This post is about the locally compactness condition of the adjunction
\[Y^{X\times Z}\simeq (Y^X)^Z.\]We know that the natural evaluation map $\operatorname{ev}:Y^X\times X\to Y$ is continuous if $X$ is locally compact Hausdorff.
Example. If $X$ is not locally compact, the evaluation map $\operatorname{ev}:Y^X\times X\to Y$ might not be continuous.
Take $X=\mathbb{Q}$ and $Y=(2,\text{discrete})$. Any compact set of $\mathbb{Q}$ cannot contain any open interval $(a,b)\cap\mathbb{Q}$, hence any open set in $\mathbb{Q}$. The basis of $Y^X\times X$ consists of the sets
\[\left(\bigcap_{\text{finite}}\langle K_i,O_i\rangle\right)\times\left((a,b)\cap\mathbb{Q}\right).\]Put $I=(a,b)\cap\mathbb{Q}\neq\varnothing$ and $K=\bigcup_iK_i$. Since $I\setminus K$ is open and nonempty, it contains a nonempty clopen set $A$. The functions in $\bigcap_i\langle K_i,O_i\rangle$ take arbitrary values on $A$. So $\operatorname{ev}$ of the nonempty basic open sets are all $2$; it can’t be continuous.
Proposition. If $f:X\times Z\to Y$ is continuous, it induces a continuous map $g:Z\to Y^X$.
Proof. For $z\in Z$, $g(z)$ is defined by the composition $X\xrightarrow{(-,z)}X\times Z\to Y$, thus is continuous. To show that $g$ is continuous, we need that $g^{-1}\langle K,O\rangle$ is open. This is of course a standard result.
To prove this, take $z\in g^{-1}\langle K,O\rangle$. For every point $(x,z)\in K\times{z}\subset f^{-1}(O)$, pick an open neighborhood $U_x\times V_x$ contained in $f^{-1}(O)$. Since $K$ is compact, there is a finite subcover $\{U_{x_k}\}k$ of $K$. Then $\bigcap_kV{x_k}$ is an open neighborhood of $z$ in $g^{-1}\langle K,O\rangle$.
The converse of the above result is not true in general if $X$ is not locally compact, as suggested by the example above. (take $Z=Y^X$.) When $X$ is locally compact Hausdorff, $f$ is given by
\[Z\times X\xrightarrow{g\times\operatorname{id}}Y^X\times X\xrightarrow{\operatorname{ev}}Y,\]thus is automatically continuous.
From here, it is a simple exercise that $\psi:Y^{X\times Z}\to (Y^X)^Z$ is a homeomorphism if $X$ and $Z$ are locally compact Hausdorff. By the proposition above, $\psi$ is a well-defined injective function without any additional assumption on $X$ and $Z$. We also showed that $\psi$ is surjective if $X$ is LCH.
A natural question arises: When is $\psi$ a continuous map? We have the following interesting answer.
Proposition. If $Z$ is Hausdorff, then $\psi$ is continuous.
Proof. Suppose $f:X\times Z\to Y$ satisfies $\psi(f)\in\langle K,O\rangle$, where $K$ is a compact subset of $Z$ and $O$ is open in $Y^X$. We want to find a neighborhood of $f$ which maps into $\langle K,O\rangle$ under $\psi$.
$O$ can be written as a union of basic open sets, a finite number of which covers $\psi(f)(K)$. In detail, this means we can find (finitely many) compact subsets $\kappa_{ij}\subset X$ and open subsets $o_{ij}\subset Y$, such that
\[U_i:=\bigcap_{j\text{ finite}}\langle\kappa_{ij},o_{ij}\rangle,\] \[O':=\bigcup_{i\text{ finite}}U_i\subset O,\] \[\psi(f)\in\langle K,O'\rangle\subset\langle K,O\rangle.\]Since $K$ is compact Hausdorff, it’s normal. Hence we can use the shrinking lemma on the following finite open cover of $K$
\[(\psi(f))^{-1}U_i,\quad i\text{ finite}\]to obtain a compact cover ${A_i}_i$ of $K$ with $\psi(f)(A_i)\subset U_i$. We now have the desired neighborhood of $f$:
\[f\in\bigcap_{i,j\text{ finite}}\langle A_i\times\kappa_{ij},o_{ij}\rangle\subset\psi^{-1}\langle K,O'\rangle.\]
Post date: 2025/02/01
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