A countable X having a closed subspace A with Cp(A) not a factor of Cp(X)

Let A be a countable space such that the function space Cp(A) is analytic. We prove that there exists a countable space X such that X contains A as a closed subset and the function space Cp(X) is an absolute Fσδ-set. Therefore, if Cp(A) is analytic non-Borel then Cp(A) is not a factor of Cp(X) and there is no continuous (or even Borel-measurable) extender e: Cp(A) → Cp(X) (i.e., a map such that e(f)¦A = f, for f ϵ Cp(A)). This answers a question of Arkhangelʹskiĭ. o construct a countable space X such that the function space Cp(X) is an absolute Fσδ-set and X contains closed subsets A with Cp(A) of arbitrarily high Borel complexity (or even analytic non-Borel).