$\operatorname{Hom}(X \times Z, Y) \cong \operatorname{Hom}(X, \operatorname{Map}(Z, Y))$ is not true in $\textbf{Top}$

I read that in the category of topological space it is not true that $\operatorname{Hom}(X \times Z, Y) \cong \operatorname{Hom}(X, \operatorname{Map}(Z, Y))$ for every $X, Y, Z$ topological spaces. In particular, I read that this does not work when $Z = \mathbb{R} \setminus \{ 1/n \mid n \in \mathbb{N}\}$. This was not explained, i.e. there was no example.

I am looking for an example or – maybe better – just a hint.

EDIT

1. $\operatorname{Map}(Z, Y)$ is endowed with the compact-open topology
2. $\cong$ means just bijection; i.e. there is already an obvious bijection as sets, but the claim is that (I think) sometimes the "output function" is not continuous

You are not explicit whether "compact" includes Hausdorff, so it is not absolutely clear what the compact-open topology is. Let us ignore this point as it is irrelevant in the sequel.

Let us first understand the exponential map on the level of sets. For sets $A, B$ let $[A,B]$ denote the set of all functions $f : A \to B$. Define $e = e^{A,B,C}: [A \times B,C] \to [A,[B,C]]$ as follows: For $f : A \times B \to C$ and $a \in A$ define a function $f_a : B \to C, f_a(b) = f(a,b)$. Then $e(f)$ is defined by $e(f)(a) = f_a$. It is easy to verify that $e$ is a bijection of sets.

Now we come to topology. Given topological spaces $X, Y, Z$, it is clear that if $f : X \times Y \to Z$ is continuous, then all $f_x : Y \to Z$ are continuous, i.e. $f_x \in \text{Hom}(Y,Z)$. If we endow $\text{Hom}(Y,Z)$ with the compact-open topology, we get the space $Z^Y = \text{Maps}(Y,Z)$. It is easy to verify that $e(f) \in \text{Hom}(X,Z^Y)$. Thus we obtain an injective function $$e : \text{Hom}(X \times Y,Z) \to \text{Hom}(X,Z^Y) .$$ It is well-known that if $Y$ is locally compact, then $e$ is bijective for all $X,Z$. At this point it becomes relevant how compact and locally compact are defined. However, for a Hausdorff $Y$ all definitions yield the same result.

A slightly more general framework assuring that $e$ is bijective for all $X,Z$ is this. The evaluation map is defined by $\Omega : Z^Y \times Y \to Z, \Omega(f,y) = f(y)$. The following are well-known:

1. Given $Y,Z$, $e$ is bijective for all $X$ iff $\Omega$ is continuous.

2. If $Y$ is locally compact, then $\Omega$ is continuous for all $Z$.

What happens if we have $Y,Z$ such that $\Omega$ is not continuous? Then for $X = Z^Y$ the map $e$ is not bijective. To see this, consider the identity map $id \in \text{Hom}(Z^Y,Z^Y)$. If $e$ were bijective, then $e(\phi) = id$ for some $\phi \in \text{Hom}(Z^Y \times Y,Z)$. By definition of $e$ we must have $\phi = \Omega$ which is a contradiction.

Thus, to show that for a space $Y$ the map $e$ is not bijective for all $X,Z$, it suffices to find $Z$ such that $\Omega$ is not continuous and take $X = Y^Z$. Clearly, such $Y$ cannot be locally compact.

Your question says that $Y = \mathbb{R} \setminus \{ \frac{1}{n} \mid n \in \mathbb{N}\}$ will do. Let us show that $\Omega$ is not continuous for $Z = I = [0,1]$.

