When is the exponential law in topology discontinuous?

Question

By $Y^X$ I mean the space of continuous functions $X \to Y$ with the compact-open topology (by compact I don't require Hausdorff. Consider the exponential law $Z^{X \times Y} \to (Z^Y)^X$ where the map is well-defined a a set map (this is not hard to see). However I don't think it's always continuous, but I don't have any examples of it being discontinuous. I know that $X$ Hausdorff implies continuous. So any counterexample must have $X$ non-hausdorff.

Answer 1

Upon further thought, I believe I can characterize all counterexamples. We define two compact splitting qualities as follows, $X$ weakly compact-splits if for every $C \subseteq X$ compact, $U_1,U_2$ open in $X$ covering $C$, there are two compact subsets of $X$, $A_1 \subseteq U_1$ and $A_2 \subseteq U_2$ such that for any two open sets $V_1,V_2$ containing $A_1,A_2$ respectively, $C \subseteq V_1 \cup V_2$. $X$ strongly compact splits if we can choose $A_1,A_2$ to cover $C$. I claim that the exponential law is always continuous iff X weakly compact splits.

Below $M(C,U)$ denotes all continuous maps sending $C$ into $U$ and $\phi$ denotes the exponential law map.

If $X$ doesn't weakly compact-split, let $Z$ be the Sierpinski space $\{0,1\}$ with open sets $\{\{\},\{0\},\{0,1\}\}$ and $Y$ be the space $\{0,1\}$ with the discrete topology. Now, consider the map $f : X \times Y \to Z$ mapping $U_1 \times \{0\} \cup U_2 \times \{1\}$ to $0$ with $U_1,U_2$ a cover of $C$ which doesn't split. Then, this continuous map thought of as in $(Z^Y)^X$ is inside the set $M(C,(M(\{0\},\{0\}) \cup M(\{1\},\{0\}))$ which is open, but I claim the pre-image is not open in $Z^{X \times Y}$. It's easy to see that because $Z$ only has $1$ non-trivial open set, $M(D,\{0\})$ forms a basis for $Z^{X \times Y}$. Now suppose $f \in M(D,\{0\}) \subseteq \phi^{-1}\left(M(C,(M(\{0\},\{0\}) \cup M(\{1\},\{0\}))\right)$. We would have that $D = D_1 \times \{0\} \cup D_2 \times \{1\}$ where $D_1$ and $D_2$ are compact in $X$. Now $f \in M(D,\{0\})$ means $D_1 \subseteq U_1$ and $D_2 \subseteq U_2$ and the subset inclusion $M(D,\{0\}) \subseteq \phi^{-1}\left(M(C,(M(\{0\},\{0\}) \cup M(\{1\},\{0\}))\right)$ implies any two open sets $V_1,V_2$ containing $D_1,D_2$ respectively must have $C \subseteq V_1 \cup V_2$. Contradiction.

Alternatively, suppose $X$ weakly compact-splits. Then we consider a $f: X \times Y \to Z$ continuous such that $f$ (thought of as in $(Z^Y)^X$) is in $M(C,\cup_{i \in I}(\cap_{j \in J_i}{M(L_{i,j},U_{i,j})}))$. We wish to show that the inverse image under $\phi$ is a neighborhood of $f$. Now because $f:X \to Z^Y$ is continuous and $C$ is compact, we can assume $I$ is finite. That is, replace $I$ with some finite subset such that $f$ is still in $M(C,\cup_{i \in I}(\cap_{j \in J_i}{M(L_{i,j},U_{i,j})}))$. Now we apply the strong compact splitting property many times to find compact sets $D_i$ indexed by $I$ such that $f(D_i) \subseteq \cap_{j \in J_i}{M(L_{i,j},U_{i,j})}$ and such that if $D_i \subseteq V_i$ for each $i$ then $C \subseteq \cup V_i$. Now we claim that in $Z^{X \times Y}$ the set $\cap_{i\in I,j\in J_i}{M(D_i \times L_{i,j},U_{i,j})}$ contains $f$ and is contained in the preimage as desired. Suppose $g : X \times Y \to Z$ is continuous and $g(D_i \times L_{i,j}) \subseteq U_{i,j}$. Then $g(D_i) \subseteq \cap_{j \in J_i}{M(L_{i,j},U_{i,j})}$ so because $g:X \to Z^Y$ is continuous, $g(C) \subseteq \cup_{i \in I}(\cap_{j \in J_i}{M(L_{i,j},U_{i,j})}$ as desired.

Now, for the example we note that $\alpha(\mathbb{Q})$ doesn't weakly compact split, so it must produce a counterexample.

New question is: does weakly compact split imply strongly compact split?

Edit: To further understand the situation, we should ask when the reverse map $(Z^Y)^X \to Z^{X \times Y}$ is continuous. Now this map may not always be defined, so we can instead ask the question with the image space being all functions from $X \times Y \to Z$ not necessarily continuous, and here I believe the answer is yes if for every compact set of $X \times Y$ and an open neighborhood there is a covering of the compact set by products of compact sets in $X$ and $Y$ which is contained in the neighborhood. This is satisfied if either $X$ or $Y$ is locally compact, meaning that we get an homeomorphism onto image of the map in the original question if $X$ is locally compact, but not necessarily a surjective map.

Answer 2

There is a paper by P. Booth and J. Tillotson, Pacific J. Math. Vol. 88. No. 1, 198 (downloadable) which discusses exponential laws $X ^{Z \times Y} \cong (X^Y)^Z$ for various function space topologies, and topologies on $X \times Y$.