Does exponential law bijective implies evaluation map continuous?

Question

Husemoller in his "Fibre Bundles" writes that the exponential law $$ \theta \colon B^{A \times X} \to (B^X)^A $$ which is always injective, is bijective if and only if the evaluation map $$ Ev \colon B^X \times X \to B$$ is continuous by an easy proof (all function spaces have the CO topology).

It's easy to prove that if $Ev$ is continuous, then $\theta$ is surjective, but I can't show the only if, i.e. if $\theta$ is bijective then $Ev$ is continuous, and I even suspect that it is wrong without additional constraints on $X$.

Is there really an "easy proof" of the reciprocal, or does Husemoller is wrong ?

Answer 1

In Engelking's General Topology book, prop 2.6.11 gives an easy proof : if $\theta$ is surjective, then for any $g \colon A \to B^X \in (B^X)^A$ there is $h = \theta^{-1}(g) \colon A \times X \to B \in B^{A \times X}$.

Now take $A \equiv B^X$ and $g \equiv Id_{B^X} \colon B^X \to B^X$, then $\theta^{-1}(Id_{B^X}) = B^X \times X \to B = Ev$ is continuous.