A counter-example for the exponentiation / adjoint of a function · DL Ferrario's Test Web Page

A counter-example for the exponentiation / adjoint of a function

The problem:

Let $X$, $Y$, and $Z$ be topological spaces. Let $Y^X$ denote the space of all continuous functions $X\to Y$, with the compact-open topology generated by the elements of the sub-basis $W_{K,U} = \{\varphi \in Y^X : \varphi K \subset U\}$, where $K$ ranges over all compact subsets of $X$ and $U$ ranges over over all open subsets of $Y$. Finite intersections of $W_{K,U}$ are a basis for the C.O. topology of $Y^X$.

For each function $f\from Z \times X \to Y$ there is a function $\hat f \from Z \to Y^X$ (the adjoint of $f$) defined as $$ \hat f(z)(x) = f(z,x) $$ for each $z \in X$ and each $x\in X$. Conversely, for any function $\hat g \from Z \to Y^X$ there is a function $f\from Z \times X \to Y$ defined as $f(z,x) = \hat f (z)(x)$.

If $f$ is continuous, then its adjoint $\hat f$ is continuous. But in general, $\hat f$ can be continuous while the corresponding $f$ is not. It turns out that if $Y$ is locally-compact Hausdorff, then the following proposition holds.

(1)
If $Y$ is locally compact Hausdorff, then $f\from Z\times X \to Y$ is continuous if and only if its adjoint $\hat f \from Z \to Y^X$ is continuous.

We give here a counter-example to proposition (1).

The rationals $\QQ$:

Let $\QQ$ be the space rationals, with the standard euclidean metric topology. Then the following proposition holds

(*)
Let $K\subset \QQ$ be a compact subset, and $A\subset \QQ$ an open subset. Then, there exists an open subset $V\subset \QQ$ such that $K\subset V$ and there exists an element $a\in A$ such that $a\not\in V$.

Proof: First, note that $A\not\subset K$. If on the contrary $A\subset K$, then there would be a sequence in $A$ converging to an irrational, and hence a sequence in $K$ without converging subsequences in $K$. So $A\not\subset K$ and hence there exists $a \in A$ such that $a\not\in K$. But then, it is easy to define an open subset $V$ containing $K$ but not containing $a$. (/Proof)

The Sierpinski Space $\mathbf{2}$:

Let $\mathbf{2}$ be the Sierpinski space. That is, $\mathbf{2} = \{0,1\}$ with topology given by its (open) subsets $\{ \emptyset, \{1\}, \{0,1\} \}$. The elements of $\textbf{2}^\QQ$ are characteristic functions of the open subsets of $\QQ$: a function $\varphi \from \QQ \to \textbf{2}$ is continuous if and only if $\varphi^{-1}(\{1\})$ is open in $\QQ$, and hence if and only if $\varphi=\chi_A$ with $A$ open in $\QQ$.

The compact-open topology on $\textbf{2}^\QQ$ is generated by the $W_{K,\{1\}}$, where $K$ ranges over all the compact subsets of $\QQ$, since the only non-trivial open subset of $\mathbf{2}$ is $\{1\}$. Being the union of a finite number of compact subsets of $\QQ$ a compact subset of $\QQ$, the set of $W_{K,\{1\}}$ is a basis (not just a sub-basis) for the C.O. topology of $\mathbf{2}^\QQ$. Also, [ \varphi \in W_{K,\{1\}} \iff \varphi^{-1}(1) \text{is an open subset containing $K$}, ] hence the elements of $W_{K,\{1\}}$ are all the characteristic functions of those open subsets of $\QQ$ which contain $K$.

The example:

Let $X=\QQ$, $Y=\mathbf{2}$, and $Z=\mathbf{2}^\QQ$. Then adjoint of the evaluation function [ \epsilon \from \mathbf{2}^\QQ \times \QQ \to \mathbf{2} ] (defined as $\epsilon(\varphi,q ) = \varphi(q)$ for each $q\in \QQ$ and each $\varphi \from \QQ \to \textbf{2}$) is the identity map [ \hat \epsilon = 1 \from \mathbf{2}^\QQ \to \mathbf{2}^\QQ, ] since by definition $\hat \epsilon( \varphi ) (q) = \epsilon(\varphi,q) = \varphi(q)$ for each $q$. Therefore

(2)
The adjoint function $\hat \epsilon \from Z \to Y^X$ is continuous.

Now, since the only non-trivial open subset of $\mathbf{2}$ is $\{1\}$,

(3)
The evaluation function $\epsilon \from \mathbf{2}^\QQ \times \QQ \to \mathbf{2}$ is continuous if and only if the subset [ \{ (\varphi,q) \in \mathbf{2}^\QQ \times \QQ : \varphi(q) = 1 \} ] is open on $\mathbf{2}^\QQ \times \QQ$.

Consider the element $\xi = (\chi_\QQ,0) \in \mathbf{2}^\QQ \times \QQ$. Since $\chi_\QQ(0) = 1$, $\xi \in \epsilon^{-1}( 1 ) $.

Let $A\subset \QQ$ be any arbitrary open subset such that $0 \in A$; let $K$ be an arbitrary compact subset of $\QQ$ (for which $\chi_\QQ \in W_{K,\{1\}} \iff \chi_\QQ K = 1$, and hence $\chi_\QQ \in W_{K,\{1\}}$). Then [ \xi \in W_{K,\{1\}} \times A~. ] Now, $ W_{K,\{1\}} \times A \subset \epsilon^{-1}(1) $ if and only if for each open subset $V$ of $\QQ$ such that $K\subset V$ and for each $a \in A$ it happens that [ \epsilon(\chi_V(a)) = \chi_V ( a ) = 1, ] that is $a \in V$. Hence, it is not true that $ W_{K,\{1\}} \times A \subset \epsilon^{-1}(1) $ if and only if there exists an open subset $V$ of $\QQ$ such that $K\subset V$ and an element $a \in A$ such that $a \not\in V$. By proposition (*), the conclusion follows.

(E)
The evaluation function $\epsilon$ is not continuous, while its adjoint $\hat\epsilon$ is continuous.