Let $X,Y,K$ be pointed spaces and $K$ locally compact Hausdorff. Let $f:X\rightarrow Y^K$ and define $g:X\wedge K\rightarrow Y$ by $g(x\wedge k)=f(x)(k)$. I want to prove that $f$ is continuous iff $g$ is continuous. The topology on $Y^K$ is the compact open topology:
Let $f:X\rightarrow Y^K$ continuous and $U\subset Y$ open: $x\wedge k\in g^{-1}(U)$ iff $f(x)(k)\in U$ iff $x\in f^{-1}(B(\{k\},U)$ where $B(A,C):=\{f:Y\rightarrow K: f(A)\subseteq C\}$ for $A$ compact and $C$ open is a subbasis element of $Y^K$. Thus there exists een open $V\subseteq X$ with $x\in V\subseteq f^{-1}(B(\{k\},U)$. Then $V\times\{k\}$ is open in $X\times K$, but is this also open in $X\wedge K$? Is this a good proof?
Suppose $g:X\wedge K\rightarrow Y$ continuous and let $B(A,U)$ be a subbasis element of $Y^K$: $x\in f^{-1}(B(A,U))$ iff $f(x)(A)\subseteq U$ iff $g(x\wedge A)\subset U$ if $x\wedge A\subset g^{-1}(U)$. Now i want to conclude by locally compactness that there exists an open $V$ with compact closure $\overline{V}$ such that $\{x\}\times A\subset \{x\}\times V\subset\{x\}\times\overline{V}\subset g^{-1}(U)$. Is this allowed? If not how to conclude continuity of $f$?
Thank you very much :)
$\require{AMScd}$ You don't need the local compactness in that direction. You have $\{x\}\times A\subseteq g^{-1}(U)$ which is open as $g$ is continuous. Then you can apply the tube lemma, which gives you an open set $V\ni x$ such that $V\times A\subseteq g^{-1}(U)$.
For the other direction: One can show, using the local compactness of $K$, that the evaluation map $$\varepsilon:Y^K×K→Y\\(h,k)\mapsto h(k)$$ is continuous. This map factors through a map $\barε:Y^K\wedge K→Y$, and the composition at the bottom row is the map whose continuity you want to show. $$\begin{CD} X\times K @>f×1>>Y^K×K @>ε>> Y \\ @VVV @VVV @|\\ X\wedge K @>>> Y^K\wedge K @>>> Y \end{CD}$$
Regarding your attempt for that direction: You cannot say that $V×\{k\}$ is open, this is usually not an open set.