Let $T\colon X\to Y$ be a continuous linear surjection between Frechet spaces. Then it is possible to use Michael's selection theorem to show that there exists a continuous (non-linear) function $g\colon Y\to X$ such that $g\circ T = {\rm id}_X$.

I am wondering whether it is possible to obtain a weak$\ast$-continuous counterpart of this theorem. Namely, what if $X$ and $Y$ are strong biduals of Frechet spaces and $T$ is weak*-continuous. Can we produce such a $g$ that would be weak$\ast$-continuous?

The relevant result I quoted is Corollary 7.1 in Bessaga and Pełczyński's, Selected topics in infinite-dimensional topology.

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    $\begingroup$ I think you mean right inverses which always exist for quotient mappings between Fréchet spaces. The trivial example of the $0$ mapping $T:\mathbb R \to \{0\}$ shows that in many cases there are no left inverses. $\endgroup$ Oct 5, 2020 at 6:41
  • $\begingroup$ I doubt that there are always weak$^*$-continuous right inverses even for duals of Banach spaces. Assume that $X$ is a closed subspace of a Banach space $Y$ and assume that the quotient (restriction) map $q:Y'\to X'$ has a weak$^*$ continuous right inverse $r$. For every sequence $f_n$ of functionals on $X$ such that $f_n(x)\to 0$ pointwise on $X$ the sequence $r(f_n)$ would be a pointwise convergent sequence of extensions to $Y$ -- I don't believe that this is always possible and I am quite optimistic that the Banach spacers on MO can provide counterexamples. $\endgroup$ Oct 5, 2020 at 10:15
  • $\begingroup$ I have overlooked the "bi" of biduals. Just for duals of Banach spaces a counterexample is the restriction $\ell_\infty' \to c_0'$ which does not have a weak$^*$-continuous right inverse. $\endgroup$ Oct 5, 2020 at 13:28


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