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$\ell$- and $t$-equivalence

Topologists are interested in the classification of topological spaces. Classifying topological spaces up to homeomorphism or homotopy type etc., is the ultimate goal for a topologist. We are interested here in classifying spaces $X$ up to homeomorphism or linear-homeomorphism type of their function spaces $C_p(X)$. In its full generality this program would be much too complicated and the complete picture is presently beyond our reach. But for function spaces of low Borel complexity some definitive results are known, and it is our aim to discuss them here.

We say that spaces $X$ and $Y$ are $\ell$-equivalent provided that $C_p(X)$ and $C_p(Y)$ are linearly homeomorphic. Notation: $X{\sim}_{\ell} Y$.

Homeomorphic spaces are obviously $\ell$-equivalent. But the converse need not be true. Let $X=[0,1]\cup [2,3]$ and $Y=[0,2]\cup\{3\}$. Then evidently, $X$ and $Y$ are not homeomorphic. However, they are $\ell$-equivalent. Indeed, define $\Phi\rightarrow C_p(X)\to C_p(Y)$ by

\begin{displaymath} \Phi(f)(y) = \left\{ \begin{array}{ll} \par f(y) & (0\le y\... ...\le 2),\ \par f(2)-f(1) & (y=3). \par\end{array} \right. \par\end{displaymath}

Then $\Phi$ is a linear homeomorphism.

We say that $X$ and $Y$ are $t$-equivalent provided that $C_p(X)$ are $C_p(Y)$ are homeomorphic as topological spaces. Notation: $X{\sim}_t Y$.

Even for simple spaces it is in general difficult to decide whether they are $\ell$- or $t$-equivalent. By Bessaga and Pe\lczynski [8] there are countable compact spaces $X$ and $Y$ for which the Banach spaces $C(X)$ and $C(Y)$ are not linearly homeomorphic. An application of the Closed Graph Theorem shows that if $C_p(X)$ and $C_p(Y)$ are linearly homeomorphic then so are $C(X)$ and $C(Y)$ (the same linear map does the job in both cases). So the examples of Bessaga and Pe\lczynski are not $\ell$-equivalent. This suggests the question whether they are $t$-equivalent. We will come back to this below.

Arhangel$'$ski{\u{\i\/}}\kern.15em [1] proved that if $X$ is compact and $C_p(X)$ is linearly homeomorphic to $C_p(Y)$ then $Y$ is compact. As a consequence, $C_p({\mathbb R})$ and $C_p({\mathbb I})$ are not linearly homeomorphic. But they are homeomorphic, as was shown by Gul'ko and Khmyleva [19].

Results in the same spirit were obtained by various authors. Pestov [28] proved that if $C_p(X)$ and $C_p(Y)$ are linearly homeomorphic then $X$ and $Y$ have the same dimension. So $C_p({\mathbb I})$ and $C_p({\mathbb I}^2)$ are not linearly homeomorphic. Observe that by the famous result of Miljutin [24], all Banach spaces $C(X)$ with $X$ uncountable and compact metrizable are linearly homeomorphic. Hence $C({\mathbb I})$ and $C({\mathbb I}^2)$ are linearly homeomorphic, but $C_p({\mathbb I})$ and $C_p({\mathbb I}^2)$ are not. For another result in the same spirit, see Baars, de Groot and Pelant [7].

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