morphisms) ? I already know that $C^{\rho}(X)$ is injective ; the missing part is the "essentiality" of the morphism $i$ (i.e. whenever $C^{\rho}(X) \xrightarrow{\ \ r \ \ } Z$ is a linear contraction with $ri$ isometric, then $r$ is itself isometric).

I've found this construction in Normed Linear Spaces by Mahlon M.Day, where it is shown that it is indeed the case when $C^{\sigma}(K)$ is itself injective (in this case the map $i$ is an isomorphism) ; this is the crucial step in Mahlon Day's proof of Kelley's theorem.

Another question : when $\phi : G \rightarrow Aut(K)$ is any morphism from a group of isometries on a normed linear space $V$ into the group of homeomorphisms of $K$, it makes sense to consider $C^{\phi}(K,V) := \lbrace f \in C(K,V) | \forall g \in G : g\circ f = f \circ \phi(g) \rbrace$. Do this spaces have a name ? Is there some literature on it ?

share|cite|improve this question

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.