13
$\begingroup$

What interests me the most is if the case of exponential law is true under the assumptions claimed for example on nlab: if $X, Y$ are Hausdorff and $Y$ is locally compact, then $F^0(X, F^0(Y, Z))\cong F^0(X\wedge Y, Z)$ using the pointed adjunction.

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

nLab reference for smash product. Note the exponential law here is stated for pointed sets.

nLab reference for exponential law 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.

Update:

There was a good point towards me that the version on nlab is different. The claim that $F^0(X\wedge Y, Z)\cong F^0(X, F(Y, Z))$ are homeomorphic under suitable assumptions.

But the map can lack to even be bijective under those assumptions. Consider $X = Y = Z = \{0, 1\}$ with discrete topology and base point $0$, then $|F^0(X\wedge X, X)| = |F^0(X, X)| = 2$ but $|F^0(X, X^X)| = 4$.

Because of this let's assume they meant the version with $F^0(X\wedge Y, Z)\cong F^0(X, F^0(Y, Z))$.

cross-posted from math.overflow

$\endgroup$
9
  • 1
    $\begingroup$ Your question about product of quotient maps is answered here: math.stackexchange.com/questions/31697/… $\endgroup$
    – freakish
    Dec 4, 2020 at 12:39
  • $\begingroup$ @freakish could you elaborate on that $\endgroup$
    – Jakobian
    Dec 4, 2020 at 12:45
  • $\begingroup$ You literally defined $p$ to be a quotient map. The only thing that I missed is that I'm not sure whether $F^0(Y,Z)$ is locally compact. But if it is, then the linked answer applies. $\endgroup$
    – freakish
    Dec 4, 2020 at 13:07
  • $\begingroup$ I know I can't deduce that $p$ is proper from assumptions, taking $X = Y = \mathbb{R}$ and $0$ as base point for both spaces, I see that $X\vee Y$ isn't compact, so $p$ isn't proper. $\endgroup$
    – Jakobian
    Dec 4, 2020 at 13:39
  • $\begingroup$ Why do you keep talking about being proper? Whitehead's theorem doesn't require that. Only local compactness matters. $\endgroup$
    – freakish
    Dec 4, 2020 at 14:33

0

You must log in to answer this question.

Browse other questions tagged .