Let $X, Y$ be metric spaces, and let $PX$ (resp. $PY$) be the set of all non-empty compact subsets of $X$ (resp. $Y$) with the Hausdorff metric. A continuous map $f\colon X\to Y$ induces a continuous map $Pf\colon X\to Y$, and therefore we have an assignment \begin{equation} \mathrm{map}(X,Y)\longrightarrow \mathrm{map}(PX,PY). \end{equation} This assigment is a continuous map with repect to the compact-open topology.

Which reference should I cite for the last assertion?

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    $\begingroup$ the following (more or less random) paper is related and one of the references that it lists might work (e.g. perhaps G. Beer, Topologies on Closed and Closed Convex Sets, Kluwer, Dordrecht (1993)). Just google hyperspace compact-open topology sciencedirect.com/science/article/pii/S0166864199002047 $\endgroup$ – Mirko Nov 28 '14 at 17:37
  • $\begingroup$ All I've found is the following article: "The embedding of a mapping space with compact open topology", by Takemi Mizokami (1998), where it is proven that the map is continuous when $map(PX,PY)$ is endowed with the pointwise convergence topology. $\endgroup$ – Federico Cantero Dec 2 '14 at 16:05

I haven't found any reference. Moreover, the fact that Mizokami's article

"The embedding of a mapping space with compact open topology", Topol. Appl. 1998

proves it for the pointwise convergence topology on the space $\mathrm{map}(PX,PY)$ together with the fact that no paper citing his paper proves my original question makes me think that it is not written in the literature. Here is a proof:

The map $\phi\colon \mathrm{map}(X,Y) \to \mathrm{map}(PX,PY)$, where both spaces are endowed with the compact-open topology, is continuous if $X$ is regular.

Proof: It suffices to prove that for each $f\in \mathrm{map}(X,Y)$ and for each subbasic neighbourhood $Q$ of $\phi(f)$, there is a subbasic neighbourhood $P$ of $f$ such that $\phi(P) \subset Q $.

Let us recall first how the topology in the righthandside is defined:

A basic open subset in $PY$ is given by a finite sequence $\langle U_i\rangle$ of open subsets $U_i$ of $Y$. A compact subspace $K\subset PY$ belongs to this open subset if $K\subset \bigcup U_i$ and $K\cap U_i\neq \emptyset$ for all $i$.

By Theorem 2.5 in "Topologies of spaces of subsets", by Ernest Michael, a compact subset ${\mathcal K}$ of $PX$ satisfies that the union $\bigcup_{K\in {\cal K}} K$ is compact as well.

Therefore a basic open subset of $\mathrm{map}(PX,PY)$ is given by pairs $W({\cal K}:\langle U_i\rangle)$ with $$ W({\cal K}:\langle U_i\rangle) = \{F:PX\to PY\mid F(K)\subset \langle U_i\rangle\} = \{F\mid F(\bigcup_{K\in {\cal K}} K)\subset \bigcup U_i, F(K)\cap U_i\neq \emptyset, \forall K\in {\cal K}, \forall i\} $$ Let us take such subbasic neighbourhood of $\phi(f)$.

  • We assert that for each $U'_i:= f^{-1}(U_i)$ there is an open subset $V:=V_i$ such that $\bar{V}_i\subset U_i$ and for each $K\in {\cal K}$ it holds that $K\cap V_i\neq \emptyset$.

    If this were not true, for each such $V$ there would exist a $K_{V}\in {\cal K}$ such that $K_V\cap V=\emptyset$. This defines a net in ${\cal K}$ which has a convergent cofinal subnet because $\cal K$ is compact. The limit point $K_0\in {\cal K}$ of this subnet satisfies that $K_0\cap V = \emptyset$ for all such $V$, and since the subnet is cofinal and $X$ is regular, it also holds that $K_0\cap U_i'=\emptyset$, but this contradicts the fact that $f(K_0)\cap U_i\neq \emptyset$.

Let us choose $V_i\subset U_i'$ with the above property. Take now a point $p_K^i\in V_i\cap K$ for each $i$ and $K$, and define $K_i$ to be the closure of $\{p_K^i\}$, which is compact because it is a closed subset of $\bigcup_{K\in {\cal K}} K$ (which is itself comapact). By construction it holds that $f(K_i)\subset \bar{V}_i\subset U_i$. Therefore the neighbourhood $$ W(\bigcup_{K\in {\cal K}} K: \bigcup_i U_i)\cap \bigcap_i W(K_i:U_i) $$ of $f$ is mapped into the neighbourhood $W({\cal K}:\langle U_i\rangle)$, and therefore $\phi$ is continuous.


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