Let $c \in I^Y$ be the constant function $c(y) = 0$. We claim that $\Omega$ is not continuous in $(c,0)$. We do it by contradiction. If $\Omega$ were continuous in $(c,0)$, then we can find $\varepsilon > 0$, compact $K_i \subset Y$ and open $U_i \subset I$ such that $c \in \langle K_i,U_i \rangle = \{ f \in I^Y \mid (f(K_i) \subset U_i \}$ and $\Omega(\bigcap_{i=1}^m \langle K_i,U_i \rangle \times U_\varepsilon) \subset [0,1)$. Here $U_\varepsilon = \{ y \in Y \mid \lvert y \rvert < \varepsilon \}$. Note that necessarily $0 \in U_i$ since $0 \in c(K_i)$. $K = \bigcup_{i=1}^m K_i$ is compact and $U = \bigcap_{i=1}^m U_i$ an open neighborhood of $0$, thus $c \in \langle K,U \rangle \subset \bigcap_{i=1}^m \langle K_i,U_i \rangle$ and $\Omega(\langle K,U \rangle \times U_\varepsilon) \subset [0,1)$.

Let $n \in \mathbb N$ such that $\frac{1}{n} < \varepsilon$. Since $(\frac{1}{n+1},\frac{1}{n})$ is closed in $Y$, the set $C = (\frac{1}{n+1},\frac{1}{n}) \cap K$ is compact, thus we find $\xi \in (\frac{1}{n+1},\frac{1}{n}) \setminus C$. Let $g : (\frac{1}{n+1},\frac{1}{n}) \to I$ be a continuous map such that $g(\xi)= 1$ and $g(x) = 0$ for $x \in C$. Extend $g$ to a continuous $f : Y \to I$ via $f(x) = 0$ for $x \notin (\frac{1}{n+1},\frac{1}{n})$. Then $f(K) = \{0\} \subset U$, i.e. $f \in \langle K,U \rangle$. We have $\xi \in U_\varepsilon$, thus $(f,\xi) \in \langle K,U \rangle \times U_\varepsilon$ and $\Omega(f,\xi) = f(\xi) = 1$. This contradicts $\Omega(\langle K,U \rangle \times U_\varepsilon) \subset [0,1)$.

Remark 1: We can generalize this as follows. Let $Y$ be a Tychonoff space (= completely regular Hausdorff space) which is not locally compact. Then the evaluation map $\Omega : I^Y \times Y \to I$ is not continuous.

To prove this, let $y_0 \in Y$ be a point which does not have a compact neighborhood. Let $c \in I^Y$ be the constant function $c(y) = 0$. We claim that $\Omega$ is not continuous in $(c,y_0)$. We do it by contradiction. If $\Omega$ were continuous in $(c,y_0)$, then we can find an open neighborhood $V$ of $y_0$ in $Y$, compact $K_i \subset Y$ and open $U_i \subset I$ such that $c \in \langle K_i,U_i \rangle$ and $\Omega(\bigcap_{i=1}^m \langle K_i,U_i \rangle \times V) \subset [0,1)$. Note that necessarily $0 \in U_i$ since $0 \in c(K_i)$. $K = \bigcup_{i=1}^m K_i$ is compact and $U = \bigcap_{i=1}^m U_i$ an open neighborhood of $0$, thus $c \in \langle K,U \rangle \subset \bigcap_{i=1}^m \langle K_i,U_i \rangle$ and $\Omega(\langle K,U \rangle \times V) \subset [0,1)$.

We have $V \setminus K \ne \emptyset$, because otherwise $V \subset K$ so that $K$ would be a compact neighborhood of $y_0$. Pick $\eta \in V \setminus K$ and let $f: Y \to I$ be a continuous map such that $f(y) = 0$ for $y \in K$ and $f(\eta) = 1$. Then $f(K) = \{0\} \subset U$, i.e. $f \in \langle K,U \rangle$. We have $\eta \in V$, thus $(f,\eta) \in \langle K,U \rangle \times V$ and $\Omega(f,\eta) = f(\eta) = 1$. This contradicts $\Omega(\langle K,U \rangle \times V) \subset [0,1)$.

Remark 2: If we take $S$= Sierpinski space, then for any Hausdorff space $Y$ which is not locally compact the evaluation map $\Omega : S^Y \times Y \to S$ is not continuous. The proof is similar as above.