Is there an operation in topology analogous to the operation of averaging over a compact subgroup in harmonic analysis?

Question

Let me start with the following

Illustration: Let $G$ be a compact group, and let $\pi:G\to H$ be its (surjective) continuous homomorphism onto a (compact) group $H$. So we can think that $H$ is the quotient group of $G$ modulo the kernel $K=\operatorname{Ker}\pi$ of the mapping $\pi$ $$ H=G/K $$ (and $\pi$ is just the quotient mapping).

The map $\pi$ generates a (linear and continuous) mapping of the spaces of continuous functions: $$ P:C(H)\to C(G). $$ This operator is a coretraction in the category of Banach spaces (and, what is the same in this situation, in the category of locally convex spaces), since it has a left inverse operator, $$ S:C(G)\to C(H),\qquad S\circ P=\operatorname{id}_{C(H)} $$ defined by the formula $$ S(f)(t)=\int_K f(t\cdot s)\ \mu_K(d s),\qquad f\in C(G) $$ where $\mu_K$ is the normalized Haar measure on $K$. (This mapping turns each function $f\in C(G)$ into a function $S(f)\in C(G)$, invariant with respect to the shifts by the elements of $K$, and this means that $S(f)$ can be considered as a function on $H=G/K$).

Now the

Question: Let $X$ be a (Hausdorff) compact space, and let $\pi:X\to Y$ be a (surjective) continuous mapping onto a (compact) metrizable space $Y$. So we can think that $Y$ is the quotient space of $X$ modulo the equivalence relation $$ x\sim x' \quad\Leftrightarrow \quad \pi(x)=\pi(x') $$ (and $\pi$ is just the quotient mapping).

The map $\pi$ generates a (linear and continuous) mapping of the spaces of continuous functions: $$ P:C(Y)\to C(X). $$

Is this operator a coretraction in the category of Banach spaces (or, what is the same here, in the category of locally convex spaces)?

In other words, does there exist an operator $$ S:C(X)\to C(Y), $$ such that $$ S\circ P=\operatorname{id}_{C(Y)} $$ ?

In my considerations the space $Y$ is metrizable, but I don't know, perhaps this condition is extra. Similarly, I don't know, perhaps the space $X$ need not to be compact, but just locally compact, or belong to some wider class.

Answer 1

As stated (i.e., without assuming metrizability of $X$), the answer is negative: Let $Y=\alpha\mathbb N$ be the Alexandrov (one point) compactification, $X=\beta\mathbb N$ the Stone-Cech compactification and $\pi:X \to Y$ the unique continuous extension to $\beta\mathbb N$ of the inclusion $\mathbb N\hookrightarrow\alpha\mathbb N$. $C(Y)$ is the space $c$ of convergent sequences whereas $C(X)=\ell^\infty$ is the space of bounded sequences and $P$ is the embedding $c\to\ell^\infty$. Phillips's theorem ($c_0$ is not complemented in $\ell^\infty$) implies that $P$ does not have a continuous linear left inverse.

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If $X$ is metrizable, such structural arguments are not very promising: On the one hand, Sobczyk's theorem ($c_0$ is complemented in every separable superspace) covers the case where $C(Y)$ is isomorphic to $c_0$ and on the other hand, Miljutin's theorem says that $C(X)$ is isomorphic to $C([0,1])$ for every uncountable compact metric space $X$.