A COUNTABLE X HAVING A CLOSED SUBSPACE A WITH C-P(A) NOT A FACTOR OF C-P(X)

Let A be a countable space such that the function space C-p(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 C-p(X) is an absolute F-sigma delta-set, Therefore, if C-p(A) is analytic non-Borel then C-p(A) is not a factor of C-p(X) and there is no continuous (or even Borel-measurable) extender e: C-p(A) -->, C-p(X) (i.e., a map such that e(f)\A = f, for f is an element of C-p(A)). This answers a question of Arkhangel'skii. We also construct a countable space X such that the function space C-p(X) is an absolute F-sigma delta-set and X contains closed subsets A with C-p(A) of arbitrarily high Borel complexity (or even analytic non-Borel).