evaluation map of the exponentials in $\mathsf{Top}$ Recall that if $\mathsf{C}$ is a category with binary product and $Y,Z\in \mathsf{C}$, a pair $(Z^Y, e)$ of an object $Z^Y$ in $\mathsf {C}$ and a morphism $e:Y^X\to Y$ is called the exponential object if  it is a universal arrow from $-\times Y:\mathsf{C}\to\mathsf{C}$ to $Z$. The notation is consistent with the case $\sf{C}=\sf{Set}$; in this case $Z^Y=\mathsf{Set}(Y,Z)$ and $e:Z^Y\times Y\to Z$ is given by $(f,y)\mapsto f(y).$
Recently I was reading this pdf by Martin Escardo and Reinhold Heckmann, which characterizes those spaces $Y$ in $\mathsf{Top}$ which are exponentiable in the sense that for every space $Z$, we can endow $\mathsf{Top}(Y,Z)$ a topology in such a way that there is a natural bijection $\mathsf{Top}(-\times Y,Z)\cong \mathsf{Top}(-,\mathsf{Top}(Y,Z))$ defined by the set-theoretic currying. By Yoneda, this is the same as saying that $\mathsf{Top}(Y,Z)$ can be topologized so that the set-theoretic evaluation map $\mathsf{Top}(Y,Z)\times Y\to Z$ is continuous and the natural transformation $\mathsf{Top}(-,Z^Y)\Rightarrow\mathsf{Top}(-\times Y,Z)$ it defines is an isomorphism.
For sure, currying is a "canonical" operation; however, I was wondering if every exponential object in $\mathsf{Top}$ arises in this form: If $Y,Z$ are topological spaces such that the exponential object $(Z^Y,e)$ exists in the sense of the previous paragraph, then, the bijection
$$\mathsf{Top}(Y,Z)\cong\mathsf{Top}(\ast\times Y,Z)\cong\mathsf{Top}(\ast, Z^Y)\cong Z^Y$$
shows that we can take as the underlying set of $Y^Z$ the set of continuous map $Y\to Z$. But I can't seem to get a clue on how $e$ is defined. So my question is:

Is it always the case that, assuming the existance of the exponential
object $(Z^Y,e)$, we can topologize $\mathsf{Top}(Y,Z)$ so that the
set-theoretic evaluation map $\mathsf{Top}(Y,Z)\times Y\to Z$ makes it
into an exponential object of $Y,Z$?

Any help is appreciated. Thanks in advance!
 A: I appreciate @Zhen Lin and @Daniel Schepler for their guidance. I will post this answer so that this question will not remain unanswered.
We start by recalling our setting: We are given two topological spaces $Y,Z$ such that the exponential object $(Z^Y,e)$ exists. As I mentioned in the question, there is a bijection
$$\varphi:\mathsf{Top}(Y,Z)\cong\mathsf{Top}(\ast\times Y,Z)\cong\mathsf{Top}(\ast, Z^Y)\cong Z^Y.$$
We want to show that $e\cdot(\varphi\times 1_Y):\mathsf{Top}(Z^Y,Y)\times Y\to Z$ is the set-theoretic evaluation. To this end, let $(f,y)\in \mathsf{Top}(Z^Y,Y).$ Note that $\varphi(f)=(fp)^{\wedge}(\ast)$, where $p:\ast\times Y\cong Y$ and $(-)^\wedge$ denotes the transpose. Thus
$$
\begin{array}
& e\cdot(\varphi\times1_Y)(f,y)&=e((fp)^{\wedge}(\ast),y)\\
&=e\cdot((fp)^{\wedge}\times 1_Y)\cdot(1_\ast\times \widetilde{y})(\ast)\\
&=(fp)\cdot(1_\ast\times \widetilde{y})(\ast)\\
&=fp(\ast,y)\\
&=f(y),
\end{array}$$
where $\widetilde{y}:\ast\to Y$ is the map corresponding to $y$, as desired.
