I split this in two parts for comparison.

**Part 1:** Exponential law in $\text{Top}$.

Let $F(X, Y)$ be the space of continuous functions from $X$ to $Y$ with compact-open topology. We denote the subbasic open set by $(A, W) = \{f\in F(X, Y): f(A)\subseteq W\}$, $A$ compact, $W$ open.

Define adjunction as $\alpha:F(X\times Y, Z)\to F(X, F(Y, Z))$ by $\alpha(f)(x)(y) = f(x, y)$. 

> $\alpha$ is continuous.

Proof: The subbasic open sets in $F(X, F(Y, Z))$ are of the form $(A, (B, W))$ where $A, B$ are compact and $W$ is open. But $\alpha^{-1}(A, (B, W)) = (A\times B, W)$ which is open in $F(X\times Y, Z)$.

Define evaluation map $e_{X, Y}:F(X, Y)\times X\to Y$ by $e_{X, Y}(f, x) = f(x)$.

> $e_{Y, Z}$ is continuous for locally compact $Y$. If $e_{Y, Z}$ is continuous then $\alpha$ is bijective.

Above is just to assure that if the map $x\mapsto(y\mapsto f(x)(y))$ is continuous then so is $(x, y)\mapsto f(x)(y)$.

> If $X, Y$ are locally compact then $\alpha:F(X\times Y, Z)\cong F(X, F(Y, Z)) $.

Proof: It suffices to show that $\alpha^{-1}$ is continuous. If $$\gamma = e_{X\times Y, Z} \circ (\alpha^{-1}\times \text{Id}_{X\times Y}):F(X, F(Y, Z))\times (X\times Y)\to Z$$ is continuous then $\alpha^{-1}$ is, because we can apply adjunction operator on $\gamma$ to obtain $\alpha^{-1}$.

Easy calculations show that $\gamma(f, (x, y)) = f(x)(y)$ and so $\gamma$ is the composition $$F(X, F(Y, Z))\times (X\times Y) \to (F(X, F(Y, Z))\times X) \times Y \to F(Y, Z)\times Y \to Z   $$ first map being the canonical homeomorphism, second and third being evaluation maps, continuous because $X, Y$ are locally compact. This proves $\alpha$ is a homeomorphism.

There is also a similar theorem which I'll mention.

> Let $Y$ be locally compact, $X, Y$ be Hausdorff. Then $\alpha$ is a homeomorphism.

Sketch of proof: It suffices to note that sets of the form $(A\times B, W)$ for compact $A, B$ and open $W$ form a subbasis of $F(X\times Y, Z)$ under our assumptions. 

**Part 2:** Exponential law in $\text{Top}^0$?

Suppose that all spaces are pointed from now on. 

Let $F^0(X, Y)$ be the subspace of $F(X, Y)$ of pointed continuous functions from $X$ to $Y$, with base point being the constant map onto the base point of $Y$. Denote the subbasic sets by $(A, W)^0 = F^0(X, Y)\cap (A, W)$ for compact $A$ and open $W$. 

Let $p:X\times Y\to X\wedge Y$ be canonical quotient map. For simplicity we'll denote $p(x, y) = x\wedge y$.

Define adjunction of pointed maps as $\alpha^0:F^0(X\wedge Y, Z)\to F^0(X, F^0(Y, Z))$ by $\alpha^0(f)(x)(y) = f(x\wedge y)$.

> $\alpha^0$ is continuous.

Proof: As before, $(\alpha^0)^{-1}(A, (B, W)^0)^0 = (p(A\times B), W)^0$, the sets of the form $(A, (B, W)^0)^0$ being subbasic.

Define the pointed evaluation map $e_{X, Y}^0:F^0(X, Y)\wedge X\to Y$ by $e_{X, Y}^0(f \wedge x) = f(x)$.

> $e_{Y, Z}^0$ is continuous for locally compact $Y$. If $e_{Y, Z}^0$ is continuous then $\alpha^0$ is bijective.

This is again just so that if $x\mapsto(y\mapsto f(x)(y))$ is continuous then also $x \wedge y\mapsto f(x)(y)$.

Finally:

> If $X, Y$ are locally compact, is $\alpha^0:F^0(X\wedge Y, Z)\cong F^0(X, F^0(Y, Z))$?

Attempt at a proof: Once again we only need to show that $(\alpha^0)^{-1}$ is continuous. Consider the map $$\gamma^0 = e_{X\wedge Y, Z}^0 \circ ((\alpha^0)^{-1}\wedge \text{Id}_{X\wedge Y}) : F^0(X, F^0(Y, Z))\wedge (X\wedge Y)\to Z $$ explicitly $\gamma^0(f \wedge (x \wedge y)) = f(x)(y)$. If $\gamma$ is continuous then by some version of pointed adjunction, so is $\alpha^{-1}$.

Once again we can decompose it into three maps, one being canonical and other two being pointed evaluation maps. 

However, the approach breaks down. Denoting $Q = F^0(X, F^0(Y, Z))$, we'd have to show that the canonical map $Q\wedge (X\wedge Y)\to (Q \wedge X)\wedge Y$ is continuous.

If we look at the statement of associativity of smash product, its proof uses Whiteheads theorem:

> If $q:X\to Y$ is quotient and $Z$ is locally compact then $q\times \text{Id}_Z$ is a quotient.

More explicitly, to show that $Q\wedge (X\wedge Y)\to (Q\wedge X)\wedge Y$ is continuous, it uses that $Q$ is locally compact by showing $\text{Id}_Q\times p$ is a quotient map. 

Now to save this proof, two things come to mind.

 1. Assume that $Q$ is locally compact. I think this is a heavy and unwieldy assumption.
 2. Assume that $p$ is a proper map. This is somewhat better, and in case when $X, Y$ are compact Hausdorff spaces, $X\wedge Y$ is also compact Hausdorff as shown here https://math.stackexchange.com/a/1645794/476484 so $p$ is proper.

Lastly, I'll mention an analogous theorem

> If $Y$ is locally compact, $X, Y$ are Hausdorff, $p$ is proper, then $\alpha^0$ is a homeomorphism.

Sketch of proof: Show that sets of the form $(p(A\times B), W)^0$ where $A, B$ are compact and $W$ is open form a subbasis of $F^0(X\times Y, Z)$.

Main reference: "Algebraic Topology" by Tammo tom Dieck, Section 2.4

Reference to the case when $X, Y$ are compact Hausdorff: "Algebraic Topology" by C. R. F. Maunder, Theorem 6.2.38 c)

[Wikipedia reference](https://en.wikipedia.org/wiki/Smash_product)

[nLab reference for smash product](https://ncatlab.org/nlab/show/smash+product). Note the exponential law here is stated for pointed **sets**.

[nLab reference for exponential law](https://ncatlab.org/nlab/show/exponential+law+for+spaces) This article lead me to the following reference, after I consulted the article with professor Zoran Škoda who wrote it:

"Lectures on Algebraic Topology. Fundamentals of homotopy theory." by M. M. Postnikov (there seems to be no English version, but there is a Russian one)

Postnikov claims that $p^*:F^0(X\wedge Y, Z)\to F^0(X\times Y, Z)$ is a topological embedding. It's clear to me that this map is continuous and injective. However, I'm not sure how to prove $p^*(A, V)^0$ is open in $p^*F^0(X\wedge Y, Z)$.

This can be found at the beggining of Chapter 4.

cross-posted from [math.se](https://math.stackexchange.com/questions/3934265/adjunction-of-pointed-maps-is-a-homeomorphism)